Download Chapter 5 ELECTRIC CURRENTS

ELECTRIC CURRENTS
ELECTRIC CURRENTS
Electric Potential Difference, Current and Resistance
5.2
Electric Circuits
5
CORE
5.1
INTRODUCTION
F
igure 501 illustrates a typical model of a metal. Metals are good electrical conductors because of the mobility of electrons.
potential end as shown in Figure 502. Electrons entering at one end of the metal cause a similar number of electrons to be displaced from the other end, and the metal conducts. Even though they are accelerated along their path, it is estimated that the drit velocity is only a small fraction of a metre each second (about 10-4 m s-1).
metal cations +
electrons
entering metal
+
+
+
+
+
+
+
Electric field
+
electrons
leaving metal
+ +
+
+
low potential (–)
(+) high potential
–
metal rod
Drift direction
Figure 501 Model of electrical conduction in metals.
Figure 502 Drift velocity of electrons.
Scientists have gathered evidence to suggest that the electrons in outer shells of metals move about freely, within a three-dimensional metal lattice of positively charged metal ions. hus, metal structure consists of positive ions in a “sea” of delocalised electrons.
An electric ield is created when a potential diference is supplied to a metal wire. hus, the average speed and average kinetic energy of the electrons increases. When they collide with positive ions in the lattice, they give up some of their energy to them. Ater this event, they are again accelerated because of the electric ield until the next collision occurs. At each stage, these collisions generate heat that causes the temperature of the metal to increase. We say that a current produces a heating efect.
Positive ions and free electrons have internal energy that depends on the temperature of the metal at that point in time, and the delocalised electrons move about at enormous speeds of about 106 m s-1, colliding with the positive ions in the metal’s lattice. However, as much as the speed is high, there is no net movement of charge unless the conductor is connected to a source of potential diference (or voltage).
When a dry cell or battery is connected across the ends of a metal wire, an electric ield is produced in the wire. he electrons drit with a drit velocity to the high positive 125
CHAPTER 5
5.1 ELECTRIC POTENTIAL DIFFERENCE,
CURRENT AND RESISTANCE
2. CORE
ELECTRIC POTENTIAL DIFFERENCE
5.1.1 Define electric potential difference.
5.1.2 Determine the change in potential energy
when a charge moves between two points
at different potentials.
he electrical potential diference between two points in a conductor carrying a current is deined as ‘the amount of electrical energy that is changed to other forms of energy when an amount of charge moves from one point to another, where P is the power dissipated in watts (W) and I is the current in amps (A)’.
he concept of electric ields will be explored more fully in Chapter 6. Here is a third way to deine potential diference in terms of electric potential.
5.1.3 Define the electronvolt.
5.1.4 Solve problems involving electric potential
difference
© IBO 2007
5.1.1 ELECTRIC POTENTIAL DIFFERENCE
In an electric circuit, charge is moved from one point in the circuit to another and in this process electrical energy supplied by a dry cell or power supply is converted to other forms of energy. For example, a ilament lamp converts electrical energy into heat and light energy. As charges move in the circuit, they gain energy supplied by the energy source.
For a charge to move from one point to another in a circuit, there has to be a potential diference between the two points because an amount of electrical energy is changed to other forms of energy as the charge moves from one point to another such as from one side of a ilament lamp to the other.
In Chapter 2, we learnt that when a force acts on a mass and moves it in the direction of the force, then work is done on the mass. Work is a scalar quantity and is deined as the force F multiplied by the distance s moved in the direction of the force. Mathematically, this was stated as:
W = F ⋅ s or W = F s cos θ
3. ‘he electrical potential diference between two points in a conductor carrying a current is deined as the work done per unit charge in moving a positive charge from a point at higher potential to a point at lower potential’.
∆E
W
∆V = ---------p- = ∆
-------q
q
where q is the amount of charge measured in coulombs C. (If work is done in moving an electron, it would therefore
move from a point at lower potential to a point at higher
potential). So as charge q moves through a potential diference ∆V, it does work equal to ∆Vq and this work is equal to the energy given to the load component in the circuit. herefore, potential diference is a measure of the energy released by an electric charge or group of charges.
he accepted unit of electric potential diference is the volt, V. It follows that it can also be measured in JC -1. Just as work is a scalar quantity, so too electrical potential diference is a scalar quantity. When we use the term voltage, we are loosely talking about electric potential diference. If the potential diference between the terminals of a dry cell is 1.5V then the energy used in carrying a positive charge of +q from one terminal to the other is 1.5 J C -1.
Doing work on an object changes its energy.
here are a number of diferent ways to deine electric
potential diference:
1. 126
he electric potential diference between two points in a conductor carrying a current is deined as ‘the power dissipated in a load per unit current in moving from one point to another’. V = P / I
he work done in transferring a charge from two points does not depend on the path along which the charge is taken because potential diference is constant between two points.
ELECTRIC CURRENTS
Suppose there is charge moving in a closed circuit as shown in Figure 503. When the charge reaches the irst load resistor at A, power is dissipated in the load resistor and the electrical energy is converted to heat energy. here is a potential (energy) diference between points A and B because work has been done on the resistor and there will be an energy diference per unit charge through the resistor. When it gets to the second load, again energy is dissipated and there is a potential diference between either side of the resistor. If the electric potential energy dissipated is the same in both of the loads, the energy diference per unit charge will be the same amount. We say the potential diference across each load resistor would be 3 volts.
In Figure 503, a charge +q moves between points A and B through a distance x in a uniform electric ield. he positive plate has a high potential and the negative plate a low potential. Positive charges of their own accord, move from a place of high electric potential to a place of low electric potential. Electrons move the other way, from low potential to high potential. In moving from point A to point B in the diagram, the positive charge +q is moving from a low electric potential to a high electric potential. he electric potential is therefore diferent at both points.
6V
A
x
Vi
In atomic and nuclear physics, we oten determine energy by multiplying charge times voltage (W = qV). Because the charge on an electron and the charge on the proton are used so frequently, a useful unit of energy is the electron-volt (eV).
One electron-volt (1 eV) is deined as the energy acquired by an electron as a result of moving through a potential diference of one volt.
Since W = q∆V, and the magnitude of charge on the electron (or a proton) is 1.6 × 10-19 C, then it follows that
W = ( 1.6 × 10
– 19
C ) × ( 1 V ) = 1.6 × 10
So that, 1 eV = 1.6 × 10
– 19
– 19
J
J
An electron that accelerates through a potential diference of 1000 V will lose 1000 eV of potential energy and will thus gain 1000 eV (1 keV) of kinetic energy.
he electron-volt is not an SI unit. In atomic physics, we ind that energy is typically a few electron-volts. In nuclear physics the energy is more likely to be measured in MeV (106 eV), and in particle physics it is typically in GeV (109 eV).
5.1.4 SOLVING POTENTIAL
I
6V
5.1.3 THE ELECTRON-VOLT (eV)
+q
DIFFERENCE PROBLEMS
V out
Example 1
B
3V
V out
0V
Calculate the work done in moving 10.0 μC through a potential diference of 150 V
Solution
Figure 503 Charge moving in a uniform electric field
In order to move a charge from point A to point B, a force must be applied to the charge equal to qE (F = qE).
