Download 32 Frequency decreases with a decrease in the tension in a string

Frequency decreases with a decrease in the tension in a string. This is because frequency is directly
proportional to the square root of tension. It is given as:
v∝
T
Hence, the beat frequency cannot be 330 Hz
Question 13 : A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound
emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat
is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?
Answer :
Ultrasonic beep frequency emitted by the bat, v = 40 kHz
Velocity of the bat, Vb = 0.03 v
Where, v = velocity of sound in air
The apparent frequency of the sound striking the wall is given as:
 v 
v′ =  −  v
 v vb 
v


= 
 × 40
v − 0.03v 
=
40
kHz
0.97
This frequency is reflected by the stationary wall (vs = 0) toward the bat. The frequency (vn) of the
received sound is given by the relation:
 v + vB 
vn = 
 v′
v 
 v + 0.03v  40
= 
 ×
v
0.97
=
1.03 × 40
= 42.47 kHz
0.97
v


= 
 × 40
v − 0.03v 
=
40
kHz
0.97
This frequency is reflected by the stationary wall (vB = 0) towards the bat.
v


= 
 × 40
v − 0.03v 
32
=
40
kHz
0.97
This frequency is reflected by the stationary wall (vB = 0) towards the bat. The frequency (v′′) of the
received sound is given by the relation:
 v + vB 
vn = 
 v′
v 
 v + 0.03v  40
= 
 ×
v
0.97
=
1.03 × 40
= 42.47 kHz
0.97
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Electrostatics
Learning Objectives :
●
Electric Charges
●
Conductors and Insulators
●
Charging by Induction
●
Basic Properties of Electric Charge
●
Coulomb’s Law
●
Forces between Multiple Charges
●
Electric Field
●
Electric Field Lines
●
Electric Flux
●
Electric Dipole
●
Dipole in a Uniform External Field
●
Continuous Charge Distribution
●
Gauss’s Law
●
Application of Gauss’s Law
●
Electrostatic Potential
●
Potential due to a Point Charge
●
Potential due to an Electric Dipole
●
Potential due to a System of Charges
●
Equipotential Surfaces
●
Potential Energy of a System of Charges
●
Potential Energy in an External Field
●
Electrostatics of Conductors
●
Dielectrics and Polarisation
●
Capacitors and Capacitance
●
The Parallel Plate Capacitor
●
Effect of Dielectric on Capacitance
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CONCEPTUAL PROBLEMS
1.
Is Coulomb’s law a universal law?
Hint. No, Coulomb’s law is not universal law as Coulombic force depends upon the medium between
charges. Moreover this law is applicable only for the point charges at rest.
2.
A negatively charged rod attracts suspended object. Can we conclude that the object is positively
charged?
Hint. No, because negatively charged rod can attract neutral objects also by inducing positive charge on
them.
3.
Vehicles carrying inflammable material usually have chains attached to them, which remain in contact
with earth during motion. Why?
Hint. Vehicles when moves through dry air will have their body charged due to friction. The accumulation
of charge can cause sparking. Thus to ground this charge, metallic chains are used which transfers
the charge on the body to vehicle to the ground.
4.
Can a body have a charge of 0.8e?
Hint. No according to quantization principle, a body can have charge, which is integral multiple of the
charge on electron.
5.
Why do quarks don’t break quantization principle?
Hint. Quarks have fractional charges but they don’t break quantisation principle because they don’t have
independent existence of their own. They always exist in combinations and the charge on the
combination is quantized.
6.
Is annihilation of electron-positron pair violation of conservation of charge?
Hint. No, annihilation is not violation of conservation of charge as sum of charge on electron and positron
is zero and the gamma rays, which are produced due to annihilation, are also neutral.
7.
How does the specific charge gets affected when the charge is moving with velocity comparable to
the velocity of light?
Hint. Specific charge of the body will decrease, because with increase in velocity charge remains constant
whereas the mass increases thus ratio of charge to mass will decrease.
8.
Two bodies are rubbed together, one gets positively and other negatively charged. Will their mass
gets affected due to charging?
Hint. Charging by rubbing occurs due to transfer of electrons from one body to the other.
Thus body which is positively charged has lost electrons thus its mass decreases, whereas the body
which has gained electrons is negatively charged and its mass increases.
9.
Similar charges repel each other. Can they attract each other?
Hint. Yes. This can happen when the charge on one body is very large as compared to the charge on the
other body, because larger charge will induce opposite charge on the smaller body and thus attracts
it.
10. A special type of rubber is used to manufacture aircraft tyres. Why?
Hint. When aircraft is moving on the runway its body gets charged due to air friction. This charge can
cause sparking. Thus to ground this charge aircraft tyres are made of conducting rubber.
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11. Two point charges q1 and q2 are such that q1q2>0. What is the nature of interaction between charges?
Hint. As the product of two charges is always positive, it implies that either both are positive or both are
negative. Thus force of interaction between them is repulsive.
12. Is electrostatic force a central force in nature?
Hint. Yes, electrostatic force is central force as it always acts along the line joining the two charges.
13. How are uniform electric field represented?
Hint. The electric lines of force for uniform electric field intensity should be straight parallel and equidistant.
14. A charge particle is free to move in an electric field. Will it always move along the electric lines of
force?
Hint. No, if the charge particle moves at certain angle with the electric lines off force then it will not move
along the straight line.
15. A sensitive instrument is to be shielded from strong electrostatic fields in environment. Suggest a
possible method?
Hint. The instrument can be shielded by enclosing it inside a metallic surface. Since no electrostatic field
can be present within the metallic surface therefore the instrument will be shielded from the electrical
influence.
16. No two electric lines of force can cross each other. Why?
Hint. The tangent to the electric lines of force gives us the direction of electrical field intensity. Therefore,
if two lines of force intersect then there will be two directions of electric field intensity at one point
which is not possible.
17. Why one can ignore the quantization of charge while dealing with macroscopic charges?
Hint. The charge on the electron is very small. If few electrons are added or removed then the change in
the magnitude of the charge is very small. Thus the charge can be assumed to be continuous.
18. A gold leaf electroscope is positively charged. An earthed metal plate is now brought near the top of
the electroscope. What is the effect on charge, potential and capacitance of the electroscope?
Hint. The charge would remain constant as the two bodies are not brought in contact. The potential will
decrease and capacitance increases.
19. A metal sphere is held fixed on horizontal insulated plate and another metal sphere is held some
distance away. If the fixed sphere is given charge, how will the other sphere react?
Hint. When the fixed sphere is given charge it induces opposite charge on the nearer end of the other
sphere and similar charge at the farther end. Thus net force between the two spheres will be attractive.
Thus free sphere will be attracted towards the fixed sphere.
20. Electrostatic force between two protons situated certain distance apart is ‘y’ Newton. What will be
the electrostatic force between the two electrons situated same distance apart?
Hint. The electrostatic force between the two electrons is also y as it depends only on the magnitude of
charges and relative distance between the charges.
21. Repulsion is surer test of electrification. Why?
Hint. Repulsion is sure test of electrification of both bodies because attraction can be between a charged
and neutral body also.
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22. Why does the printing paper gets charged when it is passed through the press?
Hint. Printing paper gets charged because of the friction between paper and machine. It can be avoided by
ionizing the region around the press so that charge on the paper gets grounded.
23. A charged insulated conductor is touched by a person standing on the insulated platform. Will the
charge on the conductor remain same or is completely discharged?
Hint. No the two will share there charge till both the conductor and the body of the man are at the same
potential.
24. A comb run through one’s hair attracts bits of paper. Why? What happens if the hair is wet or if it is
a rainy day?
Hint. This happens because of the friction between comb and hair. But if the hair is wet then comb will not
get charged due to reduced friction and it will not be able to attract paper.
25. A bird perches on high power line and nothing happens to the bird. A man standing on the ground
touches the same line gets fatal shock?
Hint. When bird perches on single bare high power line, nothing happens as circuit is not complete and
current is not flowing through his body. But when man touches the same line current flows from wire
to the ground through his body and het gets fatal shock.
26. Can ever whole charge of the body be transferred to the other body? If yes, how and if not why?
Hint. Yes, the whole charge of body can be transferred to the conducting body when smaller body is
enclosed inside larger body and they are connected by wire. This happens as charge always resides
on the outer surface of the conductor.
27. What is the relevance of large value of K [=81] for water?
Hint. Large value of K is used to explain the electrolysis phenomenon. As k for water is large thus force
between oppositely charged ions gets reduced when dissolved in water. Thus it becomes easier to
separate them.
Electric Potential
1.
Why earth is considered to be at zero potential?
Hint. Earth has very large size. Potential of earth does not change with small charge therefore it is considered
as reference level and potential taken as zero.
2.
If electric potential V within a certain region is constant, what is the nature of electric field inside the
region?
Hint. Since electric field is negative of the potential gradient therefore if the electric potential is constant
in a certain region then potential gradient or electric field intensity is zero.
3.
Can two equipotential surfaces intersect?
Hint. Equipotential surface is the one having same potential everywhere, if two potential surfaces intersect
then at the point of intersection there will be two values of the potential.
4.
If potential at a point is zero, is field at that point is zero?
Hint. No, if potential at a point is zero, field may not be zero as in the case of equatorial line of dipole
where the potential is zero but the field intensity is not zero.
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5.
Can we produce high potential on a body without getting the shock?
Hint. Yes, we can do so if we are standing on an insulating stand and we touch a high power line. This can
happen as charge or potential on the body is never dangerous the flow of charge through the body is
dangerous.
6.
What is the importance of Gaussian surface?
Hint. Gaussian surface helps us in evaluating the electric field intensity at a given point for the symmetric
charge distribution.
7.
Near the surface of earth there is an electric field of the order of 100KV/m. Why then do we not
experience shock?
Hint. Our body and earth are at the same potential therefore although there is large field there is no potential
difference between our body and earth and we will not experience shock as no flow of charge is
there.
8.
If two electrons are brought near each other, what happens to their potential energy?
Hint. If two similar charges are brought closer the stability of the system decreases because of repulsion
between them thus the potential energy
9.
Will there be effect in the potential if the medium around the charge is changed?
Hint. Yes, medium effects the potential around the charge. With increase in dielectric constant of the
medium the potential around the charge goes on decreasing.
10. What is the shape of the equipotential surface for the point charge and for the line charge?
Hint. The equipotential surface is one with same potential everywhere on it. For point charge the locus of
points having same potential will be sphere and for line charge its cylinder. Therefore equipotential
surface for point charge is sphere and for line charge it is cylinder.
11. A charge Q is placed at any point and another charge q is rotated along the semi-circular path along
P. calculate he work done against the electric field of Q?
Hint. As the directions of lines of force is radial to the path followed by the charge on the circular path.
Therefore work done in motion along circular path is always zero..
12. What is the work done by the field of nucleus in complete circular or elliptical orbit?
Hint. For the complete orbit the displacement is zero therefore work done will also be zero
13. Consider a long conductor, the middle of which is earthed. If the potential difference across the two
ends of the conductor is 220V, then what is the potential at the ends and the middle point.
Hint. As the potential of the earthed point has to be zero therefore the potential of the two ends is 110V and
–110V with mid point being at zero potential.
14. Is Gauss law valid for the gravitational fields also?
Hint. Yes gauss law is valid for gravitational field also the only difference being that the constant of
proportionality is changed from 1/4πε0 to G and the charge is replaced by mass ‘m’ inside the Gaussian
surface.
15. Does the positive and negative nature of the potential depends on the nature of test charge?
Hint. No, the potential of the body does not depend on the nature of the test charge. It depends on the
source charge. Generally if the source charge is positive the body is positively charged. If the source
charge is negative the body is negatively charged.
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16.
Electric field is discontinuous across the surface of the conductor. Is electric potential also
discontinuous there?
Hint. No electric potential is continuous. Electric field may become zero inside it but the electric potential
is constant.
Capacitance
1.
Where is the knowledge of dielectric strength helpful?
Hint. Dielectric strength is the maximum value of electric field that can be applied across an insulator
before its breakdown or the magnitude of electric field intensity across insulator at which it starts
conducting. This value helps us in determining the value of maximum potential, which can be applied
across a capacitor.
2.
The safest way t protect you from lightning is to be inside a car. Why?
Hint. The body of the car is metallic, thus it provides electrostatic shielding to the person inside it. Even if
there is charge on the body of the car, electric field inside it remains zero.
3.
A sensitive instrument is to be shielded from the strong electrostatic field in is environment. Suggest
a possible way.
Hint. To shield an instrument it must be placed inside a metallic body, because the electric field intensity
inside it will be zero.
4.
When it is lightning it is safe to be inside a house than under a tree. Why?
Hint. When we are standing under a tree we provide easy passage for the charge to flow through our body
to the ground. But when we are inside a house, the charge can pass to the ground through lightning
conductor present in the building.
5.
The discharging current in the atmosphere due to small conductivity of air is known to be 1800A on
an average over the globe. Why then does the atmosphere not discharge itself completely in due
course and become electrically neutral?
Hint. The atmosphere does not gets charged because it gets the same amount of charge by lightning all
around the globe and net charge on it almost remains same.
6.
How can whole charge of a body be transferred to another isolated conductor?
Hint. This can be done by placing a charge conductor inside hollow isolated conductor and connecting
them by a wire. As charge resides on the outer surface thus whole charge from the inner body will be
transferred to the outer body.
7.
Can we give as much charge to the capacitor as we wish?
Hint. No, the maximum charge that be given to the capacitor depends upon its capacity or the dielectric
strength of medium between the plates of capacitor.
8.
A metal foil of negligible thickness is introduced between the plates of capacitor at the center. What
will be new capacitance of the capacitor?
Hint. When a metal sheet is introduced then this system is equivalent to two capacitors each of thickness
d/2 connected in series and the equivalent capacity is C.
9.
The separation between the plates of charged capacitor is to be increased. What will be the effect on
energy in case battery is removed after charging the capacitor or battery is removed?
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Hint. When distance between the plates of capacitor is increased the capacitance of parallel plate capacitor
decreases.[a] With battery disconnected, Q remains constant thus energy will increase. [b] With
battery in connection V remains constant thus if C decreases energy will alo decreases.
10. The battery remains connected to parallel plate capacitor and dielectric slab is inserted between the
plates. What will be the effect on [a] capacity [b] charge [c] potential energy [d] electric field intensity.
Hint. If dielectric slab is inserted with dielectric constant K between the plates, then the capacitance of
capacitor will increase K times. With battery in connection V remains constant. Q =CV, will increase
K times. Potential energy [ ] will also increase K times. Electric field intensity [Q/Aε0] will decrease
K times.
11. What happen when the plates of charged capacitor are suddenly connected by metallic wire?
Hint. When we connect two plates of capacitor, charge begins to flow from higher to lower potential plate
and the energy stored in the capacitor gets converted into heat energy.
12. Why does electrical conductivity of earth’s atmosphere increases with altitude?
Hint. The electrical conductivity of earth’s atmosphere increases with altitude because of increase in number
density of charged particles and decrease in water vapours.
13. Two copper spheres one hollow and other solid are charged to same potential. Which one carries
more charge?
Hint. Charge on both the spheres will be same as capacity depends only on the radius. This happens
because charge resides only on the outer surface of conductor and not inside it.
14. Why capacitors should be handled carefully even when there is no current?
Hint. A capacitor once charged takes infinite time to discharge completely even when battery is disconnected
or there is no current. Thus if we touch the plates of capacitors we will gets shock.
15. If battery is disconnected and distance between the plates of capacitor is increased, how will the
charge, electric field, potential energy be affected?
Hint. If the distance between the plates of capacitor is increased, then the capacitance is going to decrease.
As battery is disconnected thus charge on the plates and electric field remains constant, whereas
potential energy will increase.
16. What will be effect on capacity of parallel plate capacitor when the distance between the plates is
doubled and plate area is also doubled?
Hint. As both A as well as d are doubled capacitance remains unaffected.
17. Given a battery, how would you connect the two capacitors, in series or parallel for them to store the
greater [a] total charge [b] total energy?
1
Hint. As V is constant for a given battery therefore Q = CV and U = CV 2 . As both energy and charge are
2
directly proportional to capacitance thus they will be more in parallel combination.
