26770932 Matriculation Physics Electric Current and Direct Current Circuit

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PHYSICS CHAPTER 5

CHAPTER 5: CHAPTER 5: Electric current and Electric current and

direct-current circuitsdirect-current circuits(7 Hours)(7 Hours)

PHYSICS CHAPTER 5

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DescribeDescribe microscopic model of current. microscopic model of current. Define and useDefine and use electric current formulae, electric current formulae,

Learning Outcome:

5.1 Electrical conduction (1 hour)

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dt

dQI

PHYSICS CHAPTER 5

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5.1.1 Electric current, I Consider a simple closed circuit consists of wires, a battery and

a light bulb as shown in Figure 5.1.

5.1 Electrical conduction

Area, A

eF

E

I

Figure 5.1Figure 5.1

PHYSICS CHAPTER 5

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From the Figure 5.1, Direction of electric field or electric current :

Positive to negative terminalPositive to negative terminal Direction of electron flows :

Negative to positive terminalNegative to positive terminal The electron accelerateselectron accelerates because of the electric forceelectric force

acted on it. is defined as the total (nett) charge, the total (nett) charge, QQ flowing through the flowing through the

area per unit time, area per unit time, tt.Mathematically,

t

QI

dt

dQI

OR

instantaneous currentinstantaneous current

average currentaverage current

PHYSICS CHAPTER 5

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It is a base and scalarbase and scalar quantities. The S.I. unitS.I. unit of the electric current is the ampereampere (AA). Its dimension is given by

1 ampere1 ampere of current is defined as one coulomb of charge one coulomb of charge passing through the surface area in one secondpassing through the surface area in one second.OR

AI

1s C 1second 1

coulomb 1ampere 1

Note:Note:

If the charge move around a circuit in the same direction charge move around a circuit in the same direction at all timesat all times, the current is called direct current (dc)direct current (dc), which is produced by the batteryproduced by the battery.

PHYSICS CHAPTER 5

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is defined as the current flowing through a conductor per the current flowing through a conductor per unit cross-sectional areaunit cross-sectional area.Mathematically,

It is a vector quantityvector quantity. Its unit is ampere per squared metreampere per squared metre (A mA m22) The direction of current density, direction of current density, JJ always in the same same

direction of the current direction of the current II. e.g. in Figure 5.2.

5.1.2 Current density, J

A

IJ

where current electric :Iconductor theof area sectional-cross :A

I

maxJ

0J

Area, A


PHYSICS CHAPTER 5

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In metal the charge carrier is free electronscharge carrier is free electrons and a lot of free lot of free electrons are availableelectrons are available in it.

They move freely and randomlymove freely and randomly throughout the crystal lattice structure of the metal but frequently interact with the lattices.

When the electric field is applied to the metalelectric field is applied to the metal, the freely freely moving electron experience an electric forcemoving electron experience an electric force and tend to drift drift with constant average velocity ( constant average velocity (called drift velocity) drift velocity) towards a direction opposite to the direction of the fieldtowards a direction opposite to the direction of the field as shown in Figure 5.3.

Then the electric current is flowingelectric current is flowing in the opposite direction opposite direction of the electron flowsof the electron flows.

5.1.3 Electrical conduction in metal

E

I

dv

dv


Note:Note:

The magnitude of the magnitude of the drift velocity is much drift velocity is much smaller than the smaller than the random velocities of random velocities of the free electron.the free electron.

PHYSICS CHAPTER 5

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Consider a metal rod of length L and cross-sectional area A, which is applied to the electric field as shown in Figures 5.4.

Suppose there are n free electrons (charge carrier) per unit volume in the metal rod, thus the number of free electron, N is given by

5.1.4 Drift velocity of charges, vd

E

J

I

dv

dv

L

A


V

Nn ALV and

AL

Nn nALN

PHYSICS CHAPTER 5

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The total charge Q of the free electrons that pass through the area A along the rod is

The time required for the electron moving along the rod is

Since

NeQ enALQ

t

Lv d

dv

Lt

then the drift velocity vd is given byt

QI

d

d

nAev

v

L

enALI

nAe

Iv d

JA

Iand

OR

where

electron theof charge :e

DefinitionDefinition

Density of the Density of the free electronfree electron

ne

Jv d

electron free ofnumber :neunit volumper carrier) (charge

PHYSICS CHAPTER 5

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A silver wire carries a current of 3.0 A. Determine

a. the number of electrons per second pass through the wire,

b. the amount of charge flows through a cross-sectional area of the

wire in 55 s.