Using the formula ΔW = qΔV where the potential diference,
ΔV = 150 V, we have W = F × x = E q × x
Since the force is applied through a distance x, then work has to be done to move the charge, and there is an electric potential diference between the two points. Remember that the work done is equivalent to the energy gained or lost in moving the charge through the electric ield.
∆W = q ∆V = ( 1.00 × 10
–5
C ) × ( 150 V )
= 1.5 × 10-3 J
he work done is 1.5 × 10-3 J.
127
CORE
5.1.2 CHANGE IN POTENTIAL ENERGY
BETWEEN 2 POINTS
CHAPTER 5
Example 2
ELECTRIC CURRENT AND RESISTANCE
5.1.5 Define electric current.
CORE
A 2.0 μC charge acquires 2.0 x 10-4 J of kinetic energy when it is accelerated by an electric ield between two points. Calculate the potential diference between the 2 points.
5.1.6 Define resistance.
5.1.7 Apply the equation for resistance in the
Solution
form
where ρ is the resistivity of
the material of the resistor.
Using the formula ΔW / q = ΔV = 2.0 × 10-4 J / 2.0 × 10-6 C
= 100 V
5.1.8 State Ohm’s law.
he potential diference is 1.0 × 102 V.
5.1.9 Compare ohmic and non-ohmic behaviour.
Example 3
he work done by an external force to move a -8.0 μC charge from point a to point b is 8.0 × 10-3 J. If the charge initially at rest had 4.0 × 10-3 J of kinetic energy at point b, calculate the potential diference between a and b.
Solution
Using the formula ∆V = W / q = 8.0 × 10-3 J / -8.0 × 10-6 C
= -1.0 × 102 V
he work done to have a gain in kinetic energy
= 4.0 × 10-3 J / 8.0 × 10-6 C
= 5.0 × 101 V.
ΔV =1.0 × 102 - 5.0 × 101 V
he potential diference is 5.0 × 101 V.
5.1.10 Derive and apply expressions for electrical
power dissipation in resistors.
5.1.11 Solve problems involving potential
difference, current and resistance.
© IBO 2007
5.1.5 ELECTRIC CURRENT
An electric current is a movement of electric charge that can occur in solids, liquids and gases. A steady current can be maintained when there is a drit of charge-carriers between two points of diferent electric potential. he charges responsible for the drit are:
Solids
electrons in metals and graphite, and holes in semiconductors
Liquids
positive and negative ions in molten and aqueous electrolytes
Gases
electrons and positive ions stripped from gaseous molecules by large potential diferences.
At this point, we will deine ‘the electric current as the rate at which charge lows past a given cross-section’.
q
I = ∆
-----∆t
128
It makes sense that for an electric current to low, there must be a complete circuit for it to low through. he unit of current from the above equation is the coulomb per second C s-1 and this unit is called the ampere (A).
he ampere or ‘amp’ is a rather large unit. Current is oten expressed in milliamps (mA) and microamps (µA) or even nanoamps and picoamps. Some relevant currents for situations are given in Figure 504.
A computer chip
Electron beam of a television
Current dangerous to a human
Household light bulb
Car starter motor
Lightning
10-12 – 10-6 A
10 -3 A
10-2 – 10-1 A
1 A
200 A
104 A
Figure 505 shows currents I1 and I2 lowing in the same direction in two wires. In part a, at point P on the right wire, the magnetic ield B1 is upwards (right-hand screw rule). hen using the right-hand palm rule, the direction of the force F is in an easterly direction. Similar analysis of the diagram in part b reveals that the force is in a westerly direction. Both forces are inwards, and the wires will attract each other. Check that you obtain the same result as the diagrams.
he vector diagrams and lines of force or lines of magnetic lux for currents in the same direction and in opposite directions are shown in Figures 505 (a) and (b) respectively.
I1
F
Figure 504 Some typical current situations.
Electric current is a fundamental quantity in physics, and it is the primary quantity on which electrodynamics is based. Its SI unit is the ampere (A). he ampere is deined in terms of ‘the force per unit length between parallel current-carrying conductors’. his deinition is all that is required for this course. his deinition can be extended further.
When two long parallel current-carrying wires are placed near each other, the force exerted by each will be a force of attraction or repulsion depending upon the direction of the current in each wire.
he force is attractive if the two currents low in the same direction, and the force is repulsive if the two currents low in opposite directions.
B1
I1
I2
F
F
I1
(a)
I2
Q
B2
F
B1
B2
B1
F
When a current lows in the same direction around an electric circuit, the current is said to be a direct current (dc). Dry cell and wet cell batteries supply dc. When the direction of the current changes with time it is said to be an alternating current (ac). Household electrical (nonelectronic) appliances are ac.
P
I1
I2
(b)
Figure 505 (a) and (b) Force between two parallel
current-carrying wires.
F
B2
I2
Attraction
Repulsion
F
F
B
B
current flowing into page
current flowing
into page
current flowing
out of page
Figure 506 Forces and fields between currents.
he French physicist Andre Marie Ampère showed that the quantitative relationship for the force F per unit length l between two parallel wires carrying currents I1 and I2 separated by a distance r in a vacuum was given by
F
1
1
F
--- ∝ I1 × I2 × --- ⇔ --- = kI 1 × I2 × --l
r
r
l
herefore,
µ 0 I1 I2 l
µ
F = -----0- × I1 × I2 × -l = ----------------2π
r
2πr
he constant µ0 is called the permeability of free space, is deined to equal
µ0 = 4π × 10-7 T m A-1
his last equation now allows us to inally deine the fundamental unit the ampere.
129
CORE
ELECTRIC CURRENTS
CHAPTER 5
CORE
hus one ampere is deined as that current lowing in each of two indeinitely-long parallel wires of negligible crosssectional area separated by a distance of one metre in a vacuum that results in a force of exactly 2 × 10-7 N per meter of length of each wire
conventional current electron flow
1.5 V
+
–
resistor
Example
Figure 507 Conventional and electron currents
Calculate the current lowing through a hair drier if it takes 2.40 × 103 C of charge to dry a person’s hair in 4.0 minutes.
Solution
To determine the current, given that there exists a charge of
2.40 × 103 C and that it is lowing for a period of 4.0 minutes
(= 240 seconds).
Unfortunately, this convention has been kept. Figure 507 shows a simple circuit diagram of a 1.5 V dry cell connected to a resistor. When drawing and interpreting circuit diagrams just remember that conventional current
lows from the positive to negative terminal unless you
are speciically asked for the correct electron low that
lows from the negative to the positive terminal.
5.1.6 RESISTANCE
3
his gives,
2.40 × 10
∆q
I = ------ = ------------------------- = 10 A
∆t
240
he current lowing is 1.0 × 101A.
Current is a scalar quantity but it is useful to indicate the direction of low of current.
Before the electron was discovered, the direction of the charge-carriers was already deined by scientists and engineers to be from positive to negative.
Benjamin Franklin stated that an excess of luid produced one kind of electric charge which he termed “positive”, and a lack of the same luid produced the other type of electric charge which he called “negative”. Franklin’s designation of (+) being an excess of electric charge and of (–) being a deiciency of electric charge was unfortunate.
When batteries and generators, the irst source of continuous current, were developed in the 1800s, it was assumed that electric current represented the low of positive charge as deined by Franklin. his is partly the reason that electric ields are deined in terms of a positive test charge, and the lines of electric lux being explained as going outwards from positive charges.
It is now known that an electric current is a low of electrons from negative to positive.