18. When a battery is connected across the capacitor, are the charges on the plates always equal and
opposite even for plates of unequal size?
Hint. Yes, because of charge conservation the charges on the plates of capacitor are always and opposite in
nature.
19. Can there be potential difference between two conductors of same volume carrying equal positive
charges?
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Hint. Yes, there can be potential difference, because although there volumes are equal they can have different
shapes and thus different capacities.
20. Is it possible for metal sphere of radius 1cm to hold a charge of 1C?
Hint. No, At such a high voltage electric breakdown of air occurs and charge will leak to the surroundings.
21. Why does the electric field inside the dielectric decrease when it is placed inside an external field?
Hint. The electric field inside the dielectric decreases because external field will create an induced field
inside the dielectric whose direction will be opposite to the external field. Thus, net field inside it
will decrease.
22. A parallel plate capacitor has a capacity of 6µF and when dielectric is inserted it is 60µF. What is the
dielectric constant of the medium?
Hint. The dielectric constant K = Cm/Ca, thus it is equal to 10.
23. Two spheres of different capacities are charged to different potential are then connected by conducting
wire. Will the total energy increase or decrease?
Hint. When two spheres are connected by wire, the charge flows from sphere having higher potential to
lower potential till their potential becomes same. In the process of sharing of charges energy will be
lost in the form of heat.
ACTIVITIES
I.
What is a capacitor?
Two layers of conducting material separated by a layer of an excellent insulator from what is called
CAPACITOR. The name come from the “capacity” of this three-layer device to store charge. The conducting
layers are called capacitor PLATES. The insulating layer prevents movement of charge from one plate to
the other inside the capacitor. You can make a simple capacitor by a placing a sheet of waxes paper between
two sheets of aluminum foil.
In most capacitors the plates have very large surface area - so they can store a large amount of
change. The plates are also made very thin, so that the three layers can be rolled into a cylinder and placed
inside a small can. Each plate has a screw or a wire attached to it, called a TERMINAL, which extends
outside the can and allows the plate to be connected to a circuit.
41
The “charge-holding” ability of a capacitor is called its CAPACITANCE, Capacitance is measured
in a. unit called the FARAD, named after the British scientist Michael Faraday (1791-1867).
NOTE: Sometimes capacitor plates may pick up stray charge that needs to be removed. A good way
to avoid this problem is to keep a wire connected to the terminals of your capacitor when you are not using
it. Also, after using a capacitor you will want to “neutralize” it so you can start all over again. To do this,
touch a wire simultaneously to both of the capacitor
terminals.
Discharging or Neutralizing a capacitor:
II.
What happens to charge that flows into a capacitor plate?
For this part, you will need two round bulbs, two sockets, compass, connecting wire, 3 batteries,
a battery case or three battery holders, and one of the smaller blue capacitors. In this investigation
you will charge and discharge the capacitor and hopefully learn how charge moves to accomplish
this charging and discharging.
1.
First off, it is always a good idea to check your capacitor to make
sure it is truly discharged. The best way to check this is to take one of
your connecting wires and attach the ends of this wire to the capacitor
(one alligator clip on one of the capacitor’s terminals and the other
alligator clip on the other terminal of the capacitor. Make sure the
metal ends of the alligator clips are touching the metal terminals of
the capacitor. You might see a small spark as the capacitor discharges.
You should do this step before messing around with a capacitor to
make sure that it is truly discharged.
2.
Now set up your circuit as in the following schematic—except don’t attach the wire to the negative
terminal of the battery just yet. Remember that the north arrow (red) arrow of the compass is in the
correct orientation and the compass is underneath the wire—make sure that the wire is parallel to the
compass needle. Tape the compass and hold the wire on it so it stays parallel. It’s important to get the
compass placement correct to make this work out right.
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3.
4.
5.
You’ll need to watch three things: the light bulbs, the direction of the compass deflection, and
the amount of the deflection. When you are ready, hook up the wire to the negative terminal of the
battery to complete your circuit and make your observations.
We will do this twice (or three times if needed). However, once the capacitor is charged up, you must
discharge it before it will work again. So after you make your observations, you need to repeat step
1 above to discharge your capacitor. Once it has been discharged successfully, repeat this step again
(or a third time if needed) to make sure you understand the direction of the compass rotation.
Carefully make observations concerning the light bulbs, the direction and the amount of the compass
deflection and write them down on this page somewhere.
Now we are going to use the same setup but will move the location of the compass (see the following
diagram). Remember to always have the leading edge of the wire go into the north (red) arrow of the
compass and make sure the wire is parallel and atop the compass.
Again, you want to observe the bulbs and the compass. Do this several times again. REMEMBER to
discharge your capacitor before you charge it each time.
Write down your observations concerning the light bulbs, the direction and the amount of the compass
deflection and compare this with the observations of step 3.
Now we will move the compass so it after the capacitor in the circuit. You will repeat steps 3 and 4
except the compass is now placed as shown. REMEMBER to discharge your capacitor before you
charge it each time.
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6.
7.
8.
Write down your observations concerning the light bulbs, the direction and the amount of the compass
deflection and compare this with the observations of step 3.
Final location of compass. See figure and get observations for this compass location.
Write down your observations concerning the light bulbs, the direction and the amount of the compass
deflection and compare this with the observations of step 3.
Write down your overall observations on charging the capacitor from steps 3-6:
Bulb lighting:
Direction of compass deflection:
Amount of compass deflection:
Using these observations and what we know about circuits so far, let’s see if you can draw some
arrows on the following diagram indicating the direction of CONVENTIONAL charge flow (charge
moving away from the + terminal of the battery and into the – terminal of the battery) showing how
the capacitor charges up.
Draw arrows show charge flow through each of the wires.
What evidence supports your arrow directions?
You should NOT have an arrow drawn showing charge
flow directly through the capacitor. Why not?
9.
Here are some alternative (INCORRECT) ways of showing the arrows. You should be able to provide
evidence on why these are incorrect:
Alternative 1:
Explanation of why incorrect:
small blue capacitor
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Alternative 2:
Explanation of why incorrect:
small blue eapaeit&r
Alternative 3:
Explanation of why incorrect:
small blue capacitor
Alternative 4:
Explanation of why incorrect:
small blue eapaeit&r
10.
(a)
(b)
(c)
Now we will see how and why capacitors are used in circuits. We will learn how to discharge them
through an actual circuit.
Start with a fully charged small blue capacitor. Make sure your capacitor is fully charged using the
directions on steps 2 and 3.
Set up the following circuit but don’t complete the circuit yet until you are ready to make observations:
(notice that there is no battery in this circuit)
Now complete the circuit and observe the light bulbs, compass direction during deflection, and
amount of compass deflection. You may need to do this several times to check your observations
(you’ll need to charge the capacitor each time so you can observe its discharge)
45
Observations:
(d) Now change the placement of the compass as in the following diagram (make sure it is placed
correctly under the leading edge of the wire) and watch the charged capacitor discharge now.
Observations:
(e) Now change the placement of the compass as in the following diagram (make sure it is placed
correctly under the leading edge of the wire) and watch the charged capacitor discharge now.
Observations:
11. Using these observations and what we know about circuits so far, let’s see if you can draw some
arrows on the following diagram indicating the direction of CONVENTIONAL charge flow showing
how the capacitor discharges.
Draw arrows show charge flow through each of the wires.
What evidence supports your arrow directions?
You should not have an arrow drawn showing charge flow directly through the capacitor. Why not?
46
12.
Here are some alternative (incorrect) ways of showing the arrows. You should be able to provide
evidence on why these are incorrect:
Alternative 1:
Explanation of why incorrect:
small blue capacitor
Explanation of why incorrect:
smaEbluc capacitor
13.
14.
You should have seen the capacitor lighting up the bulbs for a brief time while it is discharging.
(a) how is a discharging capacitor like a battery?
(b) how is a discharging capacitor NOT like a battery?
(c) a capacitor will not charge up automatically—you need the battery. Why?
Now, let’s see how a different capacitor might act in the circuit. Get one of the taller blue capacitors.
Charge it up using the following circuit and then discharge it using the following drawing
CHARGING CIRCUIT
Observations:
Similarities to the smaller blue capacitor:
Dissimilarities to the smaller blue capacitor:
III. Using an AIR Capacitor.
DISCHARGING CIRCUIT
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1.
2.
3.
4.
Blow through the tube at one end. What happens to the balloon in the middle?
Repeat, except this time hold your hand near the opon tube as you blow in the other end. Does any air
go in or come out? If air comes out, where did the air coming out originate?
If air must flow through both tubes to “charge” the air capacitor, must air flow through both tubes to
discharge it? “Charge” the air capacitor again and place your finger over the open tube before releasing
the tube from your mouth. What happens?
Inhale air through the tube nearest you white keeping the other tube open. Describe xvhat happens to
the balloon. Describe all the air movements that occur.
Figure shows as air capacitor with both sides open to the atmosphere through a tube in each side. There is
atmospheric pressure in each side, which we will call NORMAL air pressure.
Blow air in through the tube on one side of an air capacitor, and hold the extra air inside by closing
the tube. Draw a sketch of the air capacitor, and label the pressure in each side as NORMAL or
HIGH or LOW.
2.
Explain why the membrane between the two sides changes shape.
3.
Release both ends of the air capacitor to normal atmospheric pressure. Then place your mouth over
the tube at the other side of the air capacitor and inhale; hold the depletion inside by closing the tube.
Draw a new sketch of the air capacitor, and label the pressure in each side as NORMAL or HIGH or
LOW. Explain why the membrane between the two sides changes shape.
4.
What do your observations tell you about the comparative ability of:
(a) Above-nomial pressure to push toward NORMAL pressure?
and
(b) NORMAL pressure to push toward below-normal pressure?
Comparing the Air capacitor with the capacitor charging circuit.
1.
Rebuild the following circuit but don’t hook it up yet. Make
sure the capacitor is discharged.
48
Before you hook it up the electric pressure can be described with these words: high, normal, and low.
Look at the picture below:
Now activate your circuit and watch your light bulbs. You should see the light bulbs light, then dim,
and go out.
Now the top capacitor plate goes from normal to higher than
normal (Medium High) and the bottom capacitor plate changes from
Normal to Medium Low.
Once the capacitor gets charged, the light bulbs go out. Which
words can now describe the top and bottom plates of the capacitor?
Remembert, for the bulbs to light there might be an electric pressure
differential.
Write these two words in now on the circuit diagram below.
Discharge the capacitor when finish this part.
IV.
What happens while a capacitor charges?
For this activity you will three batteries to start, wire, a compass, two Long bulbs, and the small blue
capacitor and switch.
Charge a 25,000 (if capacitor through two long bulbs, using a 3-cell battery as shown in Figure. Use
a compass under one of the wires to monitor direction of flow.
When you have the circuit ready, close the switch and observe the light bulbs.
1. Draw arrows on Figure to show charge flow in. all parts of the circuit while the bidbs are lit.
Don’t attempt to use arrowtails to show flow rate - just show directions.
2. Figure shows the capacitor after it has been charged. Draw (+) signs by the plate that has gained
charge, and (–) signs by the plate that has lost charge.
49
3. Predict: Will the bulbs light again if you add die second 3-cell battery pack and close the circuit?
Whw or whap not?
Predict what would happen if you took your charged capacitor and added another three batteries
onto the circuit. Make a prediction if the bulbs would light again.
Now add another three batteries so you have six batteries total. (see figure on right). Make sure
to leave the switch open until you are ready.
When you are ready, close the switch and observe the light bulbs.
4. Did the bulbs light? If they did, draw arrows to show the direction of charge
flow during the second bulb lighting.
5. Did more charge go into the positive capacitor plate and out of the negative
capacitor plate? What is the evidence?
Now remove the batteries from the circuit and connect the free ends of the
wires to each other to form a closed circuit through the light bulbs. Keep the compass still under
one wire.
6. Regarding both the bulbs and the compass, what did you observe? Explain why this happened.
Discharge the capacitor when finished.
Question 1:
Two point charges qA = 3 µC and qB = –3 µC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced
by the test charge?
Answer:
(a) The situation is represented in the given figure. O is the mid-point of line AB.
B
A
+3µ C
O
–3µ C
Distance between the two charges, AB = 20 cm
∴ AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3uC charge,
3 × 10 −6
3 × 10−6
N/C
=
4πε 0 (AO) 2
4πε 0 (10 × 10−2 ) 2
E1 = along OB
Where,
ε0 = Permittivity of free space
50
1
= 9 × 109 Nm2 C–2
4πε 0
Magnitude of electric field at point O caused by –3uC charge,
3 × 10−6
−3 × 10−6
N/C
=
4πε 0 (10 × 10−2 )
4πε 0 (OB)2
E2 is along OB
∴
E = E1 + E2

3 × 10−6 
9
×
×
×
2
(9
10
)
=
[Since the values of E1, and E2 are same. the value is multiplied with 2]

(10 × 10 −2 ) 2 

= 5.4 × 106 N/C along OB
Therefore, the electric field at midpoint O is 5.4 × 106 N C–1 along OB.
(b) A test charge of amount 1.5 × 10–9 C is placed at mid-point O.
q = 1.5 × lO–9 C
Force experienced by the test charge = F
∴ F = qE
= 1.5 × 1(10–9 × 5.4 × 106
= 3.1 × 10–3 N
The force is directed along line OA. This is because the negative test charge is repelled by the charge
placed at point B but attracted towards point A.
Therefore, the force experienced by the test charge is 8.1 × 10–3 N along OA.
Question 2:
(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm.
What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of
A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance
between them is halved?
Answer:
(a) Charge on sphere A, qA = Charge on sphere B, qB = 6.5 × 10–7 C
Distance between the spheres, r = 50 cm = 0.5 m
Force of repulsion between the two spheres,
F=
qAqB
4πε 0 r 2
Where,
ε0 = Free space permittivity
51
1
9
2
–2
4πε 0 = 9 × 10 Nm C
∴
(b)
9 × 109 × (6.5 × 10−7 ) 2
F=
(0.5) 2
= 1.52 × 10–2 N
Therefore, the force between the two spheres is 1.52 × 10–2 N.
After doubling the charge, charge on sphere A, qA = Charge on sphere B, qB = 2 × 6.5 × 10–7 C = 1.3
× 10–6 C
The distance between the spheres is halved
∴
r=
0.5
= 0.25 m
2
Force of repulsion between the two spheres,
qAqB
F = 4πε r 2
0
9 × 109 × 1.3 × 10−6 × 1.3 × 10−6
=
(0.25) 2
= 16 × 1.52 × 10–2
= 0.243 N
Therefore, the force between the two spheres is 0.243 N.
Question 3: Suppose the spheres A and B in in the above numerical have identical sizes. A third sphere of
the same size but uncharged is brought in contact with the first, then brought in contact with the second,
and finally removed from both. What is the new force of repulsion between A and B?
Answer: When sphere A is touched with an uncharged sphere C,
q
amount of charge from A will transfer
2
to c. So the charge on A and C will be q/2
q
brought in contact with sphere B with charge q, total charges on the
2
system will divide into two equal halves given as,
When sphere C with charge
q
+q
3q
2
=
4
2
Each sphere will share each half. Hence, charge on each of the spheres, C and B, is
Force of repulsion between sphere A having charge
52
q
and sphere B having charge
2
3q
.
4
q 3q
×
3q 2
3q
2 4 =
=
4πε 0 r 2 8 × 4πε 0 r 2
4
9
= 9 × 10 ×
3 × (6.5 × 10−7 ) 2
8 × (0.5) 2
= 5.703 × 10–3 N
Therefore, the force of attraction between the two spheres is 5.703 × 10–3 N.
Question 4: Consider a uniform electric field E = 3 × 103 i N/C. (a) What is the flux of this field through
a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same
square if the normal to its plane makes a 60° angle with the x-axis?