(Given charge of electron, e= 1.60 1019 C)

Solution :Solution :

a. By applying the equation of average current, thus

b. Given , thus the amount of charge flows is given by

Example 1 :

A 0.3I

t

QI

t

N 191060.10.3

119 s electrons 1088.1 t

N

and NeQ

t

NeI

s 55tItQ 550.3Q C 165Q

PHYSICS CHAPTER 5

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A copper wire of radius 900 m carries a current of 17 mA. The wire contains 8.49 1028 free electrons per cubic meter. Determine

a. the magnitude of the drift velocity in the wire,

b. the current density in the wire.

(Given charge of electron, e= 1.60 1019 C)


a. By applying the equation of the drift velocity, thus

b. The current density is given by

Example 2 :

32836 m 1049.8A; 1017m; 10900 nIr

nAe

Iv d

192628

3

d1060.1109001049.8

1017

πv

17d s m 1092.4 v

and 2πrA

ern

Iv

2d

2πr

IJ

23

26

3

mA 1068.610900

1017

πJ

PHYSICS CHAPTER 5

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A high voltage transmission line with a diameter of 3.00 cm and a length of 100 km carries a steady current of 1500 A. If the conductor is copper wire with a free charge density of 8.49 1028 electrons m-3, calculate the time taken by one electron to travel the full length of the

line. (Given charge of electron, e= 1.60 1019 C)


By applying the equation of the drift velocity, thus

Therefore the time taken by one electron to travel the line is

Example 3 :

nAe

Iv d

192228d

1060.11000.31049.8

15004

π

v

14d s m 1056.1 v

and4

2πdA

edn

Iv

2d

4

dv

Lt s 1041.6

1056.1

10100 84

3

t

A; 1500m; 10 100m; 1000.3 32 ILd328 m 1049.8 n

PHYSICS CHAPTER 5

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Explain how electrical devices can begin operating almost immediately after you switch on, even though the individual electrons in the wire may take hours to reach the device.


Example 4 :

Each electron in the wire affects its neighbours by exerting Each electron in the wire affects its neighbours by exerting a force on them, causing them to move.a force on them, causing them to move.

When electrons begin to move out of a battery or source When electrons begin to move out of a battery or source their motion sets up a propagating influence that moves their motion sets up a propagating influence that moves through the wire at nearly the speed of light, causing through the wire at nearly the speed of light, causing electrons everywhere in the wire begin to move.electrons everywhere in the wire begin to move.

PHYSICS CHAPTER 5

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and use resistivity formulae,Define and use resistivity formulae,

State Ohm’s law.State Ohm’s law. Apply formulae,Apply formulae,

Learning Outcome:

5.2 Resistivity and Ohm’s law (1 hour)

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IRV

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RAρ

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5.2.1 Resistance, R is defined as a ratio of the potential difference across an a ratio of the potential difference across an

electrical component to the current passing through it.electrical component to the current passing through it.Mathematically,

It is a measure of the component’s opposition to the flow of measure of the component’s opposition to the flow of the electric chargethe electric charge.

It is a scalarscalar quantity and its unit is ohm ohm ( ) or V AV A11

In general, the resistance of a metallic conductor increases resistance of a metallic conductor increases with temperaturewith temperature.

5.2 Resistivity and Ohm’s law

I

VR

where (voltage) difference potential :Vcurrent :I

Note:Note:

If the temperaturetemperature of the metallic conductor is constantconstant hence its resistanceresistance also constantconstant.

(5.1)(5.1)

PHYSICS CHAPTER 5

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Resistivity, Resistivity, is defined as the resistance of a unit cross-sectional area the resistance of a unit cross-sectional area

per unit length of the materialper unit length of the material.Mathematically,

It is a scalara scalar quantity and its unit is ohm meterohm meter ( m m) It is a measure of a material’s ability to oppose the flow of measure of a material’s ability to oppose the flow of

an electric currentan electric current. It also known as specific resistancespecific resistance. Resistivity depends on the type of the materialtype of the material and on the

temperaturetemperature. A good electric conductorsconductors have a very low resistivitieslow resistivities and

good insulators insulators have very high resistivitieshigh resistivities.

5.2.2 Resistivity and conductivity

l

RAρ

where material theoflength :l area sectional-cross :A

(5.2)(5.2)

PHYSICS CHAPTER 5

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From the eq. (5.2), the resistanceresistance of a conductor depends on the length and cross-sectional arealength and cross-sectional area.

Table 5.1 shows the resistivity for various materials at 20 C.

Conductivity, Conductivity, is defined as the reciprocal of the resistivity of a material.the reciprocal of the resistivity of a material.

Mathematically,

It is a scalar quantityscalar quantity and its unit is 1 1 mm11.