130
Electrical resistance, R, is a measure of how easily charge lows in a material. Conductors, semiconductors and insulators difer in their resistance to current low. An ohmic material of signiicant resistance placed in an electric circuit to control current or potential diference is called a resistor.
he electrical resistance of a piece of material is deined by the ratio of the potential diference across the material to the current that lows through it.
R = V
--I
he units of resistance are volts per ampere (VA-1). However, a separate SI unit called the ohm Ω is deined as the resistance through which a current of 1 A lows when a potential diference of 1 V is applied.
Since the term resistance refers to a small resistance, it is common for a resistor to have kilo ohm (kΩ) and mega ohm (MΩ) values. ELECTRIC CURRENTS
5.1.7 FACTORS AFFECTING
Nichrome is used as the heating element in many toasters and electric radiators. he semiconductors behave in a special manner, and pyrex glass is an obvious insulator.
RESISTANCE
•
•
•
•
length
cross-sectional area
resistivity
temperature
It can be shown that when the temperature is kept constant
R = ρ ⋅ --lA
where R is the resistance in Ω, ρ is the resistivity in Ω m, l is the length of the conductor in m, and A is the crosssectional area of the conductor in m2.
Resistivity
Ωm
Silver
1.6 × 10-8
Copper
1.7 × 10-8
Aluminium
2.8 × 10-8
Tungsten
5.6 × 10-8
Constantan (alloy of copper and nickel) 49 × 10-8
Nichrome (alloy of nickel, iron & chromium) 100 × 10-8
Graphite
(3 - 60) × 10-5
Silicon
0.1 - 60
Germanium
(1 - 500) × 10-3
Pyrex glass
1012
Material
Figure 508 Resistivities for certain materials at 20°C
Example
As the length of a conductor increases, the resistance increases proportionally
R∝l
It seems logical that as the length increases so too does the resistance to the low of charge across the conductor.
As the cross-sectional area of a conductor increases, the resistance decreases proportionally.
∝
With increased cross-sectional area, there is a greater surface through which the charge can drit.
Determine the resistance of a piece of copper wire that is 10.0 m long and 1.2 mm in diameter?
Solution
he resistance, R, is given by the formula R = ρl / A, where
A = πr2.
his means that
R = (1.7 × 10-8 Ωm) (10.0 m) / π(6.0 × 10-4)2 m2 = 0.150 Ω.
he resistance of the copper wire is 0.15 Ω.
he resistivity is speciic to the type of material being used as a resistance and is afected by the nature of the delocalised electrons and the positive ions within the material. he resistivity of a material is the resistance across opposite faces of a cube with sides of one metre. It can be found that by plotting a graph of R versus (l / A), the slope of the linear graph is the resistivity ρ measured in Ω m.
Figure 508 gives the resistivities of various materials at 20o C. he irst three values show that although silver has a low resistance, it is expensive to use in electrical circuits. Copper is the preferred metal although aluminum is commonly used in electricity transmission cables because of its lower density. Constantan and nichrome are commonly used in wire form, and are termed high resistance wires. he resistance of a material increases with temperature because of the thermal agitation of the atoms it contains, and this impedes the movement of electrons that make up the current.
he increase in resistance can be shown as
R f = R 0 ( 1 + αt)
where R0 equals the resistance at some reference temperature say 0 °C, Rf is the resistance at some temperature, t °C, above the reference temperature, and α is the temperature coeicient for the material being used.
131
CORE
he resistance of a conducting wire depends on four main factors:
One interesting phenomenon of the efect of temperature on resistance is superconductivity. In 1911, H. Kammerlingh Onnes found that mercury loses all its resistance abruptly at a critical temperature of 4.1 K. When a material attains zero resistance at some critical temperature, it is called a superconductor. he possibility of having a material that has an induced electric current that lasts forever has become a topic for research physicists. Just think of the energy saving if the perfect superconductor is found that can give zero resistance at room temperature.
A typical apparatus used in the conirmation of Ohm’s Law is shown using a circuit diagram in Figure 509.
+
variable resistor
+
–
A
Ammeter
5.1.8 OHM’S LAW
he German physicist, Georg Simon Ohm (1787-1854), studied the resistance of diferent materials systematically. In 1826, he published his indings for many materials including metals. He stated that:
Provided the physical conditions such as temperature are kept constant, the resistance is constant over a wide range of applied potential diferences, and therefore the potential diference is directly proportional to the current lowing.
–
resistor
+
–
V
voltmeter
Figure 509 Typical Ohm’s law apparatus
current, I
CORE
CHAPTER 5
V∝I ⇔V
--- = constant
I
his is known as Ohm’s Law.
here are two relevant statements to be made here. Firstly, Ohm’s Law is not really a law but rather an empirical statement of how materials behave. Many materials are non-ohmic and this law is only applicable to ohmic conductors. Secondly, the law should not be written as R = V / I as this statement deines resistance. he formula is commonly written as:
V = IR
where V is the potential diference across the resistor in volts V, I is the current in the resistor in amperes A and R is the resistance in ohms Ω. When written in this form R is understood to be independent of V.
As the current moves through the resistance of a device, it loses electric potential energy. he potential energy of a positive charge is less upon leaving the resistor than it was upon entering. We say that there is a potential drop across the device.
132
potential difference, V
Figure 510 Typical Ohm’s law graph.
5.1.9 OHMIC AND NON-OHMIC
CONDUCTORS
A graph of current versus the potential diference is a straight line (see Figure 510). Devices that obey the linear relationship of the graph are said to be ‘ohmic devices’ or ‘ohmic conductors’. here are very few devices that are ohmic although some metals can be if there is no temperature increase due to the heating efect of the current. However, many useful devices obey the law at least over a reasonable range. Remember that a resistor is any device with a potential diference and is not only restricted to the typical resistor used in the laboratory. It could be any useful electrical device – a ilament lamp or globe, a diode, a thermistor. Figure 511 shows some typical V versus I graphs for some conductors. When a device does not obey Ohm’s Law, it is said to be non-ohmic. he ilament lamp is non-ohmic because Ohm’s Law requires that the temperature remains constant, and this is not the case for a globe operating.
ELECTRIC CURRENTS
5.1.10 POWER DISSIPATION IN
Example
Solution
We can transpose the formula V = IR to obtain R = V / I .
Electric power is the rate at which energy is supplied to or used by a device. It is measured in J s-1 called watts W.
When a steady current is lowing through a load such as a resistor, it dissipates energy in it. his energy is equal to the potential energy lost by the charge as it moves through the potential diference that exists between the terminals of the load.
If a vacuum cleaner has a power rating of 500 W, it means it is converting electrical energy to mechanical, sound and heat energy at the rate of 500 J s-1. A 60 W light globe converts electrical energy to light and heat energy at the rate of 60 J s -1.
So that
120 V
R = --------------- = 20 Ω
6.0 A
he resistance of the iron is 2.0 × 101 Ω.
Some common power ratings of household appliances are given in Table 512.
I
a.
b. I
c. I
V
V
V
d. I
f. I
e. I
V
g.
I
V
V
V
KEY: a ohmic conductor b. gas discharge tube
c. thermionic diode d. junction diode e. ilament lamp f. thermistor
g. dilute sulfuric acid
Figure 511 Current/voltage relationships for some devices.
133
CORE
RESISTORS
An iron draws 6.0 A of current when operating in a country with a mains supply of 120 V. Calculate the resistance of the iron?