Answer:
KH
(a) Electric field intensity, E = 3 × 103 i N/C
KH
Magnitude of electric field intensity, | E | = 3 × 103 N/C
Side of the square, s = 10 cm = 0.1 m
Area of the square, A = s2 = 0.01 m2
The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to
the plane and electric field, θ = 0°
Flux (φ) through the plane is given by the relation,
KH
θ = | E | A cos θ
(b)
= 3 × 103 × 0.01 × cos 0°
= 30 N m2/C
Plane makes an angle of 60° with the x-axis. Hence, 8 = 60°
KH
Flux,
θ = | E | A cos θ
= 3 × 103 × 0.01 × cos 60°
= 30 ×
1
= 15N m 2 /C
2
Question 5 : A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical
Gaussian surface of 10.0 cm radius centered on the charge, (a) If the radius of the Gaussian surface were
doubled, how much flux would pass through the surface? (b) What is the value of the point charge?
Answer:
(a) Electric flux, φ = –1.0 × 103 N m2/C
Radius of the Gaussian surface,
r = 10.0 cm
Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does
53
not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux
passing through the surface remains the same i.e., –103 N m2/C.
(b)
Electric flux is given by the relation,
φ=
q
ε0
Where,
q = Net charge enclosed by the spherical surface
ε0 = Permittivity of free space = 8.854 × 10–12 N–1 C2 m–2
∴ q = φε0
= – 1.0 × 103 × 8.854 – 10–12
= – 8.854 × 10–9 C
= –88.54 nC
Therefore, the value of the point charge is –8.854 nC.
Question 6: Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates
have surface charge densities of opposite signs and of magnitude 17.0 × 10–22C/m2. What is E: (a) in the
outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?
Answer: The situation is represented in the following figure.
A and B are two parallel plates close to each other. Outer region of plate A is labelled as I, outer
region of plate B is labelled as III, and the region between the plates, A and Bf is labelled as II.
Charge density of plate A, σ = 17.0 × 10–22 C/m2
Charge density of plate B, σ = –17.0 × 10–22 C/m2
In the regions, I and III, electric field E is zero. This is because charge is not enclosed by the respective
plates.
Electric field E in region II is given by the relation,
σ
∈0
Where,
E=
ε0 = Permittivity of free space = 8.854 × 10–12 N–1 C2 m–2
E=
17.0 × 10 −22
8.854 × 10 −12
Therefore, electric field between the plates is 1.92 × 10–10 N/C.
54
Question 7:
(a)
A conductor A with a cavity as shown in Fig. ...... (a) is given a charge Q. Show that the entire charge
must appear on the outer surface of the conductor, (b) Another conductor B with charge q is inserted
into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is
Q + q [Fig. (b)] (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its
environment. Suggest a possible way.
(a)
(b)
Answer:
(a)
Let us consider a Gaussian surface that is lying wholly within a conductor and enclosing the cavity.
The electric field intensity E inside the charged conductor is zero.
Let q is the charge inside the conductor and ε0 is the permittivity of free space. According to Gauss’s
law,
Flux,
KH KKH q
φ = E.ds =
ε0
Here,
E= 0
q
= 0
∈0
3
∈0 = 0
∴
q= 0
Therefore, charge inside the conductor is zero.
The entire charge Q appears on the outer surface of the conductor.
(b)
The outer surface of conductor A has a charge of amount Q. Another conductor B having charge +q
is kept inside conductor A and it is insulated from A. Hence, a charge of amount –q will be induced
in the inner surface of conductor A and +q is induced on the outer surface of conductor A. Therefore,
total charge on the outer surface of conductor A is Q + q.
(c)
A sensitive instrument can be shielded from the strong electrostatic field in its environment by
enclosing it fully inside a metallic surface. A closed metallic body acts as an electrostatic shield.
Question 8: A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7 C distributed uniformly on its
surface. What is the electric field
(a)
Inside the sphere
(b)
Just outside the sphere
(c)
At a point 18 cm from the centre of the sphere?
55
Answer
(a) Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q × 1.6 × 10–7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor,
then charges will move to neutralize it.
(b) Electric field E just outside the conductor is given by the relation,
q
E = 4π ∈ r 2
0
Where,
∈0 = Permittivity of free space
1
2
–2
4π ∈0 = 9 × 109 N m C
∴
(c)
1.6 × 10−7 × 9 × 10−9
E=
(0.12) 2
= 103 NC–1
Therefore, the electric field at a point 18 cm from the centre of the sphere is 105 N C–1.
Electric field at a point 18 m from the centre of the sphere = E1
Distance of the point from the centre, d = 18 cm = 0.18 m
q
E1 = 4π ∈ d 2
0
9 × 109 × 1.6 × 10−7
=
(18 × 10−2 )2
= 4.4 × 104 N/C
Therefore, the electric field at a point 18 cm from the centre of the sphere is 4.4 × 104 N/C
Question 9:
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be
the capacitance if the distance between the plates is reduced by half, and the space between them is filled
with a substance of dielectric constant 6?
Answer :
Capacitance between the parallel plates of the capacitor, C = 8 pF
Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air,
k=1
Capacitance, C, is given by the formula,
k ∈0 A
d
∈0 A
=
d
C=
56
...(i)
If distance between the plates is reduced to half, then new distance, d =
d
2
Dielectric constant of the substance filled in between the plates, k = 6
k ∈0 A 6 ∈0 A
=
...(ii)
d
d
2
Taking ratios of equations (i) and (ii), we obtain
C′ = 2 × 6C
= 12 C
= 12 × 8 = 96 pF
Therefore, the capacitance between the plates is 96 pF.
Question 10
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What Is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer: For the parallel combination of the capacitors, equivalent capacitor C′ is given by the algebraic
sum,
C′ = 2 + 3 + 4 = 9 pF
Therefore, total capacitance of the combination is 9 pF.
(b) Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance Cand potential difference V is given by the relation,
q = VC
...(i)
For C = 2 pF,
Charge = VC = 100 × 2 = 200 pC = 2 × 10–10 C
For C = 3 pF,
Charge = VC = 100 × 3 = 300 pC = 3 × 10–10 C
For C = 4 pF,
Charge = VC = 100 × 4 = 200 pC = 4 × 10–10 C
Question 11 : A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the sup and is
connected to another uncharged 600 pF capacitor. How much electrostatic ........ is lost in the process?
Answer
Capacitance of the capacitor, C = 600 pF
Potential difference. V = 200 V
Electrostatic energy stored in the capacitor is given by,
C′ =
E=
1
CV 2
2
57
=
1
× (600 × 10−12 ) × (200) 2
2
= 1.2 × 10–5 J
If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is
connected to it, then equivalent capacitance (C) of the combination is given by,
1
1 1
+
=
C′
C C
=
1
1
2
1
+
=
=
600 600 600 300
∴
C′ = 300 pF
New electrostatic energy can be calculated as
E′ =
=
Loss is electrostatic energy
1
× C′ × V 2
2
1
× 300 × (200) 2
2
= 0.6 × 10–5 J
= E – E′ = 1.2 × 10–5 – 0.6 × 10–5 = 0.6 × 10–5
= 6 ×10–6 J
Therefore, the electrostatic energy lost in the process is 6 × 10–6 J.
Question 12:
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of –2 × 10–9 C
from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a poi (0, 6 cm, 9 cm).
Answer
Charge located at the origin, q = 8 mC= 8 × 10–3C
Magnitude of a small charge, which is taken from a point P to point R to point Q, q1 = – 2 ×10–9 C
All the points are represented in the given figure.
58
Point P is at a distance, d1 = 3 cm, from the origin along z-axis.
Point Q is at a distance, d2 = 4 cm, from the origin along y-axis.
q
Potential at point P, V1 = 4π ∈ × d
0
1
q
Potential at point Q, V2 = 4π ∈ × d
0
2
Work done (W) by the electrostatic force is independent of the path.
∴
W = q1 [V2 – V1]


qq1
q
q
−
=
= q1 
 4π ∈0 d 2 4π ∈0 d1  4π ∈0
∴
1 1
 − 
 d 2 d1 
...(i)
1 
9
−3
−9  1
−
= 1.27 J
W = 9 × 10 × 8 × 10 × ( −2 × 10 ) 
 0.04 0.03 
Therefore, work clone during the process is 1-27 J.
Question 13 : Figure shows a charge array known as an electric quadrupole. For a point on the axis of the
quadrupole, obtain the dependence of potential on rfor r/a > > 1, and contrast your results with that due to
an electric dipole, and an electric monopole (i.e., a single charge).
Answer
Four charges of same magnitude are placed at points X, Y, Y, and Z respectively, as shown in the following
figure.
A point is located at P, which is r distance away from point Y.
The system of charges forms an electric quadrupole.
It can be considered that the system of the electric quadrupole has three charges.
Charge +q placed at point X
Charge –2q placed at point Y
Charge +q placed at point Z
Electrostatic potential caused by the system of three charges at point P is given by,
V=
1  q
2q
q 
−
+

4π ∈0  XP YP ZP 
59
=
1  q
2q
q 
−
+
4π ∈0  r + a r r − a 
=
q  r(r − a) − 2(r + a)(r − a) + r(r + a) 

4π ∈0 
r(r + a)(r − a)

q  r 2 − ra − 2r 2 + 2a 2 + r 2 + ra 
q
=
=


2
2
4π ∈0 
r(r − a )
 4π ∈0
 2a 2 
 2
2 
 r(r − a ) 
2qa 2
=
 a2 
4π ∈0 r 3  1 − 2 
 r 
Since
∴
r
>> 1
a
a
<< 1
r
a2
is taken as negligible.
r2
2qa 2
∴ V=
4π ∈0 r 3
It can be inferred that potential, V ∝
1
r3
However, it is known that for a dipple, V ∝
1
r2
1
r
Question 14: An electrical technician requites a capacitance of 2 µF in a circuit across a potential difference
of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential
difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of
capacitors.
And, for a monopole, V ∝
Answer
Tota required capacitance, C = 2 pF
Potential difference, V = 1 kV = 1000 V
Capacitance of each capacitor, C1 = lµF
Each capacitor can withstand a potential difference, V1 = 400 V
Suppose a number of capacitors are connected in series and these series circuits are connected in
parallel (row) to each other. The potential difference across each row must be 1000 V and potential difference
across each capacitor must be 400 V. Hence, number of capacitors in each row is given as
1000
= 2.5
400
Hence, there are three capacitors in each row,
60
1
1
= µF
1+1+1 3
Let there are n rows, each having three capacitors, which are connected in parallel.
Capacitance of each row =
Hence, equivalent capacitance of the circuit is given as
1 1 1
+ + + ....n terms
3 3 3
=
n
3
However, capacitance of the ciruit is given as 2 µF.
∴
n
= 2
3
n= 6
Hence, 6 rows of three capacitors are present in the circuit. A minimum of 6 × 3 i.e., 18 capacitors
are required for the given arrangement.
61
Current Electricity
Learning Objectives :
●
Electric Current
●
Electric Currents in Conductors
●
Ohm’s law
●
Drift of Electrons and the Origin of Resistivity
●
Limitations of Ohm’s Law
●
Resistivity of various Materials
●
Temperature Dependence of Resistivity
●
Electrical Energy, Power
●
Combination of Resistors — Series and Parallel
●
Cells, emf, Internal Resistance
●
Cells in Series and in Parallel
●
Kirchhoff’s Laws
●
Wheatstone Bridge
●
Meter Bridge
●
Potentiometer
CONCEPTUAL PROBLEMS
1.
The standard resistance coil are made of manganin. Why?
Hint. When current flows through a wire its temperature will increase. For alloys like manganin the
temperature coefficient of resistance is very small. Thus, even if the temperature of the coil increases
its resistance will remain constant.
2.
Does the value of temperature coefficient of resistance always positive?
Hint. No, positive temperature coefficient of resistance implies that the resistance increases with increase
in temperature. But for semiconductors and insulators the resistance decreases with increase in
temperature and their temperature coefficient f resistance is negative.
62
3.
In Wheatstone bridge method why do we break the galvanometer circuit first and then the battery
circuit?
Hint. This is done to save the galvanometer from large current, which can flow due to large induced emf in
the circuit at the time of breaking the circuit.
4.
Is Ohm’s law, universally applicable to all current carrying elements?
Hint. No, no ohm’s law can be applicable only for substances for which V-I curve is a straight line or
whose resistance remains constant with potential difference. For non- ohmic conductors like diodes
ohm’s law s not applicable.
5.
Currents of the order of 0.1A through human body are fatal what causes death: heating due to current
or something else?
Hint. When current flows through the body it has electromagnetic wave associated with it. This can interfere
with the wave connecting brain with body parts and can affect the functioning of heart. Cardiac
arrest is the cause of death in most cases.
6.
What happens if cell and galvanometer in Wheatstone bridge are interchanged?
Hint. There will be no change in the balance point condition of Wheatstone bridge even if galvanometer
and cell are interchanged.
7.
If electron drift speed is small, and the charge on electron is also very small, how can we obtain large
amount of current in the conductor?
Hint. Large currents can be obtained due to large number density of charge carriers inside the conductors,
which is of the order of 1029 m-3.
8.
A steady current flows through the cylindrical conductor is there any electric field inside the conductor?
Hint. Current flows inside the conductor when potential difference is applied across the two ends of the
conductor. Thus, if V is the potential difference applied and l is the length of the conductor then
electric ield inside the conductor is E = V/l
9.
There is an impression among people that person touching a high power line gets stuck with it. Is that
true? Explain.
Hint. No, a person will not get stuck to the wire. But when current flows through the wire it affects the
nervous system and snaps the connection between brain and body parts. It is the brain, which is not
giving an order to the hand to move it.
10.
Explain, why bending of wire does not affect the resistance?
Hint. Free electrons in the wire has small value of drift velocity and thus very low inertia. Due to it, they
are unable to move aound bends easily.
11.
A wire is carrying current. Is it charged?
Hint. No, flow of current only means that electron are moving in direction opposite to the electric field
intensity. But, number of electrons in the wire are always equal to the number of protons in the wire
Thus wire as a whole is neutral.
12.
Why the connection between resistors in Wheatstone bridge or Meter Bridge are made of thick
copper wires?
63
Hint. Connections are made of thick copper wires so that the resistance of copper wires is negligible as
compared to other resistances used in the circuit.
13.
Why balance point should be near the middle point of the meter bridge wire?
Hint. Meter bridge is based on the principle of Wheatstone bridge. When null point is obtained in the
middle the ratio arms resistances is of the same order and sensitivity of potentiometer is maximum.
14.
What happens to the drift velocity of electrons and resistance R, if the length of conductor is doubled
[keeping potential difference unchanged]?
Hint. Drift velocity is inversely proportional to the length for constant value of V potential difference.
Thus, if the length of the conductor is doubled drift velocity will be halved.
The resistance of conductor is directly proportional to the length of the conductor. Thus, if length is
doubled resistance will also be doubled.
15.
A large number of free electrons are present in metals, why there is no current in absence of potential
difference?
Hint. In absence of potential difference, the electrons are moving randomly and keeps on colliding with
each other. Thus, even if the thermal speed of the electrons is of the order of 106 m/s there drift drift
velocity is zero due to Brownian motion of electrons.
16.
When electrons drift in a metal from lower to higher potential, does that mean that all the free
electrons of the metal are moving in the same direction?
Hint. No, electrons will have resultant drift velocity in the direction from lower to higher potential. But
they still suffer collisions inside metal and with each collision the direction of motion of electron
changes.
17.
Can meter bridge be used for measuring very low resistances?
Hint. Meter bridge can’t be used for measuring low resistances because in that case we can’t neglect the
resistance of connecting wires and copper strips.
18.
A steady current flows through wire of non uniform cross-section. Explain which of these quantities
is constant along the conductor: current, current density, electric field and drift speed?
Hint. Only current is constant along the length as other quantities are inversely proportional to the crosssectional area which is variable.
19.
Why the jockey should not be pressed against the potentiometer wire?
Hint. Jockey should not be pressed against the potentiometer wire because pressing the jockey will change
the cross-sectional area of the wire and thus resistance per unit length becomes variable.
20.
Can potential difference of a cell be greater than its emf?
Hint. No, potential difference is less than emf of the cell when the cell is being discharged. But, during
charging, action of the cell potential difference s greater than the emf f the cell.
21.
If the resistance of our body is so large [of the order of 10 kilowatt], why does one experience shock
when one accidentally touches the line wire, say a 240-volt supply?
Hint. Even if the resistance is high the magnitude of current flow will be of the order of 0.024A which can
interfere with the nerve process and affects the beating of heart.
