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Electric Current
Introduction
We all must be families with the team electric current. We think who simple it is. But when even for a single
moment we try to analyze actually what it is, how is it generated then we realize that team and making it so
simpler to us. Now a days every single thing is an application of current and electricity.
Conductor for and time is called electric current
Where dq is the charge flown through a cross section area conductor in time dt. It is also called
instantaneous current I this dt - 0.
Where q is the charge flown in a finite time t. S.I unit of electric current is Ampere (A).
Illustration - 1. The current in a wire varies with time according to the relation.
I=(3+2t)A
(a) How many coloumbs of charge pars a cross section of the wi in the time interval between t=0 and t=4s ?
(b) What constant current would transport the same charge in the same time interval ?
Ans :-(a) We know that,
(b) Let the new current be I.
Current density
is defined as.
Note that it is a vector quantity.
The direction of
is same as direction of current. If a uniform current i moves over an area S and is in same
direction as area vector then,
So i is a resulting from the dot product of
and
.
Drift Velocity - It is defined as.
Where e = electron charge.
m = mass of electron
E = Electric field
= average collision time.
to the field direction. But in their way they collide with of positive ions. So they follow a very zig-zag motion,
hence length that they drift even in a long interval in small let this length be 'l' and time interval = ' '. then -
Now the electron moves under the influence of an electric field, Hence
Where
depends on temperature and material of conductor
Current and drift velocity :- We can express a relation in current and drift velocity
as,
where n = Moving charged particles per unit volume.
A = Area of cross section perpendicular to current flows.
Dumb Question - 3. What is the direction of current density ?
Ans:-
being a vector have a direction similar to
.
Dumb Question - 4. It a charge (q) is rotating with frequency f. What is the current ?
Ans :- I = qf because the units of current is amp = coloumb/s and qf constitutes this unit.
to electric field (
) with constant of proportionality being resistivity of (P) of the material, at a constant
temperature.
This relationship is called ohm's law.
Where V is the potential difference across the conductors.
placing J and E is equation (1) -
now,
this
or resistance
which is more general form of ohm's law.
Question :- Is ohm's law for all materials ?
Ans :- The current density in a conductor depends on the electric field and on the properties of the
material. In general this dependence is quiet complex But for same materials, especially metals, at a
constant temperature ohm's law holds good.
Illustration - 2. If a wire of resistance R is streched to double its length. What will be new resistance ?
Ans :- Suppose initial radius of wire be r, final be r' and new length be l' (= 2l)
Now, after streching volume of wire will not change :-
New changed Area =
New resistance
where A = initial area
Temperature dependence of resistivity and resistance.
The resistivity of a metallic conduct or nearly increases with increase in temperature.
Reason With the increase in temperature the ions of conductor vibrate with greater amplitude and the collision
between electrons and ions be come more frequent.
where P0 = resistivity at a reference temperature.
T0 = The factor is called the temperature co-efficient of resistivity.
The resistance of a given conductor depends on P, A and l. Ad temperature changes A and l also changes
but these changes are quiet small and thus
may be treated as constant.
KIRCHOFF'S LAWS
(1)Junction law
It states that the net current entering a junction is equal to net current exiting that junction.
Or, Algebric sum of the current is into any junction is zero.
In this fugure,
i1 + i2 = i3 + i4 + i5
2) Voltage/Loop Law
It states that algebric sum of the potential drop around any closed path is zero.
Sign Convention:
When we travel through a source in the direction from to+, the voltage is considered to be positive, when we
travel from + to.
For Battery
= VB - VA = + E
VB - VB = - t
For resistance
= VB - VB = - IR
VB - VA = + IR
Dumb Question - 7. What are the basic principles on which kirchoff's laws are based ?
Ans :- Kirchoff's junction rule basically supports the law of conservation of change. On the other hand,
kirchoff's second law that is kirchoff's voltage rule or kirchoff's loop rule is based on the law of conservation
of energy.
Illustration - 3. Find the current in brand BE and potential difference VBD in.
Assume the shown current distribution.
We can choose arbitarily any type of distribution as we like.
Applying kirchoff's junction law at B (or at E).
We get : I1 + I2 = I3 ................................. (1)
Applying kirchof's voltage law in the loop ABEFA: (clockwise direction)
We get :
Applying kirchoff's voltage law in BCDEB (Clockwise direction)
We get :
on solving (1), (2) and (3) :
For potential drop across VBD :
Let us search from B to D via C :
Question :route which connects B and D. It is recommended to go via shortest possible route to avoid long calculations
and maping problem more complex.