CORE
CHAPTER 5
Appliance
Blow heater
Kettle
Toaster
Iron
Vacuum cleaner
Television
Power rating
2 kW
1.5 kW
1.2 kW
850 W
1.2 kW
250 W
power is used for one hour. he consumer has to pay a certain cost per kilowatt-hour say 14 cents per kW h.
he heating efect of a current was investigated in 1841 by James Joule. He was able to demonstrate that by supplying electrical energy to a high resistance coil of wire this energy could be converted to thermal energy.
V × I × t = m × c × ∆T
Figure 512 Some possible power ratings
of household appliances.(table)
Example
Deriving expressions for
determining power
An electrical appliance is rated as 2.5 kW, 240 V.
We start with the basic deinition of power, P = W / t, then, we introduce the diferent expressions that we have already developed.
(a) (b) Using the result, W = qV and q = It ⇔ t = q / I , we have
qV
P = -------- = IV
 q---
 I
Determine the current needed for it to operate.
Calculate the energy it would consume in 2.0 hours.
Solution
Given that P = 2.5 × 103 and V = 240 V, we
use the formula, P = IV. Now,
(a)
he same result could have been obtained using the fact that the electrical energy,
W = q V = ItV
3
P
2.5 × 10
P = IV ⇒ I = --- = ---------------------- = 10.4
V
240
the current drawn is 1.0 × 101 A.
so that
(b)
ItV
W
P = ----- = ------- = IV
t
t
Next, we use the formula W = VIt, so that
3
W = ( 240 V ) × ( 10.4 A ) × ( 7.2 × 10 s )
7
= 1.8 × 10 J
However, we also have that V = IR , so that
he energy consumed is 1.8 × 107J.
2
P = I × IR = I R
We also could have used I = V / R , giving the result,
Example
2
V
P = IV = V
--- × V = -----R
R
Summary
2
W 2
P = ----- = I R = V
------ = VI
t
R
he commercial unit of electrical energy is the kilowatthour (kW h). It is the energy consumed when 1 kW of 134
A 2.5 kW blow heater is used for eight hours. Calculate the cost if electricity is sold at 12 cents per kilowatt-hour?
ELECTRIC CURRENTS
Using the fact that energy consumed (E) = power × time,
we have
Exercise
1. E = ( 2.5 kW ) × ( 8 h ) = 20 kW h
When a current is lowing through a wire attached to a dry cell
A. herefore, the Cost = (20 kW h) × $0.12 per kW h = $2.40
B. he cost to run the heater is two dollars forty cents ($2.40).
C. D. Example
2. A 1.2 kW electric water heater that is made of aluminium is used to heat 2.5 L of water from 25 °C to the boil. If the mass of aluminium water heater is 350 g and all the electrical energy is converted to heat energy, determine the time in minutes to bring the water to the boil.
3. Solution
A 1.2 kW heater delivers 1200 joules each second. 2.5 L =
2.5 kg
Q = mcΔTheater + mcΔTwater
4. = (0.35 kg × 9.1 × 102 J kg-1K-1 × 75 °C) +
(2.5kg × 4180 J kg-1K-1 × 75 °C)
Now if the heater delivers 1200 J per second, then to consume
8.08 × 105 J will take:
8.08 × 105 J
1200 J
= 673 s
60
= 11.2 min = 11 min
5.
5.00 × 102 W
2.00 × 102 W
2.00 × 103 W
1.80 × 105 W
If the power developed in an electric circuit is doubled, the energy used in one second is
A. B. C. D. = 8.08 × 105 J
15 nC
15 mC
15 µC
1.5 × 104 µC
An electric heater raises the temperature of a measured quantity of water. 6.00 × 103 J of energy is absorbed by the water in 30.0 seconds. What is the minimum power rating of the heater?
A. B. C. D. he electrical energy is used to heat the heater and the water
from 25 °C to 100 °C.
positive charges low from negative to positive terminal
positive charges low from positive to negative terminal
negative charges low from negative to positive terminal
negative charges low from positive to negative terminal
he total charge passing the same point in a conductor during 2.0 µs of an electric current of 7.5 mA is
A. B. C. D. (Speciic heat capacity of aluminium is 9.1 × 102 J kg-1K-1)
5.1
CORE
Solution
halved
doubled
quartered
quadrupled
he electron volt is deined as
A. B. C. D. the energy acquired by an electron when it passes through a potential diference of 1.0 V.
the voltage of an electron.
a fraction of the ionisation of an electron.
unit of energy exactly equal to 1.6 × 10–19 J.
135
CHAPTER 5
6. he deinition of the unit of current, the ampere, is based on
A. CORE
B. C. D. 7. Identify the charge-carriers in
a. b. c. 8. the charge per unit time delivered by an emf of 1.0 V.
the force per unit length on parallel current-carrying wires.
the force per unit length on a conductor in a magnetic ield
the charge passing a point per unit time.
b. Explain why is it that the electrons cannot travel faster in the conductor?
Explain why the electron drit produces heat?
9. Calculate the resistance of a wire if 0.5 V across it causes a current of 2.5 A to low.
10. Calculate current low through a 20 MΩ resistor connected across a 100 kV power supply.
11. A thin copper wire 200 cm in length has a 9 V dry cell connected between its ends. Determine the voltage drop that occurs along 30 cm of this wire.
12. Determine the length of tungsten wire with a diameter of 1.0 mm that is used to make a 20.0 Ω resistor.
13. A nichrome wire has a diameter of 0.40 mm. Calculate the length of this wire needed to carry a current of 30 mA when there is a potential diference of 12 V across it.
14. Explain in terms of atomic and electron movement, why resistance increases with temperature.
15. Determine how many coulombs there are when 2.0 A lows for 2.0 hours?
16. Distinguish the diference between an ohmic and a non–ohmic material.
136
Copy out and complete the table
Power p.d Current Fuse rating needed
(Watt) (Volt) (Ampere) (3,5,10,13 A)
Digital clock 4
240
Television 200
240
110 5
Hair dryer
Iron
230 4
Kettle
240 10
Appliance
18. Calculate the cost of heating the water to wash the dishes if the sink is 48 cm long, 25 cm wide and the water is 25 cm high. he tap water is at 14 °C and the inal temperature before washing up is 62 °C. Power is sold to the consumer at 14 cents per kilowatt-hour.
19. he element of an electric jug has a resistance of 60 Ω and draws a current of 3.0 A. Determine by how much the temperature of 5.0 kg of water will rise if it is on for 6 minutes.
20. Calculate the cost to heat 200 kg of water from 12°C to its boiling point if power costs 14 cents per kilowatt-hour.
21. Determine the work done in moving a charge of 10.0 nC through a potential diference of 1.50 × 102 V?
22. An electron in an electron gun of a picture tube is accelerated by a potential 2.5 × 103 V. Calculate the kinetic energy gained by the electron in electronvolts.
a length of copper wire.
an aqueous solution of sodium chloride.
the atmosphere during a lightning storm.
he speed with which electrons move through a copper wire is typically 10-4 m s-1.
a. 17. ELECTRIC CURRENTS
5.2.1 Define electromotive force (emf).
5.2.2 Describe the concept of internal resistance.
5.2.3 Apply the equations for resistors in series
and in parallel.