64
22.
It is easier to start a car engine on a warm day than on a chilly day, why?
Hint. On chilly day the temperature of a surroundings which is low results in increase in internal resistance
of the battery and decreases the current supply from the battery.
23.
Are Kirchoff’s law applicable to both a.c. and d.c?
Hint. Yes, Kirchoff’s two laws are basically law of conservation of charge and law of conservation of
energy. Thus, they are equally applicable to both ac as well as dc.
24.
On what factors does the emf of a cell depends?
Hint. Emf of a cell depends upon [a] nature of electrolyte and its concentration [b] nature of electrodes [c]
temperature of the electrolyte.
25.
If the temperature of good conductor decreases, how does the relaxation time of electrons in the
conductor change?
Hint. If the temperature of conductor decreases its resistance will decrease and relaxation time increases.
26.
Is it possible that there is no potential between the plates of the cell? If yes, under what conditions?
Hint. Yes, it is possible for terminal potential difference across the cell to be zero, if the plates of the cell
are short-circuited.
27.
What does the no deflection position in the galvanometer of potentiometer experiment tells us about
the flow of current?
Hint. When null point is obtained on the wire, then potential difference between one end of wire and null
point is equal to the emf of the cell to be measured. Thus current through galvanometer circuit is
zero.
28.
When is Wheatstone bridge most sensitive?
Hint. Wheatstone bridge is most sensitive when four resistances in the Wheatstone bridge are of the same
order or null point is obtained in the middle of meter bridge wire.
29.
On what factors, does the potential gradient of the potentiometer wire depend?
Hint. The potential difference per unit length of the wire depends on [a] current through the wire [b]
specific resistance of the wire. [c] area of cross-section of the wire.
30.
Why the current should not be passed through the potentiometer wire for long time?
Hint. For potentiometer to work properly, pd per unit length should be constant. But if the current is passed
through for a long time then the wire gets heated up and its specific resistance and resistance per unit
length [i.e. r = r/A] becomes variable.
31.
Copper wire is not used for making potentiometer wires. Why?
Hint. Copper should not be used for making potentiometer wire because temperature coefficient of resistance
for copper is large and when current flows through wire, its specific resistance and resistance per
unit length begins to vary due to heat generation.
32.
What do you understand by sensitiveness of potentiometer and how can you increase the sensitiveness
of the potentiometer?
65
Hint. Sensitivity of potentiometer is the minimum value of potential difference which it can measure or
p.d. across unit length of wire. It can be increased by [a] increasing the length of the wire [b] decreasing
the current flow through the potentiometer wire.
33.
Can you interchange the positions of the battery in the auxiliary circuit and cell whose emf is to b
determined in potentiometer circuit diagram?
Hint. No, battery and cell can’ be interchanged because the emf of the battery should always be greater
than the emf of the cell is to be measured. If the positions of two are interchanged there will be no
null point on the wire.
34.
Why do we prefer to use potentiometer for measurement of emf rather than the voltmeter?
Hint. Potentiometer is preferred for measuring the emf of the cell because it uses null deflection method
for measurement, thus it draws no current from the cell whose emf is to be measured.
35.
The emf of the driver cell should always be greater than emf of the cell to be measured. Why?
Hint. If emf of the driver cell is less than the emf of the cell to be measured because in that case fall of
potential across the length of potentiometer will be less than emf of the cell to be measured.
ACTIVITIES
I.
UNDERSTANDING CURRENTS IN SERIES AND PARALLEL CIRCUITS
Learning Objectives:
1.
To understand electric current as a flow of electrical charge.
2.
To understand that, since the total charge in a circuit is conserved, current is also conserved through
series and parallel connections.
3.
To understand that the current through two components in series is the same
4.
To understand that voltage is a cause of the flow of current.
66
Discussion:
Thinking about your observations, discuss the correctness of the following statements:
●
As the voltage supplied increases, the current through a bulb also increases.
●
The brightness of a bulb depends upon the current passing through it.
●
For a given voltage, when two bulbs are in series, the current is reduced.
●
The current passing through the battery is equal to the sum of the currents passing through the bulbs.
II. UNDERSTANDING VOLTAGE SHARING IN SERIES CIRCUITS
Learning Objectives:
1.
To understand electric current as a flow of electrical charge.
2.
To understand that, since the total charge in a circuit is conserved, current is also conserved through
series and parallel connections.
3.
To understand that the current through two components in series is the same.
4.
To understand that voltage is a cause of the flow of current.
Discussion:
Thinking about your observations, discuss the correctness of the following statements:
●
The brightness of a bulb depends upon the current passing through it and the voltage across it.
●
The sum of the voltages across the resistor and bulb is equal to the voltage supplied by the battery.
●
The voltage supplied by the battery is shared equally between the resistor and bulb.
●
The current in the circuit reduces as the total resistance is increased.
III. UNDERSTANDING THE CONCEPTS OF VOLTAGE AND CURRENT
Learning Objectives:
1.
To understand electric current as a flow of electrical charge.
2.
To understand that voltage is a cause of the flow of current and is a measure of electrical energy
change as current flows around a circuit.
67
3.
4.
To understand that a battery donates electrical energy to charge flowing through it and that components
in the circuit convert that energy into other forms.
The rate of energy conversion in each bulb depends upon the voltage across it as well as the current
through it.
Discussion:
Thinking about your observations, discuss the correctness of the following statements:
●
The ‘EMF’ is the electrical energy donated to each coulomb of charge flowing through the battery.
●
The energy bars show that the electrical energy of 1 coulomb diminishes as it passes around the
circuit.
●
The current through the bulbs is the same.
●
The rate of energy conversion in each bulb depends upon the voltage across it as well as the current
through it.
Question 1: The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4Ω.
what is the maximum current that can be drawn from the battery?
Answer : Emf of the battery, E = 12 V
Internal resistance of the battery, r = 0.4 Ω
Maximum current drawn from the battery = I
According to Ohm’s law,
E = Ir
I =
E
r
68
12
= 30 A
0.4
The maximum current drawn from the given battery is 30 A.
=
Question 2:
(a) Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the
potential drop across each resistor.
Answer
(a) Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of the
combination is given by the algebraic sum of individual resistances.
Total resistance = 1 + 2 + 3 = 6 Ω
(b) Current flowing through the circuit = I
Emf of the battery, E = 12 V
Total resistance of the circuit, R = 6 Ω
The relation for current using Ohm’s law is,
I=
=
E
R
12
=2A
6
Potential drop across 1 Ω resistor = V1
From Ohm’s law, the value of Vi can be obtained as
V1 = 2 × 1 = 2 V
... (i)
Potential drop across 2 Ω resistor = V2
Again, from Ohm’s law, the value of V2 can be obtained as
V2 = 2 × 2 = 4 V
... (ii)
Potential drop across 3 Ω resistor = V3
Again, from Ohm’s law, the value of V3 can be obtained as
V3 = 2 × 3 = 6V
... (iii)
Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V
Question 3: Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2
A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged
over the temperature range involved is 1.70 × 10–4 °C–1
Answer
Supply voltage, V = 230 V
Initial current drawn, I1 = 3.2 A
69
Initial resistance = R1, which is given by the relation,
R1 =
=
V
I
230
= 71.87 Ω
3.2
Steady state value of the current, I2 = 2.8 A
Question 4: Aheating element using nichrome connected to a 230 V supply draws an initial current of 3.2
A which settles after a few seconds toa steady value of 2.8 A. What is the steady temperature of the heating
element if the room temperature is 27.0 °C? Temperature coefficient of resistance of nichrome averaged
over the temperature range involved is 1.70 × 10–4 °C–1.
Answer
Supply voltage, V = 230 V
Initial current drawn, I1 = 3.2 A
Initial resistance = R1, which is given by the relation,
R1 =
=
V
I
230
= 71.87 Ω
3.2
Steady state value of the current, I2 = 2.8 A
Resistance at the steady state = R2, which is given as
R2 =
230
= 82.14 Ω
2.8
Temperature co-efficient of nichrome, a = 1.70 × 10–4 °C–1
Initial temperature of nichrome, T1 = 27.0°C
Study state temperature reached by nichrome = T2
T2 can be obtained by the relation for a,
a=
T2 – 27 °C =
R 2 − R1
R1 (T2 − T1 )
82.14 − 71.87
= 840.5
71 − 87 × 1.7 × 10 −4
T2 = 840.5 + 27 = 867.5 °C
Therefore, the steady temperature of the heating element is 867.5°C
70
Question 5: Determine the current in each branch of the network shown in fig.
Answer : Current flowing through various branches of the circuit is represented in the given figure.
B
( I2 – I4)
Ω
10
Ω
5
I4
I2
A
D
10 Ω
( I3 + I2 = I1)
10
Ω
5Ω
I1
E
C
5Ω
I3
( I3 + I4)
F
10 V
For the closed circuit ABDA, potential is zero i.e.,
10I2 + 5I4 –5I3 = 0
2I2 + I4 –I3 = 0
I3 = 2I2 + I4
For the closed circuit BCDB, potential is zero i.e.,
5(I2 – I4) – 10(I3 + I4) –5I4 = 0
5I2 + 5I4 – 10I3 – 10I4 –5I4 = 0
5I2 – 10I3 – 20I4 = 0
I2 = 2I3 + 4I4
For the closed circuit ABCFEA, potential is zero i.e.,
–10 + 10 (I1) + 10(I2) + 5(I2 – I4) = 0
10 = 15I2 + 10I1 – 5I4
3I2 + 2I3 – I4 = 2
71
... (1)
... (2)
... (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 3I4 + I4
– 3I3 = 9I4
– 3/4 = +I3
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
– 4I4 = 2I2
I2 = – 2I4
It is evident from the given figure that,
I1 = 73+ I2
Putting equation (6) in equation (1), we obtain
3I2 + 2(I3 + I2) – I4 = 2
5I2 + 2I3 –I4 = 2
Putting equations (4) and (5) in equation (7), we obtain
5(–2I4) + 2(– 3I4) – I4 = 2
– 10I4 – 6I4 – I4 = 2
17I4 = – 2
I4 =
−2
A
17
Equation (4) reduces to
I3 = –3(I4)
 −2  6
= −3   = A
17
17
I2 = –2(I4)
 −2  4
= −2   = A
17
17
I2 – I4 =
4  −2  6
−  = A
17  17  17
I3 + I4 =
6  −2  4
+  = A
17  17  17
I1 = I3 + I2
Therefore, current in branch AB =
4
A
17
72
... (4)
... (5)
...(6)
... (7)
In branch BC =
6
A
17
In branch CD =
−4
A
17
In branch AD =
6
A
17
 −2 
In banch BD =   A
 17 
Total current =
4 6 −4 6 −2 10
+ +
+ +
= A
17 17 17 17 17 17
Question 6:
(a) In a metre bridge [Fig.], the balance point is found to be at 39.5 cm from the end A, when the resistor
Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a
Wheatstone or meter bridge made of thick copper strips?
(b) Determine the balance point of the bridge above if X and Vare interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge?
Would the galvanometer show any current?
Answer
A metre bridge with resistors X and Y is represented in the given figure.
(a)
Balance point from end A, /1 = 39.5 cm
Resistance of the resistor Y = 12.5 Ω
Condition for the balance is given as,
X
100 − I1
=
Y
I1
100 − 39.5
× 12.5 = 8.2 Ω
X=
39.5
73
Therefore, the resistance of resistor X is 8.2 Ω.
The connection between resistors in a Wheatstone or metre bridge is made of thick copper strips to
minimize the resistance, which is not taken into consideration in the bridge formula.
(b) If X and Kare interchanged, then 4 and 100-/i get interchanged. The balance point of the bridge will
be 100-4 from A.
100–/1 = 100 – 39.5 = 60.5 cm
Therefore, the balance point is 60.5 cm from A.
(c) When the galvanometer and cell are interchanged at the balance point of the bridge, the galvanometer
will show no deflection. Hence, no current would flow through the galvanometer.
Question 7: A storage battery of emf 3.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc
supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What
is the purpose of having a series resistor in the charging circuit?
Answer :
Emf of the storage battery, E - 8.0 V
Internal resistance of the battery, r = 0.5 Ω
DC supply voltage, V = 120 V
Resistance of the resistor, R - 15.5 Ω
Effective voltage in the circuit = V1
R is connected to the storage battery in series. Hence, it can be written as
V1 = V – E
V1 = 120 – 8 = 112 V
Current flowing in the circuit = I, which is given by the relation,
V1
I=
R+r
=
112
112
=
=7A
15.5 + 5 16
Voltage across resistor R given by the product, IR = 7 × 15.5 = 108.5 V
DC supply voltage = Terminal voltage of battery + Voltage drop across R
Terminal voltage of battery = 120 – 108.5 = 11.5 V
A series resistor in a charging circuit limits the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Question 8:
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω are joined in
series to provide a supply to a resistance of 8.5 Ω. What are the current drawn rrom tne supply ana its
terminal voltages
(b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω. What
maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?
74
Answer
(a) Number of secondary cells, n = 6
Emf of each secondary cell, E – 2.0 V
Internal resistance of each cell, r – 0.015 Ω
series resistor is connected to the combination of cells.
Resistance of the resistor, R – 3.5 Ω
Current drawn from the supply = I, which is given by the relation,
nE
R + nr
6×2
=
8.5 + 6 × 0.015
12
= 1.39A
=
8.59
Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A
I =
(b)
Therefore, the current drawn from the supply is 1.39 A and terminal voltage is 11.87 A.
After a long use, emf of the secondary cell, E = 1.9 V
Internal resistance of the cell, r = 380 Ω
Hence, maximum current =
E 1.9
=
= 0.005A
r 380
Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is required to
start the motor of a car, the cell cannot be used to start a motor.
Question 9:
Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the infinite network
shown in Fig. 3.32, Each resistor has 1 Ω resistance.
Answer :
The resistance of each resistor connected in the given circuit, R = 1 Ω
Equivalent resistance of the given circuit = R
The network is infinite. Hence, equivalent resistance is given by the relation,
∴
R = 2+
R
(R + 1)
75
(R)2 – 2R – 2 = 0
R=
=
2± 4+8
2
2 ± 12
= 1± 3
2
Negative value of R cannot be accepted. Hence, equivalent resistance.
R = (1 + 3) = 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio,
12
= 3.72 A
3.23
Question 10: Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a
1.5 V cell. The balance point of the cell In open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the
external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine
the internal resistance of the cell.
Answer :
Internal resistance of the cell = r
Balance point of the cell in open circuit, I1 = 76.3 cm
An external resistance (R) is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, I2 = 64.3 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,
 I1 − I 2 
R
r= 
 I 2 
=
76.3 − 64.8
× 9.5 = 168Ω
64.8
Therefore, the internal resistance of the cell is 1.68Ω.
76
Electromagnetic Induction
and Alternating Current
Learning Objectives :
●
Electromagnetic induction
●
Faraday’s laws
●
Induced emf and current
●
Lenz’s Law
●
Eddy currents
CONCEPTUAL PROBLEMS
1.
Lenz’s law is a consequence of the law of conservation of (a) Linear momentum (b) Charge
(c) Angular momentum (d) Energy (e) None of the above
Hint. According to Lenz’s law, the induced current will always oppose the change which causes it. Indeed,
the induced current has to oppose the change because, otherwise the change which causes the current
will persist and the current will continue to flow once it is started. You will then have a supply of
energy (in the form of electric current) without any external agency doing any work. This will then
violate the law of conservation of energy, which is impossible. So, Lenses law holds good in accordance
with the law of conservation of energy. Option (d) is the correct answer.
2.
The electrical entity inductance can be compared to the mechanical entity (a) Energy (b) Impulse
(c) Momentum (d) Torque (e) Inertia
Hint. The correct option is (e). Inertia in Mechanics is the property by which a body (or, a mechanical
system) tries to oppose any change in its state of rest or of uniform motion. Inductance is the property
by which an electric circuit tries to oppose any change of current flowing in it. So, inductance and
inertia are comparable. The above two questions high light two basic points. Now consider the
following:
3.
A bar magnet is released into a copper ring which is directly below it. What about the acceleration
of the magnet? Greater than ‘g’ or equal to ‘g’ or less than ‘g’?