Internal resistance of a battery
The potential difference across a real source in a circuit is not equal to the e.m.f of the cell,
Dumb question :- Why this happens ?
Solution :- The reason is that charge moving through the electrolyte of the cell encounters same resistance
known as, internal resistance of the cell.
It is denoted by r. so, there is a potential drop across the ends of the e.m.f source.
Potential difference (V) across the terminals of a battery For an e.m.f source, the potential changes will be obtained as illustrated below :-
Special Cases :1) It current flows in opposite direction then,
V = E + ir
2) V = E if current through the cell is zero.
3) V = 0 if the cell is short circuited.
Dumb question - 8.
Why V = 0 if the cell is short circuited ?
Ans :-
NOw in this case applying kirchoff's voltage loop law,
Ilustration - In the given circuit
E1 = 10 V, E2 = 8 V, r1 = r2 = 2
.
Applying kirchoff's voltage law in the circuit (moving anticlockwise)
Gruping of Resistances :
1) In series : Req = R1 + R2 + R3 + ...................... Rn
Equivalent resistance is defined as :
where V = Voltage of battery
I = Current following through battery.
Using kirchoff's loop rule in clockwise direction we get:
Thus, Req across AB = R1 + R2 + R3 further if battery has any internal resistance it will be added to give total
resistance.
2) In parallel Req = R1 + R2 + R 3 + ........... + Rn
Here also,
Suppose, in the given figure we have to find Req across AB :
Assuming the following distribution of current as shown.
Applying kirchiff's Junction Law at A.
(Because potential drop across each are same being in parrallel.)
Ans:- Resistance AB and BC are in series.
So, their net resistance =
We reduced, the fugure now to :
Further resistance AC and AB are II
So, their net resistance =
So, new diagram :
Resistance across AC and CD in series so, their
equivalent
Finally we have
So, we have
Req across AD = 3
.
Heating effects of current:When a constant current i flows for time t, then the amount of charge that flows is :
Q = It ..................................... (1)
Electrical Energy that is delivered = W
= Q.V
= VIt.
Where V = Potential across conductor in which current is flowing.
Power supplied to the current =
If e.m.f. is E, as in the undrlying circuit :
I is the current flowing in the circuit here assumed to be anti clockwise direction.
Now, multiplying both sides by I.
Where E.I is the rate of conversion of chemical energy to electrical energy.
I2R is the power supplied to external resistance R.
2
I r is dissipated power as a result of resistance due to internal resistance of the battery.
Joule's Effect
Whenever a current flows through a conductor heat is in it. This effect is commonly called Joule's heating
effect.
Powers is always dissipated in a resistance with this sate the heat produced in time t is,
Now, Power = P = I2R.
Heat Induced = H = I2Rt
Where if H is in joules
Then i is in am press.
R is in ohms.
T is in seconds.
Dumb Question - 9. Why is this heat produced in the resistance whenever a current passes through it ?
Ans :- Suppose the potential difference between the ends of a resistor is V and current I is flowing through it.
Work done by the electric field on free electrons in time t is given by :
Now, this work done by the field is converted into thermal energy of the resistor and thus, heat is dissipated
through it.
Illustration - 6. An electron bulb B1(220V, 60W) is connect in series with another electric bulb B 2(220V, 45W)
When they both are connect in series to 220V supply which bulb will glow more ?
Ans :- As
Let resistance of first bulb be R1 and second be R2.
Similarly,
[where V =220V, P1 = 60w]
let power developed across B1 be P1' and B2 be P2'.
As in series current across both will be equal.
thus, bulb rated 220V and 45w will glow more.
Dumb Question - 10. What do a bulb stated 220V and 60W conveys ?
Ans :- It conveys that if that bulb is connected to a battery giving a potential difference of 220V then the
power output from that bulb will be 60W. That is, it will give out 60V of energy, which is converted to heat
energy per unit time.
Condition for maximum power output from the battery
ie Internal resistance = External resistance.
Suppose the arrangement shown in a circuit.
By kirchoff's laws,
Power output of resistance = P
r=R
For maxima
=0
Thus, the power output is maximum net external resistance equals internal resistance of the battery.
Dumb Question - 11. In the above derivation whose power output is maximum ?