5.2.4 Draw circuit diagrams.
5.2.5 Describe the use of ideal ammeters and
ideal voltmeters.
5.2.6 Describe a potential divider.
5.2.7 Explain the use of sensors in potential
divider circuits.
5.2.8 Solve problems involving electric circuits.
© IBO 2007
5.2.1 ELECTROMOTIVE FORCE
In the previous section, potential diference was deined in terms of the amount of work that has to be done per unit charge to move a positively-charged body in an electric ield.
W
∆V = ∆
-------∆q
or
∆W = ( ∆q ) ∆V
Note: It is common place to use the expression V = W / It
as opposed to ΔV = ΔW / IΔt and q = It as opposed to
q = IΔt. hat is, to leave out the ‘change in’ (or ‘Δ’) notation
when solving problems.
he terms emf, potential diference and voltage are commonly interchanged in talking about electricity. he term electromotive force should not be used as it is not a force at all but rather a potential energy diference. Again, there are historical reasons for the use of the term in the irst place.
In the true sense, electromotive force (emf) is the work
per unit charge made available by an electrical source. For the simple circuit in Figure 507 , the emf of the dry cell is 1.5 V. However, when a voltmeter is used to test the potential diference across the resistor the reading on the voltmeter is found to be less than 1.5 V. his is due to the cell not being an ‘ideal dry cell’ because it has some internal resistance.
For the moment, let us say that emf is the energy supplied per unit charge, and potential diference is the energy
released (dissipated) per unit charge.
Example
A battery supplies 15.0 J of energy to 4.00 C of charge passing through it. Determine the emf of the battery
From the present deinition of electric current, we have,
∆q
I = ----∆t
∆q = I∆t
or
By substituting for q in the top equation, we end up with
∆W = I × ∆t × ∆V
or
∆W
∆V = ------I∆t
Solution
Using the formula, V = W / q, we have, V = 15.0J / 4.00 C
= 3.75 V
he emf of the battery is 3.75 V.
herefore it can be stated that
‘Potential diference in external circuits is the power, (F), dissipated (released) per unit current’
Either deinition for potential diference is acceptable. However, the above-mentioned ties voltage and current together.
137
CORE
5.2 ELECTRIC CIRCUITS
CORE
CHAPTER 5
The major sources of εmf are:
IR is known as the terminal voltage.
1. ELECTROMAGNETIC
when a coil of wire is rotated in a magnetic ield, an induced current is produced. Power stations use generators to produce a current.
hat is,
2. CHEMICAL
oxidation-reduction reactions transfer electrons between chemicals. Dry cells, fuel cells and batteries are examples.
3. PHOTOELECTRIC EFFECT
electrons are emitted from certain metal surfaces when high frequency light is shone on their surfaces. hese photocells are used in watches, clocks, automatic doors.
4. PIEZOELECTRIC EFFECT
certain crystals can produce a charge on one side when placed under stress. If one side of the crystal is charged and the other not, a potential diference exists across the crystal. his is used in crystal microphones.
5. THERMOELECTRIC EFFECT
when two pieces of certain metals are wound together and one end is heated while the other end is cooled, a current is produced. hermocouples can be used as temperaturemeasuring devices.
5.2.2 TERMINAL VOLTAGE, EMF AND
IR = emf – Ir
All sources of emf have some internal resistance and it is a factor in determining how useful a source of emf is.
A dry cell being a primary cell releases its energy and becomes “dead”. When it is just about inoperable, the build–up of the products of the relevant oxidationreduction reaction increases the internal resistance beyond its normal value. he emf available does not decrease from giving the maximum number of joules per coulomb of charge. However, when the emf is used up in overcoming the potential diference across the internal resistor, the terminal voltage will drop to zero. Batteries are generally regarded as being a secondary cell because they can be recharged many times.
If a series of potential diference readings measured with a voltmeter across a variable resistor R are taken concurrently with the current lowing in the circuit measured with an ammeter for each variable resistance, and a graph of voltage versus current is plotted, from the relationship
Terminal voltage = V = emf –Ir
he y–intercept of the graph will be the value of the emf, and the negative gradient is the internal resistance.
INTERNAL RESISTANCE
When electrons low around a circuit, they gain potential energy in the cell and then lose the energy in the resistors. In a previous section, it was stated that the emf is the energy supplied per unit charge, and the potential diference is the energy released (dissipated) per unit charge in a circuit.
V1
Example
A dry cell has an internal resistance of 1.50 Ω. A resistor of 12.0 Ω is connected in series with the dry cell. If the potential diference across the 12.0 Ω resistor is 1.20 V, calculate the emf of the cell.
RΩ
I
Solution
ε .m .f
V2
+– rΩ
internal resistance of cell
We irst need to determine the current lowing through the
system. his is done by using the formula, V = IR from
where we obtain, I = V / R.
Figure 514 Internal resistance circuit.
herefore, we have that
In the circuit in Figure 514, the total energy supplied is determined by the value of the emf. he total energy released is equal to the potential diference across resistor R plus the potential diference of the internal resistance r.
emf = V1 + V2 = IR + Ir = I ( R + r )
138
ELECTRIC CURRENTS
I1
V
1.20 V
I = --- = ----------------- = 0.100 A
R
12.0 Ω
I
I2
I3
CORE
Next, we use the formula, emf = I(R + r), where the internal
resistance, r = 1.5 Ω.
hat is, emf = I(R + r) = 0.100 A × (12.0 Ω + 1.50 Ω)
Figure 515
Kirchoff’s current law.
= 1.35 V
In Figure 515, we have that I = I1 + I2 + I3 + …..
he emf of the cell is 1.35 V.
his is based on the Law of Conservation of charge.
5.2.3 D.C. CIRCUIT ANALYSIS
Electricity is transmitted to household localities using alternating current ac as the voltage can be stepped-down to the local supply voltage before entering houses. Some household appliances such as lights, stoves, water heaters, toasters, irons operate using ac. However, electronic devices such as stereos, computers and televisions operate using direct current dc. he power supply is stepped–
down to say 12 V by a transformer inside the appliance, and a diode is inserted (diodes are devices that will only allow current to low in one direction). Direct current is produced.
Kirchof ’s voltage law – loop rule;
‘In a closed loop, the sum of the emfs equals the sum of the potential drops’.
In the Figure 516, we have that V = V1 + V2 + V3 + …..
V1
V2
V3
R1
R2
R3
I
V
An electric circuit has three essential components
Figure 516 Kirchoff’s voltage law.
1. A source of emf.
2. A conducting pathway obtained by conducting wires or some alternative.
his is a statement of the law of conservation of energy. ‘Energy supplied equals the energy released in this closed path’.
3. A load to consume energy such as a ilament globe, other resistors and electronic components.
Series circuits
A current will always take the shortest path in the circuit. If for some reason the current inds a way back to the source without passing through the essential components, a short-circuit occurs. Fuses are used in appliances to stop damage due to short-circuits.
In the mid-nineteenth century, G.R. Kirchof (1824-1887) stated two simple rules using the laws of conservation of energy and charge to help in the analysis of direct current circuits. hese rules are
In a series circuit:
1. All the components have only one current pathway.
2. All components have the same current through them.
3. he sum of the potential drop across each component is equal to the emf of the cell.
Kirchof ’s current law–junction rule states that;
‘he sum of the currents lowing into a point in a circuit equals the sum of the currents lowing out at that point’.