Hint. The acceleration is less than ‘g’ since the falling magnet will generate an induced current in the
copper ring and the induced current will oppose the motion of the magnet.
77
4.
A jet plane is flying horizontally at a speed of 1800 km/hour. What is the potential difference
developed between the tips of its wings if the wing span is 25m? Earth’s magnetic field at the
location is 0.4 gauss and the angle of dip is 30°.
(a)
25mV
(b) 250mV
(c) 500mV
(d) 2.5V
(e)
5V
Hint. The motional emf developed between the tips of the wings is given by V = BvLv where Bv is the
vertical component of the earth’s magnetic flux density, L is the distance between the tips of the
wings (wing span) and ‘v’ is the velocity. [Note that this emf is produced because of the cutting of the
vertical field lines and this is why we use the vertical component of the field]. We have Bv = B sin
30 = 0.4×10–4×½ = 0.2×10–4 tesla. Also, L = 25m and v = 500m/s. The emf then works out to be
0.25V = 250mV. [Note that gauss is the cgs unit of magnetic flux density which is often used. One
tesla = 104 gauss]. Let us consider another question involving Faraday’s disc, the first electric generator:
5.
A circular copper disc 10cm in diameter rotates 1800 times per minute about a central axis at
right angles to the plane of the disc. A uniform magnetic field of 1 tesla is applied perpendicular
to the plane of the disc. The voltage induced between the centre and the edge of the disc is
(a)
0.235V
(b) 0.47V
(c) 2.35V
(d) 4.7V
(e)
zero
Hint. The motional emf induced when a conductor moves perpendicular to a magnetic field is the product
of the area swept per second (by the conductor) and the magnetic field. Therefore, in the present
case, induced voltage = area swept per second by a radius × B = n πr2B = (1800/60) × π × (0.05)2 ×
1
= 0.2356V. The correct option therefore is (a).
Let us modify this question as follows
6.
A copper rod 10cm long rotates 1800 times per minute about an axis passing through one end
at right angles to the rod. A uniform magnetic field of 1 tesla is applied perpendicular to the
plane of rotation. The voltage induced between the ends of the rod is:
(a)
0.235V
(b) 0.942V
(c) 2.35V
(d) 4.7V
(e)
Zero
Hint. This mcq is similar to the previous one. The voltage induced will be four times the previous value
since the area swept is four times. The correct option is (b).
Now let us modify the above ‘rod problem’ as follows:
7.
An aluminium rod 10 cm long rotates 1800 times per minute about an axis passing through its
centre at right angles to it. A uniform magnetic field of 1 tesla is applied perpendicular to the
plane of rotation. The voltage induced between the ends of the rod is:
(a)
0.235V
(b) 0.47V
(c) 2.35V
(d) 4.7V
(e)
Zero
Hint. This is a simple question but there is chance of committing a mistake! The correct option is neither
(a) nor (b). The potential at the ends (with respect to the mid point of the rod) will be the same
(0.235V) so that the potential difference between the ends will be zero [Option (e)].
You should note that the motional emf is generated because of the shifting of mobile charge carriers
due to the Lorentz force.
78
7.
A copper disc of radius 0.1m is rotated about its centre with 20 revolutions per second in a
uniform magnetic field of 0.1T with its plane perpendicular to the field. The emf induced across
the radius of the disc is
(a)
π /20volt
(b) π /10volt
(d)
10π
π millivolt
(e) 2π
π millivolt
(c) 20π
π millivolt
Hint. The emf induced = n πr2B = 20π(0.1)2 ×0.1 = 0.02 π volt = 20π milli volt [Option (c)].
8.
A varying magnetic flux linking a coil is given by φ = Xt2. If at time t = 3s, the emf induced is
9V, then the value of X is
(a)
0.66Wb.s-2
(b) 1.5Wb.s-2
(d)
–1.5Wb.s-2
(e) –0.33Wb.s–2
(c)–0.66Wb.s-2
Hint. The induced emf = –dΦ/dt. Therefore, –2Xt = 9 when t=3s so that -6X = 9, from which X = -1.5. The
correct option is (d).
ACTIVITIES
I.
Use the apparatus shown below to show how a current can be induced in a wire using magnetism.
Explore the factors that affect the size of the induced emf and current.
Straight wire in a field
Apparatus
● Steel yoke
● Magnadur magnets (2) (as in Westminster Electromagnetic kit)
● 26 swg PVC-covered copper wire
● Wire strippers
● Cathode-ray oscilloscope or centre-zero galvanometer (Philip Harris Y19360/7 or similar)
What to do:
Connect a long wire to a sensitive galvanometer or cathode-ray
oscilloscope. Note the effect on the meter or CRO as you do each of the
following:
- Move the wire into a strong magnetic field as shown above
- Hold the wire at rest within the field.
- Remove the wire from the field.
- Repeat, but move the wire more quickly or more slowly.
- Reverse the field and repeat.
- Make a loop in the wire and pass it over one of the magnetic poles so that two thicknesses of wire
move through the field together. Keep the wire still and move the magnet.
79
II. Working in pairs undertake the following tasks:
Experiment (a)
Wind the copper wire into a coil of 20 turns. Connect it to the meter. Hold the coil still and push the
North pole of the magnet into the coil. Record what happens. Now try the same with the South pole.
Try moving it fast and then slowly. Now hold the magnet still and move the coil. Watch carefully as
the meter movements will be small. Record the size and direction of the current on the meter in the
Worksheet table.
Experiment (b)
Connect the wire to the meter as shown in the diagram. Move it downwards between the two magnets,
record what happens on the meter. Now move it upwards. Now move the wire sideways between the
magnets and then from end to end. Watch carefully as the meter movements will be small. Record the
size and direction of the current on the meter in the
Worksheet table. How do these readings compare with those you got in the first experiment?
Experiment (c).
Repeat experiment (a) but this time use the large mounted coil.
Demonstration
Induced voltage can be demonstrated by connecting a solenoid to a galvanometer and moving a bar
magnet in and out of the solenoid.
The induced voltages result in currents that can be detected by the galvanometer. Note the results of
different actions of the magnet. When the magnet is pushed into or pulled out of the solenoid, a
current is induced. When the magnet is moved one way (e.g., into the coil), the needle deflects one
way; when the magnet is moved the other way (out of the coil), the needle deflects the other way. Not
only can a moving magnet cause a current to flow in the coil, the direction of the current depends on
how the magnet is moved. However, if the magnet is at rest inside the solenoid, no current is produced
in the coil. Consequently, only changing magnetic fields will induce currents.
Class Discussion
You can conclude from these observations that a changing magnetic field will induce a voltage in the
coil, causing a current to flow. It follows that if the magnetic flux through a coil is changed, a voltage
will be produced. This voltage is commonly referred to as the induced voltage. The term voltage,
although common, is an historical term, and students say that a voltage is induced.
There are three possible ways to change magnetic flux (remember that magnetic flux is defined as
Φ = BAcosè):
1. Change the magnetic field
2. Change the area of the loop
3. Change the angle between the field and the loops
Demonstration
Allow a moving magnet to induce a current in a coil of wire. Connect the coil to a galvanometer and
move a bar magnet in and out of the centre of the coil. Note the deflection of the galvanometer needle.
Describe what happens when you move the magnet more or less quickly, and when you put the North
or South pole into the coil first. Then, try other relative motions and positions, including some where
80
the axes of the magnet and the coil are perpendicular. Does it make a difference if you move the coil,
rather than the magnet, in the same relative motion?
III. Induction in an Aluminum Can
Teacher’s Guide
In this activity, Lenz’s Law is demonstrated. Lenz’s Law states that an induced electromotive force
generates a current that induces a counter magnetic field that opposes the magnetic field generating
the current. In this activity, an empty aluminum can floats on water in a tray, such as a Petri dish.
Students spin a magnet just inside the can without touching the can. The can begins to spin.
Understanding what happens can be explained in steps:
• First, the twirling magnet creates an alternating magnetic field. Students can use a nearby compass
to observe that the magnetic field is really changing.
• Second, the changing magnetic field permeates most things around it, including the aluminum can
itself. A changing magnetic field will cause an electric current to flow when there is a closed loop
of an electrically conducting material. Even though the aluminum can is not magnetic, it is metal
and will conduct electricity. So the twirling magnet causes an electrical current to flow in the
aluminum can. This is called an “induced current.”
• Third, all electric currents create magnetic fields. So, in essence, the induced electrical current
running through the can creates its very own magnetic field, making the aluminum can magnetic.
Now the aluminum can’s induced alternating magnetic field interacts with the twirling magnet’s
alternating magnetic field. This interaction spins the can since the magnetic forces are opposed, as
Lenz’s law states. Because the can sits on top of water, the friction between the can and the
surface is very small, and only the magnetic forces act on the can while the students twirl the
magnet .
Materials
●
Small aluminum can
●
Container (in which to float aluminum can)
●
Spill tray
●
Small cylindrical magnet (neodymium works best)
●
Thread
●
Tapwater11
Procedure
1. Have enough materials on hand for student groups of two. Cat food cans work well. Just be sure they
are made of aluminum. Soft drink cans do not work well. Their walls are too thin and sharp edges on
cut cans poses a safety risk. Petri dishes make good float containers, and Styrofoam meat trays will do
as spill trays.
2. Have students fill the container in which they will float the aluminum can as full as they can with
water. They should form a meniscus at the top of the container so that the aluminum can floats above
the top of the container.
3. The length of thread should be tied to the center of the magnet so that it hangs straight. When you
dangle the magnet into the can, the magnet should also be at least one inch shorter than the diameter
81
of the can so that it does not bang into the sides of the can while students attempt to spin it.
4. As you dangle the magnet from your hand, when you twist the thread between your fingers, the
magnet should spin. Students may need to practice this so that they can successfully do it for the
activity.
5. When the spinning magnet is lowered into the can, an electrical current is induced in the aluminum –
which is a conductor. This electrical current itself creates a magnetic field in the metal of the can that
opposes the magnetic field of the spinning magnet. These opposing fields cause the can to spin.
6. This is a demonstration of an electromagnetic effect known as Lenz’s Law. Explicitly state Lenz’s law
to the class.
Teacher Answer Key
(1) No. Aluminum is not a magnetic metal; (2) The compass needle moves and/or the can is spinning;
(3) The can slowly spins; (4) The answer to how this experiment works can be found above. We recommend
this question be used to determine if further discussions or explanations are needed with the class.
Meniscus Petri dish Aluminum can
Magnet (suspended on thread, inside the can) thread suspended from your hand
IV. Mutual induction
One of Faraday’s major discoveries was mutual induction . This phenomenon has many practical
benefits. It is this principle that is the basis of the transformer and generator. The current in the
first coil produced a current in the second coil even though they were not physically connected.
A coil of wire wrapped around an iron core is the basis for an electromagnet. When the coil is
connected to a DC source the iron rod becomes magnetised.
The electromagnet used here has a loop of iron bent in a U-shape and at each end there are a
number of turns of copper wire around it. The ends of the coil have terminals that can be used to
connect the coils to external circuits.
Mutual induction in a pair of electromagnets
The figure below shows the same type of arrangement except
that the electromagnets are arranged vertically core to core.
When the AC source is switched on the lamp will glow.
The top circuit has induced a current in the lower one and
so the lamp glows.
82
Practical advice
If time permits, I recommend that this is a class practical, but it could be a demonstration. These classic
demonstrations are a little remote from the context, but they are nevertheless useful and thought-provoking
and provide a useful lead in to a quantitative treatment of electromagnetic induction.
There is quite a lot of theory at this point, involving the key ideas of flux and flux linkage and leading
up to Faraday’s law. To introduce Lenz’s law, an energy argument is used; we recommend discussing with
students the consequences if the induced emf was in the reverse direction.
The case of an isolated straight wire moving in a magnetic field can be treated as a special case of
Faraday’s law (i.e. rate of change of flux linkage), rather than in terms of ‘cutting’ field lines, since the
latter approach obscures the fact that there is really only one law underlying electromagnetic induction,
irrespective of how it is brought about.
Fleming’s right-hand rule could make an appearance here, following an argument in terms of Lenz’s
law. One way to remember which hand to use is to note that (in the UK) motors drive on the left, so the lefthand rule applies to motors – and the other hand is for dynamos.
83
Optics
Learning Objectives :
●
Reflection of light by spherical mirrors
●
Refraction
●
Total internal reflection
●
Refraction at spherical lenses surfaces and by Lenses
●
Refraction through a Prism
●
Dispersion by a prism
●
Some natural phenomena due to sunlight
●
Optical Instruments
●
Huygens Principle
●
Refraction and reflection of plane waves using Huygens Principle
●
Coherent & Incoherent addition of waves
●
Interference of light waves and Young’s experiment
●
Diffraction
●
Polarisation
CONCEPTUAL PROBLEMS
1.
How the use of goggles enables an under-water swimmer to see clearly inside the lake. Which goggles
are better-made of glass or polaroids?
Hint. In water is goggles are not used, swimmer’s eyes are in direct contact with water. As
µ water > µair, the image will not be focussed on retina. If goggles are used, eye ball is in contact with
air in front so, reparative index will of remain same & swimmer can see clearly inside the lake.
If goggles are of colored glass then certain amount of incident light is absorbed and hence the object
appears dimbut polaroids absorb only that polarised light which produces dazzling in the eye.
2.
Does an air bubble inside water behave as a convex lens in air?
Hint. No, air bubble inside water behaves as a concave lens in air.
84
µ1
µ2
Consider two rays of light from an object ‘0’ in water. On entering the soab bubble, the rays deviate
from the normal. On reaching the outer surface, the rays bend towards the normal.
As rays are divergent so the air-bubble behaves as a concave lens in air.
3.
If a transparent sheet of refractive index µ and thickness t is introduced in front of one of the slits.
What will be the change in fringe patern?
Hint. After introducing the transparent sheet of ref. index µ, and thickness, t the optical path length of this
path will become µt instead of t ∴ ∆x′ = (µ –1)t
Without film the position of fringe is y = ;
New position of same fringe is y′ =
D
[ ∆x + (µ − 1)t]
d
∴ Lateral shift of fringe is y0 = y′ – y =
As β =
∴
4.
D
[ ∆x]
d
y0 =
D
(µ − 1)t
d
λD
d
∴
D β
=
d λ
β
(µ − 1)t
λ
As it is independent of n ∴ each fringe or entire fringe pattern is shifted by y0.
The shifting is towards that side where transparent sheet is introduced.
A slit or an aperture diffracts light. Even then we say light travels in a straight line and ray optics is
valid – why?
Hint. For single slit, half-angular width of central maxima is θ ; sin θ =
In travelling a distance, z, the diffracted beam acquires a width
The width will become more than width of slit
i.e.
zλ
a2
≥ a when z ≥
= z F . Freshel distance
a
λ
85
zλ
a
λ
a
For human eye a = 3 × 10–3m. l = 5 × 10–7m
a2
= 18m . It means width of beam due to diffraction does not become more than 3 mm unless
zf =
λ
beam travels more than 18 m so we can neglect broadning of beam by diffraction.
5.
Does diffraction pattern limit the resolving power of an instrument?
Hint. When two objects lie close by, their diffraction patterns may overlap so their images can no longer be
identified separately. Hence their resolution becomes difficult. Thus resolving power is limited. (R.P.
α
1
)
angular separation
6.
For a simple microscope, the angular size of object equals the angular size of the image yet it offers
magnification. How?
Hint. One can keep the small object much closer to the eye than 25 cm and hence have it subtend a large
angle. The image is at 25 cm which is magnified. Without microscope we have to keep the object at
25 cm to see.
Activities
I.
II.
III.
IV.
Take any plastic bottle filled with water and place it in front of your any page of your text book.
Observe the shape of letters.
Which alphabetical letters are same & which of letters are different?
Take a glass bowl and put a coin in it. Now fill it with water. Can we see the coin from the top of glass
bowl?
Can we see the coin from the sides of the bowl?
Take an empty plastic water bottle. Cut its upper portion & make a hole on its side. Fit a glass pipe or
a plastic straw in that hole. Fill the bottle with water and point a laser torch inside the water. Observe
where laser hight is visible.
Take two polaroids. Hold one polaroid facing a chart or any picture on class-wall. Now rotate second
polaroid in front of first polaroid & observe the change.