Ans :- Energy released from resistor equals (energy provided by the battery) - (work done by the battery).
Thus, If power output of battery is maximum the power output of resistor will also, be maximum.
GROUPING OF CELLS
a) Series Grouping.
Suppose n cells of e.m.f. E and internal resistance r connected in series :
Then net e.m.f. = nE.
Total resistance = nr + R.
Thus, net Current =
Illustration - 7. In a series of N cells how many should be reserved so that current becomes
earlier value ?
of the
Ans :- Initial current =
Let x cells be reserved.
= (N - x) E - nE.
Total resistance = Nr + R.
New current =
= 2N - 4x = N
4x = N
x=
.
(b) Parallel grouping :-
Applying Kirchoff's Law through branch :
(c) Series grouping with different batteries :-
Eeq = E1 + E2 + E3 +.................... + En
Req = r1 + r2 + .................................. + rn
(d) Parellel Grouping with different batteries :-
Number of rows are m and numbers of cells - in each rows in n :So, it is equivalent -
Applying Kirchoff's laws :-
Dumb question :- How is total emf mE ?
Solution :- 'm' . 'E' rated batteries are in series so net emf becomes mE.
Dumb Question :- How is net resistance
?
Solution :- 'm' batteries in each row means 'mr' resistance becomes
.
Illustration - 8. Find the equivalent emf and resistance of a single battery which is equivalent to a
combination of al these batteries in the figure.
Ans :- For parallel combination :-
Now, this is in series with 6V battery -
Thus, net e.m.f. = (6 -3)V
And internal resistance = 2
So, the equivalent circuit is -
a) Galvanometer :- It can detect very small currents as it has negligible resistance.
The corresponding potential difference for full scale deflection is :V = ig.G
b) Ammeter :IT IS ALWAYS CONNECTED IN SERIES WITH THE BRANCH 'I' HAS TO BE MEASURED !
It is a corrent measuring instrument. A galvanometer can be converted into a ammeter by connecting a
small resistance S (called shunt) in parallel with it.
Thus, S(I - Ig) = G X Ig
Dumb Questions :- Why is S in parallel ?
<>Solution :- As S is in parallel the current (I - Ig) will be very high while Ig will be lo. So that the potential
difference VA - VB does not change much.
Dumb Questions :- Why is Ig ?
<>Ans :- Ig is that current which gives full scale deflection in galvanometer.
TO THE BRANCH WHOSE BE MEASURED !
A voltage measuring device is called a voltmeter. It measures the potential difference between two points. A
galvanometer can be connecting a high resistance (R) in series with it.
Dumb Questions :- Why is RV high ?
Solution :- If a large register (RV) is applied in parallel to two points then it doesn't draw any current as R V is
high. So Ig is very low. And hence the current IAB doesn't change much.
Dumb Questions :- What is the ideal resistance for as ideal ammeter and ideal voltmeter ?
Ans :- For ideal ammeter R = 0
For ideal voltmeter R =
Illustration - 9. A galvanometer having a coil resistance of 50
gives a full scale deflection when a current
of 0.5mA is passed through it. What is the value of the resistance which can convert this galvanometer into
ammeter giving full scale deflection for a current of SA ?
Ans :- As we derived earlier :
Now, Ig = .5 10- 3A.
I = 5 A.
G = 50
As expected it is coming out to be very low.
In ideal ammeters this shunt resistance is 0.
Illustration - 10 :- How can we make a galvanometer with G = 20
and Ig = 1mA in to a voltmeter with a maximum range of ISV ?
Ans :- Using
.
we have,
.
As expected it is very high.
In ideal voltmeter it tends to infinity.
* Tip :- The basic Aim of using such devices is that they don't affect the current much when used. And hence
the voltmeter is such that when it is used, it doesn't draw much current so that potential difference remains
the same and is such that it also doesn't let the voltage to da between the two points where it is installed.
(and hence doesn't change the current).
WHEATSTONE'S BRIDGE
In this system suppose galvanometer shows null (zero current.
Then VBC = 0
VB - VC = 0
VB = VC
I1R1 = I2R3 .......................................... (1)
Similarly form other ends
I1R1 = I2R4 .......................................... (2)
Dividing (1) and (2) :
And, At this point the bridge is said to be balanced.
Dumb Questions - 14. Why on value of current through galvanometer to be zero VBC = 0 ?