139
CHAPTER 5
So that in this instance, we have that
I = 4V / 8Ω = 0.5 A
he current lowing is 0.5 A.
he potential drop across each resistor, Ri , is
given by, Vi = IRi .
So that for the 6 ohm resistor, the potential
diference (or potential drop) is given by
V1 = IR1 = 0.5 × 6 = 3 V.
Similarly, for the 2 ohm resistor, the potential
diference (or potential drop) is given by V2 =
IR2 = 0.5 × 2 = 1 V.
From Kirchof ’s laws, we have that
I = I1 + I2 + I3 + … and V = V1 + V2 + V3 + …..
(c)
V1 V2 V3
V1 + V2 + V3 + …
V
R = --- = -------------------------------------------- = ------ + ------ + ------ + …
I
I
I
I
I
But, the terms in this sum are equal to R1, R2, R3, … (i.e., R1 = V1 / I, R2 = V2 / I…).
So that,
he potential diferences in the 6 Ω and the 2 Ω resistors are
3 V and 1 V respectively.
R eff = R 1 + R 2 + R 3 + …
he total or efective resistance Ref of a series circuit is equal to the sum of the separate resistances.
Example
From the diagram given in the Figure below of a potential divider, calculate
(a) (b) (c) V1
V2
6W
2W
4
Potential
CORE
hen, from Ohm’s law
3
2
DV = 3
1
1
DV 2 = 1
A
the efective resistance of the circuit
the current lowing
the potential diference across each resistor
displacement (from A)
B
Figure 518 Potential – displacement graph.
V1
V2
Figure 518 shows the corresponding potential/displacement
graph for the circuit in the example above.
6Ω
2Ω
Notice how we refer to the potential diference, ∆ V{F} as
opposed to simply V {F}
I
4V
Parallel circuits
In the parallel circuit in Figure 519
Solution
(a)
here is more than one current pathway.
140
All components have the same potential diference across them.
2. he sum of the currents lowing into any point is equal to the sum of the currents lowing out at that point.
Using our rule (from above), we have that
the efective resistance is given by
R eff = R 1 + R 2 = 6 Ω + 2 Ω = 8 Ω
(b)
1. hat is, the efective resistance is 8 Ω.
We can determine the current by making use
of the formula, I = V / Ref.
ELECTRIC CURRENTS
V
Example
R1
I2
V
R2
I3
V
R3
I
From the diagram given in the Figure below, calculate
(a) (a) (a) V
Rn
In
the efective resistance of the circuit,
the current lowing in the main circuit,
the current in each resistor.
I1
+
Figure 519
V
–
CORE
I1
3Ω
I
I2
A parallel circuit
6Ω
From Kirchof ’s laws, we have that
I = I 1 + I2 + I 3 + …
and
V = V1 = V 2 = V3 = …
6V
Solution
From Ohm’s law, we have V = IR ⇔ V / R = I or 1 / R = I / V
Meaning that,
(a) I1 + I2 + I3 + …
I1 I2 I3
1
--------- = --------------------------------------- = ---- + ---- + ---- + …
V V V
R eff
V
1
1 1
--------- = --- + --- = 3--- ⇒ R eff = 2
R eff
3 6
6
However, V = V1 = V2 = V3 =…., so that
I1 I2 I3
1
1
1
1
--------- = ------ + ------ + ------ + … = ----- + ------ + ----- + …
R1 R 2 R3
V1 V2 V3
R eff
Using 1 / Ref = 1 / R1 + 1 / R2 , we have,
he efective resistance is 2 Ω.
(b)
We can now determine the current lowing
through the system:
V
V = IR eff ⇒ I = --------R eff
he efective resistance Ref of a parallel circuit is less than the sum of the separate resistances.
6V
I = --------- = 3 A
2Ω
he current lowing in the main circuit is 3 A.
(c)
Using Vi = IiRi , we have:
For the 3 ohm resistor; I = 6 / 3 = 2A
For the 6 ohm resistor; I = 6 / 6 = 1A
he currents in the 3 Ω and the 6 Ω resistors
are 2 A and 1 A respectively.
141
CHAPTER 5
RESISTORS IN COMPOUND
(c)
he current in the 4 Ω resistor is 2 A.
he potential diference across the 4 Ω
resistor = IR = (2 A) . (4 Ω) = 8 V
he potential diference across the parallel
network = 12 V – 8 V
=4V
he current in the network resistors is given by,
CIRCUITS
CORE
Example
V V
I = I1 + I2 = ------ + -----R 1 R2
From the diagram given in the igure (below), calculate
I1
4Ω
3Ω
I
I
I2
I
I = 4--- A + 4--- A = 1.33 A + 0.67 A
3
6
6Ω
12 V
he current in the 4 Ω resistor is 2 A. he currents in the 3 Ω
and the 6 Ω resistors are 1.33 A and 0.67 A respectively.
5.2.4 ELECTRIC CIRCUIT SYMBOLS
(a) (b) (c) the total resistance of the circuit
the current lowing in the main circuit
the current in each resistor
Solution
(a)
For the parallel circuit we have,
1 1
1
--------- = --- + --- = 3--- ⇒ R eff = 2
3 6
R eff
6
So that , Ref = 2Ω
hus the total resistance (of the parallel and series arms),
RT, is given by
RT
= 2Ω
+
he circuit symbols shown in Table 522 are used for drawing circuit diagrams in this course. his standard set of symbols is also provided in the data booklet that is used during examination sessions. You should use these symbols in place of any other that you may have learnt.
5.2.5 AMMETERS AND VOLTMETERS
Galvanometers are used to detect electric currents. hey use a property of electromagnetism – a coil with a current lowing in it experiences a force when placed in a magnetic ield. Most non-digital ammeters and voltmeters consist of a moving coil galvanometer connected to resistors. Digital meters are becoming more common, and the digital multimeter can act as an ammeter, a voltmeter or an ohmmeter.
4Ω
A voltmeter
= 6Ω
1. is always connected across a device (in parallel).
2. has a very high resistance so that it takes very little current from the device whose potential diference is being measured.
3. has a high resistor (a multiplier) connected in series with a galvanometer.
4. an ideal voltmeter would have ininite resistance with no current passing through it and no energy would be dissipated in it.
hat is, the efective resistance is 6 Ω.
(b)
To determine the current, once again we use,
V = IR, so that,
V
12
I = ------ = ----- = 2 A
RT
6
he current lowing in the main circuit is 2 A.
142
ELECTRIC CURRENTS
wires crossing (not joined)
cell
battery
lamp
a.c. supply
switch
ammeter
voltmeter
A
V
galvanometer
resistor
variable resistor
potentiometer
heating elemen t
fuse
transformer
oscilloscope
CORE
joined wires
Table 522 Common circuit symbols
Example 1
A galvanometer has a resistance of 1.0 × 103 Ω (mainly due to the resistace of the coil), and gives a full-scale delection (fsd) for 1.0 mA of current. Calculate the size of a multiplier resistor that would be needed to convert this to a voltmeter with a fsd of 10.0 volts.
Solution
he Figure below demonstrates a possible set up.
I
RG
i
I
G
I -i
RS
143
CHAPTER 5
Because the resistors Rg and Rs are in series then the potential diference across the resistors will be 10 volts. herefore, i (Rg + Rs) = 10
CORE
0.001 × (1000 + Rs) = 10 In electronic systems, it is oten necessary to obtain smaller voltages from larger voltages for the various electronic circuits. a potential divider is a device that produces the required voltage for a component from a larger voltage. It consists of a series of resistors or a rheostat (variable resistor) connected in series in a circuit. A simple voltage divider is shown in Figure 525.