It is True?
❖
❖
❖
❖
❖
❖
The path of light rays is reversible.
If i = 0, r = 0, i.e., the ray incident in the direction of normal is reversed in direction.
Deviation produced on one reflection δ = (180° – 2i) or (180° – 2r).
Also δ = 2θ, where θ is the glancing angle i.e., the angle made between the incident ray and the
surface of plane mirror.
Focal length of a plane mirror is infinity.
Power of plane mirror is zero.
86
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
Radius of curvature of a plane mirror is infinite.
Linear magnification produced by a plane mirror is unity.
Deviation produced by two inclined plane mirrors is 2θ if θ is acute or (360-2θ) if θ is obtuse. The
deviation is independent of angle of incidence.
The image of the point source is always formed by a plane mirror, provided that the source is placed
anywhere in front of the mirror.
The image can be seen from any position, provided the line joining the eye to the image cuts the
surface of the mirror.
Objects except the self luminous ones become visible due to the light diffusively reflected from their
surface.
Twilight is an example of diffused reflection of sun light from clouds, dust particles and other
floating particles in the air.
Whenever the image is real it is inverted and that whenever the image is virtual it is erect.
Image formed by a concave mirror may be real or virtual. The virtual image is bigger than object.
Convex mirror always form a virtual image smaller than the object. The image always lie between
the pole and the focus.
In a concave mirror the minimum distance between the object and its real image is zero.
When a light ray travels from denser to rarer medium, it deviates away from normal white if it is
travelling from rarer to the denser medium, it is deviated towards the normal.
Refractive index is the optical property of the medium.
Refractive index of vacuum is one while the refractive index of air is 10003, but for simplicity it
is taken to be one.
The value of µ depends on the nature of the two mediums, colour of light and temperature of the
two mediums.
A+B
as λ increases µ decreases. That is why
λ2
Refractive index of the medium decreases on increasing temperature.
According to µR < µV Formula µ =
Unless mentioned, all refractive indices are measured with respect to air.
If the ray travels from vacuum to a medium then µ is known as the absolute refractive index of that
medium.
µ =
I II
Wavelength in I medium (λ1 )
Velocity of light in I medium (V1 )
& IµII
Wavelength in II medium (λ 2 )
Velocity of light in II medium (V2 )
µg =
1
g µw
❖
Since the path of light rays is reversible
❖
‘µ’ is maximum for diamond.
If i = 0, r= 0 i.e. if a ray is incident normally, then it is refracted into the II medium undeviated.
If a beaker is filled with immiscible transparent liquids of refractive indices µ1, µ2,..µ3... filled up
to heights d1, d2, d3,....., then the apparent depth is
❖
❖
w
87
d1 d 2 d 3
+
+ + ...
µ1 µ 2 µ 3
❖
If the travelling microscope is focussed on the mark on the bottom of an empty beaker and its reading
is xa, the beaker is now filled with any transparent liquid and the mark is again focussed, the reading
is x2 ,finally some Lycopodium powder is sprinkled on the top of the liquid surface, and the microscope
is again focused on the powder, the reading is x3. Then the refractive index of the liquid is
µ=
❖
x 3 − x1
x3 − x 2
If a transparent rectangular block of thickness t is placed in the path of divergent light rays coming
from an object 0, then the on observing from the other side, the
Eye
0
Eye
O′
I
I′
sh ift
(a)
(b)
object appears nearer by the distance Fig (a) =
❖
❖
❖
❖
❖
t(µ − 1)
µ
If the light rays are convergent then the on observing from the other side, the image appears
t(µ − 1)
µ
away from block. Fig(b).
At sunset and sunrise, sun appears above horizon while actually it is below horizon, due to atmospheric
refraction.
Rivers appear shallow, coin in a beaker filled with water appears raised, pencil appears broken, due
to refraction.
he duration of day appears to be increased by nearly 4 minutes due to atmospheric refraction.
If a light ray travels from a rare medium of refractive index µ1 to a dense medium of refractive index
µ2 and the reflected and refracted rays are at right angles to each other then,
µ2
µ1 = tan i where i is angle of incidence
❖
❖
❖
❖
The value of critical angle depends on the (i) refractive indices of the two media (ii) colour of light
and (iii) the temperature of medium.
Critical angle increases on increasing temperature.
1
. The critical angle for red rays is maximum and for violet rays it is minimum.
µ
Q µR < µV
If i < C, then the phenomenon of refraction takes place.
C<l<
88
❖
if i > = C, then the refracted ray travels along the boundary separating the two media.
if i > C, the ray is totally internally reflected into the first medium.
The maximum value of angle of refraction is 90°.
Brilliance of diamonds is due to their high refractive index, low critical angle and hence easy total
internal reflection.
The air bubbles in glass paper weight appears silvery white, due to total internal reflection.
A test tube blackened from outside appears silvery white due to total internal reflection when dipped
in water.
Prism has a property to refract the ray of light towards the base.
If A > 2C, then there will be no emergent ray from the prism.
For maximum deviation, either i1 = 90° or i2 = 90°.
Dispersion takes place because the light rays of different colours are deviated at different angles,
so an angular separation is produced between them.
Optical centre may lie within or outside the lens.
In case of concavo-convex and convexo-concave lens, the optical centre lies outside the lens
in case of plano-concave or plano-convex lens, the optical centre lies on the surface of the lens.
❖
An air bubble in water behave as a concave lens.
❖
If a lens of focal length f is divided into two equal halves as shown in figure (a), then the focal length
of each half becomes 2f
If the lens is cut into two halves by a plane parallel to principal axis, then the focal length of each
part will remain f. (See fig. b)
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
❖
(a)
The minimum distance between the object and its real image is 4f.
(b)
Area of image
Area of image
, also m2 =
where m is linear magnification.
Area of object
Area of object
When a convex and a concave lens of same focal lengths are combined then the power of will be
zero and its focal length will be infinite. The system will behave as a plane glass plate.
Magnification in area =
h
where h = height of the light ray from principal axis.
f
If the plane surface of a plano-convex lens of focal length f is silvered, then it will behave as the
concave mirror of focal length f/2.
Deviation produced by a lens is δ =
If the plane surface of a plano-concave lens of focal length f is silvered, then it will behave as the
convex mirror of focal length f/2.
If the convex surface of the plano-convex lens is silvered, then
1
f lens
= +
1
1
−
( − R lens ) ( − R mirror )
89
❖
❖
Very small objects subtend small visual angle due to their smallness. We can increase the visual angle
by bringing these objects closer to the eye, but we cannot do so beyond a certain limit (25 cm)
because then the objects will not be seen distinct. If with the help of proper lenses, a large image
of small object is formed, then this image will subtend a large visual angle at the eye and the object
will appear large Microscope is based on this principle.
Objective” of large aperture is taken so that it may collect sufficient light and form bright image of
very distant objects.
❖
Eyepiece of small aperture is taken so that the whole light may enter the eye.
❖
If a person can see up to a distance x and wants to see the object placed at distance y then focal
xy
length of lens to be used f =
x−y
Colour Blindness is caused due to the non-excitement of any retinal cone corresponding to its colour,
e.g. if the blue retinal cone of a person is not excited, then the person will not be able to differentiate
blue colour. This defect cannot be remedied
❖
❖
The magnifying power of reflecting telescope is ‘m’ = R (Radius of curvature of mirror)
❖
Colour of transparent objects depends on (λ). The colour of light transmitted by them e.g.. blue glasstransmits blue colour and absorbs all others, (b) The colour of incident light e.g. a blue glass seen
in red light will appear black.
❖
Colour of opaque objects depends on (a) the colour of light reflected by it, e.g., a green leaf appears
green because it reflects green colour and absorbs ail others, (b) The colour of incident light, e.g
the green leaf seen in red light will appear black.
❖
A sustained interference pattern cannot be obtained by using two independent source of light, it is
because of the following reasons:
(a) Two independent sources cannot emit waves continuously.
(b) The waves emitted by two independent sources do not have same phase or a constant phase
❖
Young’s experiment demonstrates both the diffraction of light waves at the slits and the interference
between the light waves emerging from the slits.
❖
The maximum intensity is obtained at places where the phase difference between the interfering
waves is 2λ or the path difference in, nλ(0, λ, 2λ...). The path difference between consecutive
maxima is λ
❖
The minimum intensity is obtained at places where the phase difference between the interfering
waves is (2n –1)π or the difference is (2n – 1) λ/2 (i.e., λ/2,3λ/2, 5λ/2). The path difference between
successive minima is λ
❖
If the width of the slit is gradually increased, the contrast between the fringes become poorer and
poorer. Ultimately the fringes disappear, leaving behind a region of uniform illumination.
❖
If both slits are illuminated by white light, then central white fringe is obtained, having on both sides
a few coloured fringes and then a uniform illumination.
❖
Malus law holds for a combination of reflecting planes, nicols, polaroids etc, but not for Pile of
plates.
90
❖
The angle between plane of vibration and direction of propagation of wave is 0°.
❖
The angle between plane of polarisation and the direction of propagation of wave is 0".
❖
Angular half width of central maximum is = λ/a and the width of central maximum is = 2λ/a.
❖
The width of a central or a secondary maximum is directly proportional to Therefore greater the Wave
length of the light used, the greater is the width of the maximum.
❖
The width of a secondary maximum is inversely proportional to a, the width of the slit. Therefore,
as the width of the slit is increased the secondary maximum will become narrower. In case of a
sufficiently wide slit, the secondary maximum becomes invisible and only the central maximum is
obtained which corresponds to the sharp image of the slit. In other words, the distinct diffraction
pattern will be obtained only in case of a narrow slit.
❖
The intensity of a secondary maximum is less than that of central maximum because, the intensity
of central maximum is due to wavelets from all parts of the slit, while the first secondary maximum
is due to wavelets from one-third part of the slit only (first two parts send wavelets in opposite phase),
the second secondary maximum is due to wavelets from one-fifth part only first four parts send
wavelets in opposite phase) and so on. Hence, the intensity of secondary maxima goes on decreasing.
❖
Difference between interference and diffraction is that interference is the superposition effect between
the wavelets starting from two coherent sources while the diffraction is the superposition effect
between the wavelets starting from the single wave front.
Q1. One face of a rectangular glass-plate 6 cm thick is silvered. An object held 8 cm in front of the first
face forms an image 12 cm behind the silvered face. Find the µ of glass.
Sol. Let the apparent position of the silvered suface = x cm.
As object distance = Image distance
∴
x + 8 = 12 + 6 – x
⇒
x = 5 cm
µ=
R ea l
P o sitio n
A p p are nt
P o sitio n
x
I
Real depth
6
= = 1.2
Apparent depth 5
O
8 cm
12 cm
6 cm
Q2. An object is placed at a fixed distance from a screen. When convex lens is placed between the object
& the screen at a distance of 15 cm from the object, then an image is formed on the screen having a
size 5 times. The size of object calculate the focal lenght of the lens and distance between object and
screen. How much the lens be displaced towards the screen so that again a distinct image of the object
be formed on the screen? What will be the magnification of this image?
Sol. m =
v T
= =5
u 0
∴ 5=
v
⇒ v = 75cm
15
91
∴ Distance between object and screen = 75 + 15 = 90 cm
1 1 1 1
1
6
= − =
−
=
⇒ f = 12.5cm
f v u 75 −15 75
In the displaced positions of lens, object and imaged stance are interchanged ∴ displacement
x = v –u = 75 – 15 = 60 cm.
Also m1m2 = – 1 or m2 = −
1
1
=−
m1
5
Q3. The focal lengths of the objective and eye-piece of a telescope are 50 cm & 5 cm respectively. Least
distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale placed at a
distance of 200 cm away from objective. Calculate separation between objective and eye-piece and
magnification.
Sol. f0 = 50 cm, fe = 5 cm & u0 = 200 cm
For objective
1
1
1
−
=
f0
V0 u 0
⇒
1  1 
1
−−
=
 ⇒ V0 = 66.66 cm
V0  200 
50
For rye piece
1
1
1
−
=
fe
Ve u e
⇒
1
1
1
= − +
⇒ ue = 4.17 cm
25 u e
5
∴ Length of telescope = 66.66 + 4.17 = 70.8 cm
m = m0 × me =
200
(−25)
×
= −2
3(−200)  25 
 − 
6
Q4. A thin film of transparent material, µ = 1.25 is coated on a glass µ = 1.5. White light falls normally on
the film. In the reflected light complete destructive interference is observed at 6000 A° and constructive interference at 7000°. Find the film thickness.
Sol. Interfering rays are reflected from the boundary of an optically denser medium each time ∴ a path
λ
is introduced each time.
difference of
2
For constructive interference
2µt cos r = nλ
For normal incidence cos r = 1
∴
2µt = nλ or 2 × 1.25 × t = 7000 n
...(1)
For destructive interference
λ
∴
2µt = (2n − 1)
2
92
since the wavelength is less this time, the order should be one more than constructive interference.
∴
⇒
From equation (1) & (2)
6000
2
2 × 1.25 × t = (2n + 1) 3000
2 × 1.25 × t = [2(n + 1) − 1]
...(2)
7000n = 3000 (2n + 1)
⇒
n= 3
Q5. A plano convex lens when silvered on the plane side behaves like a concave mirror of focal length
60 cm. However when silvered on the convex side, it behaves like a concave mirror of focal length
20 cm. Find the refractive index of lens.
Sol. When plane surface is silvered
r
r
⇒ 120 =
µ −1
µ −1
Rm = 2f m =
When curved surface is silvered
Rm = 2 × 20 =
⇒
r
µ
r
µ
3
120
× ⇒µ=
=
2
µ −1 r
40
4
7

Q6. Water  µ =  in a tank is 18 cm deep. Oil of µ = lies on water making a convex surface of

3
4
R = 6cm. consider oil act as a thin lens. An object S is placed 24 cm above water surface. The location
of image is at x cm above the bottom of the tank. Find the value of x.
Sol.
S
Applying lens formula
24 cm
µ 2 µ1
µ 2 − µ1
−
=
V u
R
74
1
7 4 −1
−
=
V −24
6
A ir
6 cm
R=
S
x
18 cm
⇒ V = + 21 cm. This image distance will act as (virtual) a object distance in the tank.
Inside tank
⇒
∴
4
7
3 − 4 = ∞
V ′ 21
V′ = 16 cm
x = 2 cm
93
Q7. AB & CD are two slabs : The medium between the slabs has µ = 2. Find the minimum angle of
incidence of θ so that ray is totally reflected by both the slabs.
Sol.
θ
A
µ1 = 2
B
i
µ2 = 2
C
µ3 = 3
D
µ 
1
Critical angle ie1 = sin −1  1  = sin −1
= 45°
 µ2 
2
µ 
3
= 60°
Critical angle ie2 = sin −1  3  = sin −1
2
 µ2 
∴
imin = 60° for both the slabs.
94
Electronics Devices
Learning Objectives :
●
●
●
●
●
●
●
●
●
Classification of Metals, Conductors and Semiconductors
Intrinsic Semiconductor
Extrinsic Semiconductor
p-n Junction
Semiconductor diode
Application of Junction Diode as a Rectifier
Special Purpose p-n Junction Diodes
Junction Transistor
Digital Electronics and Logic Gates
CONCEPTUAL PROBLEMS
1.
The resistance of the pn junction is low when forward biased and is high when it is reverse biased.
Explain.
Hint. The resistance of pn junction is low when it is forward biased because force acts on charge carriers to
move them across the junction whereas the force in reverse biasing acts in opposite direction which
opposes the motion of charge carriers across the junction.
2.
How will you test whether a transistor is spoiled or in working order?
Hint. a working transistor has low resistance when it is forward biased whereas it has high resistance when
it is reverse biased. But if the transistor is spoiled the resistance is low for both forward and reverse
biasing.
3.
Would you prefer to use a transistor as common base or common emitter configuration.
Hint. We prefer to use the common emitter configuration of the transistor because the current gain is more
in that configuration.
4.
Name the p-n junction diode which emit spontaneous radiation when forward biased. How do we
choose the semiconductor, to be used in these diodes, if emitted radiation lies in visible region.