Ans :- Current flows between two points only if there is any potential difference between those points. Here,
in this case as current through BC is zero, this suggests that potential of B equals the potential of C. Hence
VB = VC or VB - VC = 0 or, in other words VBC = 0.
So if between B and C, another resistance is placed, then no current flows through that resistance.
d) Potentiometer :It is a device used to measure potential difference with high accuracy, without drawing any current from the
unknown source. [* Basic Figure of a potentiometer : ??]
Principle of Potentiometer
It is based on the principle, that a current carrying wire has potential difference across its ends proportional
to its length with a uniform cross sectional area.
It can be used for various purpose :
(1) To find e.m.f. of an unknown battery :
First a known battery (Ek) is placed as :
Suppose the null point is at l1
Then, Ek = i(l l1) .............................................(1)
Dumb Questions - 15. What is this null point ?
Ans :- On a close observation we can see that potentiometer is similar to wheatston's bridge.
At null point current through galvanometer equals 0.
Now, the known battery is replaced by unknown battery (Eu).
Now, Eu = I(l l2) .............................................. (2)
From (1) and (2)
2. Find internal resistance(r) of an unknown battery.
, where R is a known resistance.
For any circuit for battery : V = E -ir = iR
......................................... (A)
Where R = external resistance.
r = internal resistance.
We connect a battery of e.m.f. E and internal resistance r as:
Now, E = ill1 ................................................. (1) Now a known resistance R is connect across the terminals
as :
Now,
............................................. (2)
from (1) and (2)
............................................. (3)
Now, placing value of (3) in (A) :
Question :- Find current in all branches
Ans :- Assume the current distribution as shown :
Applying kirchoff's loop law in the three loops we get :
2l - Si1 - 6(i1 + i2) - i1 = 0 ......................................... (1)
S - 4i2 - 6(i1 + i2) - 8(i1 + i3) = 0 ................................(2)
2 - P(i2 + i3) - 16i3 = 0 ............................................. (3)
On simultaneously solving there three equations we get :
Dumb Question - 16. In branch AC which will we Lonsides i1, i2 ?
Ans :- We will Lonsides the current i1 + i2 in AC as we have Lonsidered that in those two loops different
currents i1 and i2 are flowing now, their common edge will have the sum according to the direction of two
currents. This is called principle of superposition.
E2 :- Two resistance have temprature coefficient 1 and 2 and R01 an R02 the resistance at 00C. Find
eq if they both are connect (a) in series and (b) in parallel ?
Ans :- For series :
At 00C Req = R01 + R02 ................................................ (1)
At any temperature t0C
Req = R01 + R02
Req = R01[1 +
[R01 + R02][1 +
1(t
eqt]
- 0)] + R02[1 +
= R01[1 +
1t]
2(t
- 0)]
+ R02[1 +
2t]
In Parallel :
At t0C
Using Binomial Expansion :
E3. A rod length L and cross sectional area A lies along the axis between x = 0 and x = L.
Its resistivity is given by : P(x) = P0
.
The end x = 0 is at a potential V0 and x = L at V = 0.
(a) Find net resis tance of the rod.
(b) The electric potential V (n) ?
Ans :- (a)
At any distance n suppose an elementary particle dn be considered such that dn
Its resistance
As R =
0.
Thus, R =
(b) We know that,
Potential difference = i dR = dV
04. A battery of e.m.f. 1.4V and internal resistance 2
is connected to a 100
resistor through an
ammeter. The resistance of te ammeter is
.
If ammeter rods .02 A. What is the resistance of the Voltmeter ?
Ans :- Circuit diagram is :
Now first finding total resistance of the circuit.
where G = Resistance of voltmeter.
Now, Current =
In the given circuit 5
resistor develops 45 J/s Power due to current flowing through it. Calculate the heat
developed/second through 2
resistor ?
Ans :- Assume the shown current distribution after writing equivalent circuit :
Now, as power through 5
resistance is 45 J/S.
Applying Kirchoff's loop rule in the loop.
Thus, net current is 4A through 2
resistor.
Thus, power lost from this resistance = (2) X 42 = 32 J/S.
Question :Connected is series and have different internal resistances r 1 and r2( >l1). Find an external resistance R at
which the potential difference across the terminals of one of the sources becomes equal to D ?
Ans :- V = E - ir E and i for both the sources are equal. Therefore, potential difference V will be zero for a
source having greater internal resistance, that is, r2.
.