0.001Rs = 9
Rs = 9000 Ω
he multiplier resistor = 9.0 × 103 Ω
An ammeter
1. 5.2.6 A POTENTIAL DIVIDER
is always connected in series with a circuit.
2. V1
has a very low resistance compared with the resistance of the circuit so that it will not alter the current lowing in the circuit.
3. has a low resistor (a shunt) connected in parallel with a galvanometer.
4. would ideally have no resistance with no potential diference across it and no energy would be dissipated in it.
V
R
V2
Figure 525 A simple voltage divider
Using Ohm’s Law, V1 = IR1 and V = I(R1 + R2 ).
V1 / V = IR1 / I(R1 + R2). herefore,
Example
V1 = R1 / (R1 + R2) × V
his is known as the potential divider equation.
A galvanometer has a resistance of 1.0 × 103 Ω (mainly due to the resistace of the coil), and gives a full-scale delection (fsd) for 1.0 mA of current. Calculate the size of a shunt resistor that would be needed to convert this to a meter with a fsd of 5.0 A.
Example 1
In the potential divider shown below, calculate:
(a) (b) (c) Solution
the total current in the circuit
the potential diference across each resistor
the voltmeter reading if it was connected between terminals 2 and 6.
he Figure below demonstrates a possible set up.
I
i
RG
G
12 V
I
I -i
2Ω
2Ω
2Ω
2Ω
2Ω
2Ω
RS
Because the resistors Rg and Rs are in parallel then the potential diference across the resistors is the same. herefore, iRg = (I – i) Rs.
Rs = (i / I – i) Rg = [10-3 A / (5 – 0.001) A] × 1000 Ω = 0.20 Ω
he shunt resistor = 0.20 Ω
144
1
2
3
4
5
6
7
ELECTRIC CURRENTS
5.2.7 SENSORS IN POTENTIAL DIVIDER
Solution
(b)
(c)
he total resistance = 12 Ω. From Ohm’s law
V = IR and I = V / R
I = 12 V / 12 Ω = 1 A
he 12 V is equally shared by each 2 Ω
resistor. herefore, the potential diference
across each resistor = 2 V. Alternatively,
V = IR = 1 A × 2 Ω = 2 V
Between terminals 2 and 6 there are 4
resistors each of 2 Ω. herefore, the potential
diference the terminals
= 4 × 2 V = 8 V.
Because resistance is directly proportional to the length of a resistor, a variable resistor also known as a potentiometer or colloquially as a “pot” can also be used to determine the potential diference across an output transducer (device for converting energy from one form to another) such as a ilament lamp in Figure 527.
A number of sensors (input transducers) that rely on a change in resistance can be used in conjunction with potential dividers to allow for the transfer of energy from one form to another. hree such sensors are:
•
•
•
the light dependant resistor (LDR)
the negative temperature coeicient thermistor (NTC)
strain gauges.
A light dependent resistor (LDR) is a photo-condutive cell whose resistance changes with the intensity of the incident light. Typically, it contains a grid of interlocking electrodes made of gold deposited on glass over which is deposited a layer of the semiconductor, cadmium sulide. Its range of resistance is from over 10 MΩ in the dark to about 100 Ω in sunlight. A simple LDR and its circuit symbol are shown in Figure 528.
If the pointer was at A then the potential diference would be zero as there is no power dissipated per unit current in the potentiometer (the load), and there would be no output voltage. However, if the pointer is moved up to two-thirds the length of the potentiometer as in the igure, then the output voltage across the ilament lamp would be ⅔ × 6V = 4V.
Figure 528 A light dependent resistor
potentiometer
here are only two ways to construct a voltage divider with an LDR sensor – with the LDR at the top, or with the LDR at the bottom as shown in Figure 529.
6V
I
A
I
V
V in
Vi
Figure 527
V out
Laboratory potentiometer
Pots have a rotating wheel mounted in plastic and they are commonly used as volume and tone controls in sound systems. hey can be made from wire, metal oxides or carbon compounds.
V out
0V
Figure 529 (a) and (b)
0V
two ways to construct an L.D.R.
Light dependant resistors have many uses in electronic circuits including smoke detectors, burglar alarms, camera 145
CORE
CIRCUITS
(a)
CHAPTER 5
10
0000
Resistance Ω
Resistors that change resistance with temperature are called thermistors (derived from thermal resistors).
hey are made from ceramic materials containing a semiconductor the main types being bead and rod thermistors. he NTC (negative temperature coeicient) thermistor contains a mixture of iron, nickel and cobalt oxides with small amounts of other substances. hey may have a positive (PTC) or negative (NTC) temperature coeicient according to the equation:
50
20 40 60 80 100
Temperature / ˚C
Figure 531
A PTC thermistor
R f = R 0 ( 1 + αt)
where R0 equals the resistance at some reference temperature say 0 °C, Rf is the resistance at some temperature, t °C, above the reference temperature, and α is the temperature coeicient for the material being used.
200
Resistance Ω
CORE
light meters, camera aperture controls in automatic cameras and controls for switching street lights of and on.
he circuit symbol for a thermistor is shown in Figure 530.
20 40 60 80 100
Temperature / ˚C
Figure 532 An NTC thermistor
Figure 530 The circuit symbol for a thermistor
For a NTC thermistor, the resistance decreases when the temperature rises and therefore they can pass more current. his current could be used to to operate a galvonometer with a scale calibrated in degrees as used in electronic thermometers or car coolant system gauges. hermistors are also used in data-logging temperature probes but the analogue signal has to be converted to digital signal using an analogue to digital converter (ADC).
With normal resistors the resistance becomes higher when the temperature increases and they therefore have a small positive temperature coeicient. Figures 531 and 532 demonstrate how the resistance changes with temperature for both types of thermistors.
146
An electronic thermometer can be made using an NTC thermistor as shown in Figure 533.
200 Ω
milliammeter
with fsd
0 – 10 mA
Figure 533 An electronic thermometer
When a metal conducting wire is put under vertical strain, it will become longer and thinner and as a result its resistance will increase. An electrical strain gauge is a device that employs this principle. It can be used to obtain information about the size and distribution of strains in structures such as metal bridges and aircrat to name but two. A simple gauge consists of very ine parallel threads of a continous metal alloy wire cemented to a thin piece of paper that are hooked up to a resistance measuring device with thick connecting wires as shown in Figure 534. When it is securely attached on the metal to be tested, it ELECTRIC CURRENTS
will experience the same strain as the test metal and as this happens the strain gauge wire become longer and thinner and as such the resistance increases.
4. In the circuit below, a heater with resistance R is connected in series with a 48 V supply and a resistor S.
48 V
CORE
structure under strain
thin threads
of wire
If the potential diference across the heater is to be maintained at 12 V, the resistance of the resistor S should be
connecting lead
Figure 534 A resistance measuring device
A. B. C. D. 5.2.8 SOLVE PROBLEMS INVOLVING
ELECTRIC CIRCUITS
5. Many problems have been done with their worked examples. Now try to solve the following exercise.
Exercise
circuit X
In a television tube the picture is produced in a luorescent material at the front of the picture tube by
A. B. C. D. 3. 6. a larger potential diference is used to run it
its resistance wire ilament is longer
more electric current lows through it
it has a higher amount of inert gas in it
circuit Z
lowest → Y Z Z X X Y Z Y highest
X
Y
Z
X
he graph below shows the current/voltage characteristics of a ilament lamp.