Hint. The p-n junction that emits radiation when forward biased is called photodiode. In case of gallium
arsenide the emitted radiations lies in the visible region.
95
5.
In a transistor forward bias is very small as compared to reverse bias. Explain why?
Hint. If the emitter voltage is large, then the number of charges drifting from emitter to collector through
base becomes very large. It results in lot of heat generation which can damage the transistor. But if
collector reverse bias is large charges may drift quickly to collector but there is small and no heating
effect is produced.
6.
What is the difference between the transistor as amplifier and the step up transformer?
Hint. The difference between the transistor as an amplifier and the transformer is that in the transistor
increase in voltage does not imply the decrease in current whereas in transformer if the magnitude of
the voltage increases there is a corresponding decrease in current.
7.
Why is the base of transistor of made thin in comparison to the emitter or collector region.
Hint. The base of transistor is made thin in comparison to the collector and the emitter region because the
recombination in the base region should be small.
8.
What is the effect of temperature on the conductivity of a semiconductor?
Hint. The conductivity of the semiconductor increases with the increase in temperature because as the
temperature increases more and more covalent bonds break resulting in the release of charge particles.
9.
In a transistor base is very lightly doped, why?
Hint. In transistor the base is lightly doped so that the recombination in the base region should be small
and the magnitude of output current is large
10. Why does the thickness of depletion region in a pn junction diode increases with increase in reverse
bias?
Hint. When a pn junction is formed, a small potential difference is set up across the depletion layer. But
when it reverse biased the charges move away from the junction thus increasing the width of the
depletion region.
11. Explain why the transistor starts working immediately on switching on whereas vacuum tube circuits
take some time before they starts working?
Hint. Vacuum tube circuits are based on the heating effect of the current and thermionic emission takes
place from cathode when it is heated. Thus, it takes some time to start as cathode can’t be heated
instantly.
12. Why is the depletion region formed at the pn junction?
Hint. The depletion region is formed because of the recombination of electrons and the holes at the junction
13. What will happen if collector as well emitter are forward biased?
Hint. In this case transistor will work as two pn diodes and it can not work as amplifier or oscillator.
14. What is Fermi level and what is Fermi energy?
Hint. Fermi energy level is the highest energy level occupied by the electrons at zero-kelvin and the energy
of that level is fermi energy.
15. When semiconductor junction diode is formed electrons should flow from n to p region but all the
electrons do not do so? Explain why?
Hint. When electrons are transferred from p type to n type, the n type semiconductor gets positively charged
and p type gets negatively charged which creates potential difference across junction. This potential
difference after some time prevents the flow of electron from n type to p type.
96
16. Is the number of free electrons and holes equal in extrinsic semiconductors. Are they charged?
Hint. No, the number of electrons and holes are not equal in extrinsic semiconductors. Both n type and p
type semiconductors are electrically neutral.
17. Is the junction diode linear or a non linear circuit element?
Hint. Junction diode is a non linear circuit element because the V-I curve is not a straight line
18. What will happen if both emitter and collector junction are reverse biased
Hint. In this situation no current will flow in the semiconductor because majority carriers cannot move
across the emitter base or base collector junctions.
19. How will you detect intensity of light using diode?
Hint. Light intensity is proportional to the intensity of current which can be measured by using a photodiode
in reverse biasing.
20. Why gallium arsenide solar cells are preferred over silicon solar cells?
Hint. Gallium arsenide solar cells are used because they can operate with visible energy whereas the
silicon diodes works with infrared energies.
21. Define input and the output resistance of a transistor?
Hint. Input resistance of transistor is the ratio of change in base voltage to the change in base current at
constant collector voltage .
22. What is the phase relation between the input and the output signal in an amplifier?
Hint. Input and output current are in phase in common base transistor amplifier and in transistor as common
emitter the input and the output are out of phase by 1800
23. What is zener breakdown and what is the zener breakdown voltage?
Hint. Zener breakdown takes place when a strong reverse bias electric field is applied across the
semiconductor. The covalent bonds in the structure break simultaneously resulting in large increase
in the magnitude of the current. The voltage at which zener breakdown occurs is zener breakdown
voltage
24. What is the avalanche breakdown voltage of a junction diode?
Hint. When the reverse bias voltage is increased to a large value large number of covalent bonds break
near the junction resulting in large reverse currents flowing in the transistor. The voltage at which
breakdown occurs is avalanche breakdown voltage
25. What will happen if the input circuit is reverse biased and the output circuit is forward biased?
Hint. In this case the collector starts acting as emitter and the emitter as collector with transistor functioning
as usual.
26. What type of charge carriers flow during reverse biasing of the diode?
Hint. In reverse biasing the charge carriers flowing in a semiconductor are minority charge carriers.
27. Transistor is a temperature sensitive device. Explain.
Hint. Transistor is a temperature sensitive device because the number of charge carriers increases with the
increase in current but if the current increases to a very large magnitude it may damage the transistor.
28. Why an extrinsic semiconductor gets permanently damaged if temperature is increased beyond a
certain limit?
97
Hint. If temperature is increased beyond a limit large number of covalent bonds break in the structure
resulting in release of charge carriers and making it highly conducting
29.
Explain why input resistance of transistor is low and the output resistance is high.
Hint. Input resistance of the transistor is low because input circuit is forward biased whereas the output
circuit is reverse biased.
30.
How can you increase the current gain in an amplifier circuit?
Hint. Current gain can be increased by decreasing the doping level in the base and keeping it thin.
31.
What is the potential barrier of pn junction in a silicon transistor?
Hint. Potential barrier for silicon transistor is of the order of 0.3V and for germanium transistor it is of the
order of 0.7V.
32.
For what particular application the common base configuration is preferred over the common emitter
configuration?
Hint. Common base is preferred over common emitter if we want the voltage amplification without the
phase difference between input and the output.
33.
By increasing load resistance can we increase or not the gain of transistor indefinitely.
Hint. No, because increases RL will decrease the net output voltage. If the output voltage becomes less
than input then it cannot act as amplifier.
34.
In the depletion region of pn junction what are the charge carriers in its unbiased state?
Hint. In depletion region there are no charge carriers present
35.
What type of feedback is required in transistor as an oscillator?
Hint. In transistor as an oscillator positive feedback is required i.e. emf is induced in such a way that if
current in output increases emf is induced in input to support the forward bias and if the current in
output decreases emf is induced in input to oppose the forward bias.
36.
Can we measure the potential barrier of a pn junction by putting voltmeter across is?
Hint. In the depletion layer, there are no free charges present, thus it offers infinite resistance to the flow of
current through it. Therefore potential barrier across a p-n junction cannot be measured using voltmeter.
37.
Fill in the blanks
[1] When p-n junction if forward biased, then the motion of charges across the barrier is due to……..
and when it is reverse biased then the motion of charge carriers is due to……..
[2] An ideal pn junction diode conducts, when __________ and does not conduct when ______
Hint. [1] diffusion, drift [2] forward biased, reverse biased
38.
State two disadvantages of semiconductor devices.
Hint. [1] The semiconductor devices can’t withstand high temperature and an get damaged
[2] the semiconductor devices gets damaged by power surge and can’t withstand high power.
39.
NAND and NOR gate may be considered as digital building blocks. Why?
Hint. The repeated use of NAND or NOR can produce all 3 basic gates i.e. OR, AND and NOT. Thus they
are called basic building blocks.
98
Activities
I.
A Simulation of electron migration
Query: How are pn junctions formed?
Hypotheses: (Student generated list)
Teacher’s note: If necessary lead students to the idea that electrons migrate towards the
more positive area.
Materials: (2 students per group)
2 - 10 cm x 30 cm cards with the following design:
1 - 10 cm × 10 cm card with the following design:
This card represents the electromotive source, such as a battery.
21 - 2 cm (outside diameter) washers. Pennies may be used.
Procedure:
1.
2.
3.
Fill one of the cards with the washers or pennies. This card now represents the n-type material. The
washers and pennies represent the excessive electrons from the n-dopant.
Card 2, “free” of electrons, represents the p-type material. It has an excess of “holes” from the pdopant.
Join the n-type and p-type materials by allowing the common borders to touch.
Move the first column of “electrons” to fill the first column of available holes on the p-type material.
What you have formed is known as the pn junction and the depletion zone. When the pn junction is
formed the migration of electrons stops.
Discussion Questions:
1.
2.
3.
4.
5.
Draw a scaled replica of the pn device and label the location of the pn junction.
Why did the electrons of the n-type material moved to the “holes” of the p-type material?
Why does the migration of electrons stop?
How did the migration of electrons change the physical characteristic of the n-type and p-type
materials?
How would you best describe the pn junction?
99
6.
7.
Why is the pn junction also known as the depletion zone?
What would be needed to continue electron migration across the depletion zone?
Answers to Questions:
1.
2.
3.
4.
5.
6.
7.
II.
Electrons have a natural tendency to flow towards the more positive area.
Migration ceases when the available carriers have formed the depleted “barrier” or depletion zone.
The n-type material now contains “holes” and carries a positive (+) charge and the p-type material
now contains electrons and carriers a negative (–) charge. Forming a semi-conductive device in a
state of equilibrium.
The pn junction is the immediate area of the n-type : p-type border where migrating electrons left
“holes” in the n-type material and filled the immediate “holes” of the p-type material with electrons.
The immediate area of the pn junction is known as the depletion zone or region because it is free of
available carriers.
An electric potential known as the barrier potential exists across the depletion zone. This potential is
material specific. Silicon’s barrier potential is ~0.7v and Germanium’s barrier potential is ~0.3v. To
continue the migration of electrons across the depletion zone, an outside EMF that has a larger
electric potential than the barrier potential must be used.
This lesson is used to illustrate how two differently doped materials that are electrically neutral can
be chemically joined to form a simple electronic semi-conductive device. This device is the foundation
of the semiconductor industry as it has the capabilities of regulating the electric current and the
direction that the current flows.
Students will model the formation of forward and reversed biased pn junctions.
Students will recognize the effect that forward and reversed biased pn junctions have on current
flow.
Materials:
Use the same materials as in investigation 1. Also add to this the cutout of the EMF source.
Two lengths of string, each 30cm long.
Scotch Tape.
Procedure A: For the forward biased pn junction.
1.
Attach a length of string to each pole of the EMF source. The string represents the wire leads that are
attached to the EMF.
2.
Setup a pn junction that is three columns wide on each side.
3.
Attach the negative lead to the n-type material and the positive lead to the p-type material. You now
100
have a closed circuit. (Remember that electrons flow from the negative terminal towards the positive
terminal.)
4.
On the p-type card, move the outermost column of “electrons” of the depletion zone towards the
positive lead wire. Remove these electrons from the board. (This represents the electrons passing
through the positive lead and moving towards the EMF positive lead and back to the battery where
they will be recycled back to the negative side of the EMF source, the negative lead.
5.
On the n-type card, shift all columns of “electrons” one step towards the pn junction. Fill the open
holes with the “electrons” that have passed through the wire leads. Note the width of the depletion
zone.
6.
Draw a scaled representation.
7.
Repeat steps 4, 5 and 6.
Procedure B: For reverse biased pn junction.
1.
Setup as in steps 1 and 2, from procedure A.
2.
Attach the negative lead, from the EMF, to the p-type material. Attach the positive lead to the n-type
material. (Again remember that electrons flow from the negative pole to the positive pole of the
battery or EMF source.)
3.
Move all columns of “electrons” on the n-type card one step towards the positive lead. (One column
of “electrons” will move off of the board. These electrons represent the electrons that are moving
through the positive lead towards the EMF source.)
4.
Fill the next open column of “holes”, on the p-type card, with the “electrons” that have passed
through the wire leads. Note the width of the depletion zone.
5.
Draw a scaled representation.
6.
Repeat steps 3-5.
Discussion Questions:
1.
2.
3.
4.
Why does the depletion zone shrink in a forward biased pn junction?
How does a forward biased pn junction change the conductivity of the device? Please explain your
answer.
Why does the depletion zone enlarge in a reverse biased pn junction?
How does a reverse biased pn junction change the resistivity of the device? Please explain your
answer.
Drawings for Simulation and Answers to Questions.
Drawings for forward biased pn junction. (Left side: n-type, Right side: p-type)
Before EMF is added.
101
After EMF is applied 1.
After EFM is applied 2.
In both cases, the EMF, electromotive force, is pushing electron current through the n-type material
towards the p-type material and to the positive side of the EMF. The net voltage, from the EMF, provides
and excessive amount of electrons on the n-type material large enough that it can overcome the barrier
potential of the pn junction. This causes the electrons and “holes” to migrate towards the pn boarder. This
action reduces the depletion zone and increases the conductivity of the device.
Drawings for reverse biased pn junction. (Left side: n-type, Right side: p-type)
Before EMF, voltage, is added.
After EMF is applied 1.
After EMF is applied 2.
102
In this case the EMF is moving positive electric charges in the p-type material away from the junction.
The availability of excess electrons at the negative side of the EMF allows the electrons to migrate further
into the p-type material making the P-type material even more negative. The electrons in the n-type material
are attracted toward the positive side of the EMF source and as a result the n-type material becomes even
more positive. In this case the EMF is moving negative electric charges in the n-type material away from
the junction. Since the charge carriers are being pulled further apart the depletion zone has widened and the
resistivity has increased.
Answers to questions:
1.
When the negative and positive potentials (poles) of the EMF are applied to the n-type and p-type
materials, respectively, the negative potential of the EMF pushes the electrons in the n-type material
towards the junction. The holes, of the p-type material, are also pushed towards the junction by the
positive potential of the EMF. The combined effects cause the width of the junction to be reduced.
2.
With this reduction of the pn junction and assuming that the applied voltage is greater than that of the
barrier potential, the electrons in the n-type material will gain enough energy to break through the
depletion zone. When this occurs, the electrons are free to fill the “holes” in the p-type material and
conduction occurs.
3.
When the negative and positive potentials (poles) of the EMF are applied in reverse on the pn device,
the electrons in the n-type material move towards the positive terminal of the EMF and the “holes”
in the p-type material migrate towards the negative terminal of the EMF. In this situation, the depletion
zone has widened.
4.
With an increase of the width depletion zone, the electrons have a formidable barrier that prevents
ready migration across the junction. This barrier increases the resistivity of the junction and the
conductivity of the device is close to zero.
III.
The Diode
Query:
What does a diode do?
Hypothesis:
(Student generated. If necessary, lead students to the idea that diodes may determine
the direction of the flow of electrons.)
Materials:
Three diodes: 2 standard diodes and an LED
A 470 Ω resistor
A battery pack. (9v)
Multi-meter or ammeter
A proto-board
Insulated bell wire
Switch
Note: If multi-meters or ammeters are not available, use LED’s (Light emitting diodes)
If proto-boards are not available, use wire leads with ‘alligator’ clips to make
connections.
103
Procedure:
1.
Set up a simple circuit as illustrated.
2.
3.
4.
Close circuit by closing the switch. Record the reading on the meter.
Open the circuit by opening the switch.
Reverse the terminal attachment, of the diode, in the circuit. To do this, attach the wire leads to the
opposite electrodes. Close the circuit and record the reading on the meter.
Open the circuit.
Replace diode with another diode and repeat steps 2 – 5. Repeat this process with a light emitting
diode (LED).
5.
6.
Data:
Diode
Meter Reading (mA)
D1
D1 (reversed)
D2
D2 (reversed)
LED
LED (reversed)
Conclusion:
Discussion Questions
1.
What is a diode?
104
2.
Are diodes unidirectional in regards to current flow? Please explain your answer.
3.
What physical characteristic, of a diode, identifies the direction of the current flow?
4.
Draw the schematic symbol for a diode and label the terminals.
5.
Using the appropriate schematic symbols, draw the circuit used for the experiment.
Switch :
Battery (EMF):
Meter:
Resistor:
Answers to Questions
1.
A diode is a two terminal electronic device that acts as a one-way conductor.
2.
Yes, from the data, the diodes conduct electricity only when they have the proper polarity applied. In
some cases, the initial setup of the diode prevented a current flow and by simply switching the
terminal positions current was allowed to flow.
3.