01. Six equal resistances of 4
each are connect as shown. Find Req about PQ.
Ans :- Finding Req about PQ.
The Circuit is equivalent to :
Now, current through AB is zero, as it is a balanced Wheat Reduces Stone Bridge.
Thus, the circuit reduces to
Thus,
.
02. Find the potential difference Va - Vb in the circuit shown in the figure :
Ans :- Assume the shown current distribution :
Applying kirchoff's voltage law in abcda :
E1 + I1R1 = I2R3 ..................................(1)
Applying kirchoff's voltage law in abfca :
I2R3 + (I1 + I2)R2 = E2 .................................(2)
Using (1) and (2):
Thus, Va - Vb = I2R3
[Using (3) and (4)]
Question :-
Certain circuit :
Find
(a) Power dessipated in 6
(b) VC - VB.
resistance.
(c) VA - VD ?
Ans :- (a) Power in 6
= I2 R
(b) Now, Applying Kirchoff's Voltage law at E.
Current through 1
resistor = 9A.
.
(c) VD - 12 + 5 + 28 - 8 - 4 = VA
VD - VA = - 33V
VA - VD = 33V
M2: Find equivalent battery having same e.m.f. and equivalent resistance of a battery equivalent to V1 and
V2 having internal resistances r1 and r2 in the adjoining figure .
Ans (1) Potential difference across the terminals of the battery is equal to e.m.f. when current drawn from
the battery is zero. Assuming no current is coming out.
So, Current distribution is :
Thus, VA - VB = Net equivalent E.M.F. = V1 - ir1(V2 - ir1)
(2) As r1 and r2 are in parrallel.
Equivalent internal resistance = r =
.
Question :Electrical circuit is shown in figure. Calculate the potential difference across the resistor of 400 ohms as will
be measured by the voltmeter V of resistance 400
either by applying Kirchoff's laws or otherwise.
Ans :- This circuit is equivalent to :
Which in turn is equivalent to :
As, it is a balance Wheat Stone's Bridge as
.
of 100
between A and B can be ignored.
The voltage will read the potential difference across Q.
First we find that i1 = i2 =
.
Potential difference across resistance Q = Q X i1
.
Therefore, Voltmeter will read
V.
Question :Charging and 100 V after charging. When charging below the current was 5A. What is the current at the end
of the charging if the internal resistance is equal and constant to 8
in the whole process.
Ans :- The voltage supplied is constant for both times and equal V given by -
Let the current finaly (at the end of charging br I)
H -1: A conductors has a temperature independent resistance R and its total heat capacity is C. At t = 0 it is
connect to d.c. voltage source V. Find T(t) assuing power dissipated into surrounding space to very as q =
k(T - T0) where k is constant, T0 is surrounding temperature (and equals conductor's initial temperature) ?
Ans :- Energy supplied by source/unit time = energy lost in environment/unit time + energy used in raising
temp/unit time
On solving :
.
Dumb Question - 11. How we get the initial equation of the above question ?
Ans Initial equation states :
Energy supplied by source/unit time = energy lost to surroundings/unit time + energy used in raising
temp/unit time
On close observation one finds that it is just the Law of Conservation of Energy.
M - 2. What amount of heat will be generated in a resistance R due to a charge q passing through it if the
current in the coil :
(a) Decreases down to zero uniformly during a time internal to ?
Ans (a) Drawing corresponding i - t graph :
i0 can be found by the information that charge q = area it
At any time t assume a small increament in time dt such that dt
dH = Heat produced in this time interval = i2Rdt
Total Heat produced =
.
0.
Dumb Question - 18. How charge = Area under it graph ?
Ans :- We know that Current = I =
Thus, dQ = I dt
Integrating both sides
[Assuming at t = 0, Q = 0].
Now,
is nothing but area under the It graph.
H - 3. A voltmeter of resistance R1 and an ammeter of resistance R2 are connected in series across a
negligible resistance battery. When a resistance R is connected in parallel to voltmeter, ammeter increases
its reading to three times and voltmeter's reading reduces by one third. Find R 1 and R2 in terms of R ?
Ans :- Case I :
In this case E = I(R1 + R2).
Case II :
Now, as ammeter reads 3 times so, net current 3i is drawn from the battery.
Current in voltmeter resources to
Remaining
passes through R.
VC - VD =
R1 = 8R.
Applying Kirchoff's Voltage Rule in ABFGA :