24
an electrical discharge
a beam of positive ions
an electrolytic deposition of metal atoms
a stream of electrons
A 100 W light globe gives out a brighter light than a 60 W globe mainly because
A. B. C. D. A. B. C. D. 1 Ω
3 Ω
6 Ω
9 Ω
16
3
Current A
2. circuit Y
Which one of the following shows the resistances of the circuits in increasing order of magnitude?
hree identical resistors of 3 Ω are connected in parallel in a circuit. he efective resistance would be
A. B. C. D. R/2
R/4
R/3
3R
he diagrams below show circuits X, Y and Z of three resistors, each resistor having the same resistance.
5.2
1. R
S
thin paper
8
0
0
2
4
6
Voltage /V
8
10
147
CHAPTER 5
he resistance of the ilament at 3.0 V is
CORE
A. B. C. D. 7. 0.25 Ω
250 Ω
4000 Ω
8000 Ω
L1
3 : 2
2 : 3
1.5 : 1
2 : 9
1 Ω
3 Ω
6 Ω
9 Ω
L2
S
L3
When switch S is closed
A. B. hree identical resistors of 3 Ω are connected in parallel in a circuit. he efective resistance would be
A. B. C. D. 9. hree identical lamps L1, L2 and L3 are connected as shown in the following diagram.
Metal X has half the resistivity of metal Y and length three times that of Y. If both X and Y have the same surface area, the ratio of their resistance RX / RY is:
A. B. C. D. 8. 12. L1 and L3 brighten and L2 goes out
all three lamps glow with the same brightness
L2 brightens and L1 and L3 remain unchanged
L1 and L3 go dimmer and L2 goes out
C. D. 13. Two resistors are connected in parallel and have the currents I1 and I2 as shown in the diagram
If 18 J of work must be done to move 2.0 C of charge from point A to point B in an electric ield, the potential diference between points A and B is:
I1
A. B. C. D. I2
R1
R2
10. 0.1 V
9.0 V
12 V
20 V
he fundamental SI unit – the ampere – is deined in terms of:
A. B. C. D. potential diference and resistance
the time rate of change of charge in a circuit
the force acting between two current carrying wires
the product of charge and time
A. B. C. D. 14. 11. A 100 W light globe gives out a brighter light than a 60 W globe mainly because
A. B. C. D. 148
a larger potential diference is used to run it
its resistance wire ilament is longer
more electric current lows through it
it has a higher amount of inert gas in it
If the efective resistance of the circuit is R then
R I1/R1 R R = = = = (R1 + R2) / R1R2
I2 / R2
R1R2 / R1 + R2
I (R1 + R2) / R1R2
he following circuit was set up to determine the internal resistance r of a dry cell. he load resistor was varied from 100Ω to 150Ω, and the current in the circuit was measured using an ammeter. he two respective values of the current are given in the circuits.
ELECTRIC CURRENTS
2.2 mA
A. B. C. D. 1.5 mA
A
r
17. 150 Ω
1.5 mA
A voltmeter connected between the points X and Y should read:
Determine the equivalent resistance when 12 Ω, 6 Ω and 4 Ω are placed in
(a) (b) A
0 V
3 V
6 V
9 V
series
parallel
18. Calculate the work done in moving a 12.0 µC through a p.d of 240 V.
19. he diagram shows resistances joined in a compound circuit.
r
he internal resistance r of the dry cell is
I1
A. B. C. D. 15. 3.0 Ω
7.1 Ω
58.8Ω
250 Ω
2Ω
I
20 Ω
I
I2
4Ω
12.2 V
his question concerns the electric circuit below
(a) (b) I1
2Ω
(c) 4Ω
(d) (e) A. B. C. D. 2 A
4 A
6 A
9 A
A photo-electric cell draws a current of 0.12 A when driving a small load of resistance 2 Ω. If the emf of the cell is 0.8 V, determine the internal resistance of it under these conditions.
21. In terms of emf and internal resistance, explain why is it possible to re-charge a nickel-cadmium cell while normal dry cells have to be discarded once they are lat.
22. When a dry cell is connected to a circuit with a load resistor of 4.0 Ω, there is a terminal voltage of 1.3 V. When the load resistor is changed to 12 Ω, the terminal voltage is found to be 1.45 V. Calculate
Consider the circuit below that contains a 15 V battery with zero internal resistance.
1.5 V
4Ω
X
Y
4Ω
Determine the total resistance of the circuit.
Calculate current lows through the 2.0 Ω resistor.
Deduce the potential diference across the 20.0 Ω resistor.
Determine is the potential diference across the 6.0 Ω resistor.
Calculate is the current through the 4.0 Ω resistor.
20. he value of current I1 is
16. 6Ω
(a) (b) the emf of the cell.
the internal resistance of the cell.
149
CORE
100 Ω
CHAPTER 5
23. CORE
24. 25. An electrical appliance is rated as 2.5 kW, 240 V
(b) 27. (b) (c) An iron draws 6.0 A of current when operating in a country with a mains-supply of 240 V. Calculate the resistance of the iron.
(a) 26. (a) Calculate the current lowing through a hair drier if it takes 2.40 × 103 C of charge to dry a person’s hair in 3.0 minutes.
Calculate the current it needs to draw in order to operate.
Determine how much energy would be consumed in 2 hours.
A 2.5 kW blow heater is used for eight hours. Calculate the cost of running the blow heater if the electricity is sold at 15 cents per kilowatt-hour.
(d) Complete the last column for the inverse of the current giving the correct unit.
Plot a graph of R against 1 / I
Describe the relationship that exists between the resistance and the current.
Determine the electromotive force of the dry cell
30. Starting from the laws of conservation of energy and conservation of charge, derive a formula for calculating the efective resistance of two resistors in parallel.
31. he diagram shows a typical circuit.
1.0 Ω
B
2.0 Ω
A
C
0.5 Ω
1.0 Ω
D
he circuit below refers to the following questions:
24.0 Ω
1.0 Ω
4.0 Ω
R
1.5 V
V
I
3.0 Ω
1.0 Ω
100.0 V
(a) 35 A
(b) (i) (ii) Determine the current lowing through R, and the value of resistor R.
Deduce the reading on the voltmeter V.
(c) (d) 28. (i) (ii) 29. Describe the meaning of the « 12 V » on a 12 volt car battery.
A 14 V car battery drops to 12 V when supplying a current of 5.0 A. Determine the internal resistance of the battery.
A circuit was set up to investigate the relationship between the current I through a resistor and the magnitude of the resistance R while a constant electromotive force was supplied by a dry cell. he results of the investigation are given in the following table:
R ± 0.5Ω
2.0
6.0
12
16
18
150
I ± 0.1 A
5.0
1.7
0.83
0.63
0.56
32. Determine the efective resistance of the whole circuit.
Determine the currents lowing in each network resistor.
Determine the potential diferences VAB and VAD.
Determine the potential diference between B and D.
Determine the resistance of the LDR in the diagram below if a current of 4.5 mA is lowing in the circuit.
1 kΩ
9V