LEDs have terminals of different lengths. The long lead is the anode (+) and short terminal is the
cathode (–). Others have a terminal with a flattened portion, this is the cathode (–) with the other
being the anode (+). Or the casing has a flattened side, here the terminal that is closes to the flattened
side is the cathode (–). Other diodes have (+) and (–) markings to signify specific electrodes as either
the anode (+) or the cathode (-). The most common marking is a band near the cathode end of the
diode. When placing a diode into a circuit, attach the device leads so that the cathode is attached to
the most negative potential and the anode to toward the positive potential. This will keep the diode
forward biased.
4.
Schematic showing forward biased diode:
5.
105
Note:
→
The arrows (→
→) indicate the direction of electron current flow.
In many electronics and physics textbooks, the current flow is shown going in the opposite direction.
The current moving from the (+) terminal to the (-) terminal is known as conventional current flow.
A diode is an electrical component that controls the direction of the flow of electricity.
IV.
Learn how a diode works.
Materials
This lesson requires the following classroom materials:
●
An LED (light emitting diode)
●
9V battery
●
Resistors of various values
Procedures
The following two experiments demonstrate that the diode conducts only in one direction. Engineers use
this property to design complex circuits in which diodes control the flow of electrons.
Forward-biased diode
● Connect one terminal of the resistor to the negative terminal of the battery.
● Connect the other terminal of the resistor to the negative terminal of the diode (the shorter leg of
the diode).
● Connect the positive terminal of the diode (the longer leg of the diode) to the positive terminal of
the battery.
● The LED should light up if the connections were made correctly. This circuit is shown in the
schematic diagram and in the picture of the breadboard below.
106
IV.
How to Transistors Works
A bipolar junction transistor has three terminals - Base, Collector, and Emitter corresponding to the
three semi-conductor layers of the transistor. The weak input current is applied to the inner (base)
layer. When there is a small change in the current or voltage at the inner semiconductor layer (base),
a rapid and far larger change in current takes place throughout the whole transistor.
co lle cto r
b ase
ern itter
Pictured above is a schematic diagram of the more common NPN transistor. Below is an illustration
of the same transistor using water rather than electricity to illustrate the way it functions:
The illustration shows pipework with three openings B (Base), C (Collector), and E (Emitter). The
reservoir of water at C is the supply voltage which is prevented from getting though to E by a plunger. If
water is poured into B, it pushes up the plunger letting lots of water flow from C to E. If even more water
is poured into B, the plunger moves higher, and the flow of water from C to E increases.
Therefore, a small input current of electricity to the Base leads to a large flow of electricity from the
Collector to the Emitter.
Numericals
Example 1 : The V-I characteristic of a silicon diode is shown in the Fig. Calculate the resistance of the
diode at (a) ID = 15 mA and (b) VD = –10 V (check the resistance of diode in forward and reverse bias)
107
Solution : Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing
through the origin, we can calculate the resistance using Ohm’s law.
(a) From the curve, at I = 20 mA, V= 0.8 V, I = 10 mA, V = 0.7 V
rfb = ∆V/∆I = 0.1V/10 mA = 10 Ω
(b) From the curve at V = –10 V. I = –1 µA.
Therefore,
rrb= 10V/1(µA= 1.0 × 107Ω
Example 2 : In a Zener regulated power supply a Zener diode with Vz= 6.0 V is used for regulation.
The load current is to be 4.0 mA and the unregulated input is 10.0 V. What should be the value of series
resistor Rs?
Solution : The value of Rs should be such that the current through the Zener diode is much larger than the
load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e.,
Iz = 20 mA. The total current through Rs is therefore, 24 mA. The voltage drop across Rs is 10.0 – 6.0 =
4.0 V. This gives Rs = 4.0V/(24 × 10–3) A = 167 Ω. The nearest value of carbon resistor is 150 Q.. So. a
series resistor of 150 Q Is appropriate. Note that slight variation in the value of the resistor does not matter,
what Is important is that the current IZ should be sufficiently larger than IL.
Example 3. In Fig. , the VBB supply can be varied from 0V
to 5.0 V. The Si transistor has βdc = 250 and RB = 100 kΩ, Rc = 1 KΩ, VCC = 5.0V. Assume that when the
transistor is saturated, VCE = 0V and VBE = 0.8V. Calculate (a) the minimum base current, for which the
transistor will reach saturation. Hence, (b) determine V1 when the transistor is ‘switched on’, (c) find the
ranges of V1 for which the transistor is ‘switched off and ‘switched on’.
Solution :
Given at saturation VCE – OV, VBE – 0.8V
VCE = VCC – ICRC
IC = VCC /RC = 5.0V/1.0kΩ = 5.0 mA
Therefore IB = I IC / β = 5.0 mA/250 = 20µA
The input voltage at which the transistor will go into saturation is given by
VIH = VBB = IBRB + VBE
= 20µA × 100 kΩ + 0.8V = 2.8V
108
The value of input voltage below which the transistor remains cutoff is given by
VIL = 0.6V, VIH= 2.8V
Between 0.0V and 0.6V, the transistor will be in the ‘switched off state. Between 2.8V and 5.0V, it
will be in ‘switched on’ state.
Note that the transistor is in active state when IB varies from 0.0mA to 20mA. In this range, IC = βIB
is valid. In the saturation range, IC ≤ βIB.
Example 4 : What is the effect of negative feedback in oscillator?
Solution : In an oscillator, the feedback is in the same phase (positive feedback). If the feedback
voltage is in opposite phase (negative feedback), the gain is less than 1 and it can never work as oscillator.
It will be an amplifier with reduced gain. However, the negative feedback also reduces noise and distortion
in an amplifier which is an advantageous feature.
109
Appendix
Life of Science Articles
2.
3.
The science teachers have been facing great difficulty in maintaining the stock of science materials.
Majority of them do not know the life of various science equipments. To help, the teachers, the
“Science Branch News Bulletin” has reproduced the copy of the circular issued by the Director
of Education order No. F-4 (52)/-61 Edn (P) dated 2nd August, 1962 which is as follows :
The Director of Education is pleased to accept the recommendation of the Committee, appointed
by him, to fix the life of furniture articles and other non-consumable articles (science equipments
etc.) used in the Directorate of Education, Delhi/and the Govt./Aided schools in the Union Territory
of Delhi, vide order no. F(52)/61-Edn({) dated 1.9.1961 and accordingly fixes the life of each
item, as shown in the Annexures “A”, B, C, D, E, F, H and I to the extent mentioned therein.
In case there is any other item, the life of whih has not been fixed, the same may kindly be intimated
to this Branch urgently, so that the same may also be considered.
In the case of schools, which are/were in tents, the life of the articles, to be condemned, may
be reduced by 25 p.c. of the life, as fixed in the attached Annexures, provided the condemnation
Board are satisfied for the proper use of the articles in question. A certificate to this effect must
be endorsed on he lists. (This para was subsequently added vide letter no. F.4(52)/61-Edn(P), dated
December, 1962).
Annexure ‘A’ (Years)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
Almirah wooden
Almirah Iron
Black Board
Black Board Stand
Benches
Buckets (Tin)
Bicycle
Bicycles stand (iron)
Bicycle stand (wooden)
Chairs wooden seat
Chairs iron/steel seat
20
50
3
3
5
2
8
10
3
5
10
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
110
Chairs cane seat
Cash Box (wooden)
Cash Box (iron)
Chauki (Takhat)
Desk Single Shift
Desk Double Shift
Stool
Durries
Rack (wooden)
Rack (iron)
Officers Table
5
5
25
10
7
5
3
5
10
25
20
23. Teachers Table
7
35. Notice Board
5
24. Office Table
10
36. Tray (wooden)
5
25. Library Table
10
37. Tray (iron)
5
26. Physics Table
10
38. Paper Stand
7
5
27. Chemistry Table
8
39. Foot Rest
28. Domestic Science Table
8
40. Hat Hanger & Looking glass
10
10
41. Confidential Box
10
2
42. Teapoy wooden
10
10
43. Carpet
10
32. Newspaper Stand
5
44. Trunk
10
33. Waste Paper Basket (Tin)
5
45. Map Stand
5
34. Waste Paper Basket (wooden)
3
46. Cash Safe
50
29. Biology Table
30. Table Cloth
31. Screens
Annexure ‘B’ Physics Apparatus (Non-consumable)
1.
Balance (Spring)
5
21. Screen (Glass)
2
2.
Balance (Physical)
5
22. Lens Stand (wooden)
2
3.
Weight Boxes
5
23. Optical bench (wooden)
2
4.
Boyle’s Law Apparatus
5
24. Spectomdeter
5
5.
Vernier Callipers
5
25. Wire Gauge Stand Iron
5
6.
Fortin’s Barometer
5
26. -do-Wooden
3
7.
Metallic Cylinder
5
27. Travelling Microscope
8.
Metal Sphere
7
28. Copper Calorimeter
5
9.
Metres rod (wooden)
1
29. Hypsometer (Copper)
5
10. S.G. Bottle
5
30. Thermometer
1
11. Spherometer
5
31. Max. Min. Thermometer
5
12. Screw Gauge
5
32. Magnet (bar)
5
33. Compass needle
3
13. StopWatch
10
10
14. Inclined plane
5
34. Compass (for lines of force)
5
15. Gravesand’s apparatus
5
35. Deflection Mangnetometer
5
16. Young’s modulus
5
36. Ammeter
10
17. Concave Mirror
2
37. Voltameter
10
18. Convex Lens
5
38. Galvanometer
7
19. Glass Prism
5
39. Accumulator
2
20. Glass Slab
5
40. Laclanche Cell
2
111
41. Electric Bell
2
64. Tunning Forks
2
42. Electrophorus
5
65. Resonance apparatus
5
43. Gold leaf electroscope
5
66. Stove (Oil)
5
44. Glass rod
1
67. Binoculars
10
45. Ebonite rod
1
68. Soldering Rod (Fire)
5
46. Silk and Cat Skin pieces
1
69. Solder (Electric)
2
47. Proof plane
2
70. Graduate cylinder
2
48. Slide wire Bridge
5
71. Glass Plate Machine
5
49. Potentiometer
5
72. Spirit Level
5
50. One Way and two way keys
5
73. Battery Clamps
3
51. Resistance box
5
74. Siren
7
52. Rheoslat
5
75. Hydrometer
5
53. Resistance Coil
5
76. Lactometer
5
54. Stading Key (two keys)
5
77. Drawing Board
2
55. Tangent Galvanometer
7
78. Barometer Tube
2
56. Induction coil
5
79. Photographic Camera
20
57. Torch lamp holder
2
80. Telescope
20
58. Switches
1
81. Newton’s Disc
10
59. cutout fuses
1
82. Pin hole camera
10
60. Pliers
5
83. Microscope
30
61. Spirit Lamp
4
84. Epidiascope
20
62. Tripod Stand
5
85. Radio set
10
63. Retort stand and clamps
5
86. TV. Set
10
●
University Physic : H.D. Young, M.W.
Zemansky and F.W. Sears, Narosa Pub. House.
●
Physics - Foundations and Frontiers : George
Gamow and J.M. Ciearland, Tata Mcgraw Hill.
112
113
2a
2b
Name
3
Date of
purchase
4a
Normal
in
building
4b
used in
tents/
double
Shifts
Total Life in lived
8
required
Pagein stock
Total
9
required to
be written off
No./quantity
10
article in
Stock
Balance of
serviceable
11
stock/property
Register
Rate as per
Consumable/Non-Consumable Articles To be Written Off
1
No. No.
the Article
No. & Name of
5
life of the
Articles
Fixed
Vide letter
No.
accordance
of 4(52) 61
Edn
Dated
2.8.82
12
Total cost of the
articles required
to be written off
List A (Whose Life Has been Laiddown) (Unserviceable Articles)
Name of the School——————
PROFORMA OF CONDENMATION
13
Remarks if any
affecting the life
of the articles
not covered by
other Columns
6
Whether the
articles
mentioned in
Col. 9
have
completed
their life in
accordance with
the life
mentioned
in Co. 5
7b
Page
14
of the head of the
Insititution
Recommendation
7a
S.No.
S. No & Page in
the
Property Register
List of Non-consumable
114
Certified that in the above statement only those items have been included whose life has been fixed vide letter No. F4 (52)761
Edn 2-8-62 as amended.
Certified that the articles mentioned in Col. 9 has already completed their life as fixed vide letter No. F(52)/61 Edn. 2-8-62
amended.
Item Nos. ----------- have been physically verified to be unserviceable and are recommended for being written off.
Certified that the articles contained in the list have been checked from the stock register and the date of the purchase price etc.
have been certified.
3.
4.
5.
6.
Member of the Condemnation Board
Certified that the no. sanction in respect of the articles in question was received by the school concerned in the part was any
of these articles includes in list pending sanction.
-------- there has not been any negligence leading to breakage of or damage to the articles on the part of any officer concerned
with the charge of the articles,
Certified that the articles mentioned in the Col 9 have become unserviceable due to normal wear and tear certificate further that
2.
1.
Certificate :
Suggseted Readings
●
Factes of Physics : A conceptual Approach, West Publishing Company.
●
Fundamentals of Physics : David Halliday, Robert Resnick, Jearl Walker, Asian Books Pvt. Ltd.,
New Delhi.
●
University Physics : H.D. Young, M.W. Zemansky and F.W. Sears, Narosa Pub. House.
●
Physics - Foundations and Frontiers : George Gamow and J.M. Clearland, Tata Mcgraw Hill.
●
College Physics : R L Weber, K.V Manning, M.W. White & G.A Weygard, Tata Mcgraw Hill.
●
College Physics : Reymond A. Sarvey and Jerry S. Fanghan Harcourt Brace & Co. Principles
of Physics : Raymond A. Serway & John W. Jewett, Jr.
●
The Elements of Physics : I.S Grant & W R. Phillips.
●
Physics can be fun : Y. Perelman, Mir Publishers.
●
Advanced level Physics M. Nelkon & RParker, Arnold - Heinemann.
●
Success in Physics - Tom Duncan, John Murray Publications Ltd.
●
Success in Electronics - Tom Duncan : John Murray Publications Ltd.
●
Concepts of Physics - H.C. Verma : Bharti Bharti Bhawan Publishers.
●
3000 Solved Problems in Physics : Alvin Halpern.
●
Schaum’s solved problem series : Tata Mcgraw Hill.
●
Mcgrwaw Hill’s Dictionary of Physics.
●
Physics - Resnick and Halliday
Net Resources
The following is just a suggestive list:
●
www.physicsclassroom.com
●
www.learn4good.com/kinds/high_school_science_physics
●
www.hsphys.com
●
faraday.physicssuiowa.edu/resource.html
●
www.library.aucktand.ac.nz/subjects/physics/phymeta.com
●
http://en.wikipedia.org/wiki/Optics
●
http://www.ee.umd.edu/~taylor/optics.htm
115
●
http://www.walter-fendt.de/ph14e/mwave.htm
●
http://www.school-for-champions.com/science/ac.htm
●
http://www.aiiaboutcircuits.com/vol_2/cht_1/1.html
●
ed.org/Education Resources/HighSchooi/Electricity/altematingcurrent.htm
●
http://theory.uwinnJnpeg.ca/physics/bohr/node1.html
●
http://en.wikipedia.org/wiki/Atomic_physics
●
http://en.wikipedia.org/wiki/Wage-particle_duaiity
●
L11/L8/de_Broglie_Wages/de_brog!ie-waves.htm!
●
http://physics.nrnt.edu/~raymond/ciasses/ph.13xbook/node189.html
●
http://www.search.com/refernce/Geiger%2DMarsden_experiment?redir=1
●
http://einstein.stanford.edu/content/faqs/maser.html
●
http://en.widipedia.org/wiki/Electromagneticjadiation
●
http://www.eo.ucar.edu/rainbows
●
http://www.atoptics.co.uk/bows.htm
●
http://astro.nineplanets.org/bigeves.html
●
http://users.ece.gatech.edu/~alan/ECE3080/Lectures/ECE3080-L-1Introduction%20to%20Electronic%20Materials%20Pierret%20Chap%201%20and%202.pdf-Georgia
Tech
●
www.pptsearch365.com/diode-properties-ppt.html
●
ebookfreetoday.com/bulk-semiconductor~0~ppt.html
●
downppt.com/ppt/semiconductor.html
and many more which you may get on the internet
116