Material Resistivity, ( m)Silver 1.59 108

Copper 1.68 108

Aluminum 2.82 108

Gold 2.44 108

Glass 10101014

Table 5.1Table 5.1

ρσ

1 (5.3)(5.3)

PHYSICS CHAPTER 5

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Two wires P and Q with circular cross section are made of the same metal and have equal length. If the resistance of wire P is three times greater than that of wire Q, determine the ratio of their diameters.


Given

Example 5 :

lllρρρ QPQP ;

QP 3RR andA

ρlR

3P

Q d

d

Q

QQ

P

PP 3A

lρ

A

lρ and

4

2πdA

2Q

2P

43

4

πd

ρl

πd

ρl

OR3

1

Q

P d

d

PHYSICS CHAPTER 5

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When a potential difference of 240 V is applied across a wire that is 200 cm long and has a 0.50 mm radius, the current density is 7.14 109 A m2. Calculatea. the resistivity of the wire,b. the conductivity of the wire.


a. From the definition of resistance, thus

b. The conductivity of the wire is given by

Example 6 :

I

VR

m 1068.1 8 ρ

whereA

ρlR

JA

V

A

ρl

91014.7

24000.2

ρ

1178

m 1095.51068.1

1

σ

m; 1050.0m; 00.2V; 240 3 rlV29 mA 1014.7 J

and JAI

ρσ

1

PHYSICS CHAPTER 5

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States that the potential difference across a metallic the potential difference across a metallic conductor is proportional to the current flowing through it if conductor is proportional to the current flowing through it if its temperature is constant.its temperature is constant.Mathematically,

Ohm’s law also can be stated in term of electric field E and current density J.

Consider a uniform conductor of length l and cross-sectional area A as shown in Figure 5.5.

5.2.3 Ohm’s law

(5.4)(5.4)

IV

where conductor a of resistance :R

where constantTThen

IRV

Figure 5.5Figure 5.5E

IA

l

PHYSICS CHAPTER 5

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A potential difference V is maintained across the conductor

sets up by an electric field E and this field produce a current

I that is proportional to the potential difference. If the field is assumed to be uniform, the potential difference

V is related to the field through the relationship below :

From the Ohm’s law,

EdV ElV

IRV JAI where

A

ρlJAEl

A

ρlR and

ρJE

σρ

1

and

OR σEJ (5.5)(5.5)

PHYSICS CHAPTER 5

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Figures 5.6a, 5.6b, 5.6c and 5.6d show the potential difference V against current I graphs for various materials.

V

I0

Gradient, m

= R

Figure 5.6a : metalFigure 5.6a : metal

V

I0

Figure 5.6b : semiconductorFigure 5.6b : semiconductor

PHYSICS CHAPTER 5

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V

I0

Figure 5.6c : carbonFigure 5.6c : carbon

V

I0

Figure 5.6d : electrolyteFigure 5.6d : electrolyteNote:Note: Some conductors have resistances resistances which depend on the depend on the

currentscurrents flowing through them are known as Ohmic conductorsOhmic conductors and are said to obey Ohm’s lawOhm’s law.

Meanwhile, non-ohmic conductorsnon-ohmic conductors are the conductors where their resistance depend only of the temperatureresistance depend only of the temperature.

PHYSICS CHAPTER 5

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A copper wire carries a current of 10.0 A. The cross section of the wire is a square of side 2.0 mm and its length is 50 m. The density of the free electron in the wire is 8.0 1028 m3. Determinea. the current density,b. the drift velocity of the electrons,c. the electric field intensity between both end of the wire,d. the potential difference across the wire,e. the resistance of the wire.

(Given the resistivity of copper is 1.68 108 m and charge of

electron, e= 1.60 1019 C)Solution :Solution :

a. The current density is given by

Example 7 :

;m 100.8m; 100.2A; 0.10 3283 naIm 50l

A

IJ 2aA and

2a

IJ

2623

mA 105.2100.2

0.10

J

PHYSICS CHAPTER 5

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d. By using the equation of drift velocity, thus

c. The electric field intensity is

;m 100.8m; 100.2A; 0.10 3283 naIm 50l

nAe

Iv d

192328d1060.1100.2100.8

0.10

v

14d s m 1095.1 v

and2aA

ena

Iv

2d

JE 68 105.21068.1 E

1C N 042.0 E

PHYSICS CHAPTER 5

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d. By applying the relationship between uniform E and V, hence

e. From the ohm’s law, therefore

;m 100.8m; 100.2A; 0.10 3283 naIm 50l

ElV

50042.0V

V 1.2V

IRV

R0.101.2

21.0R

PHYSICS CHAPTER 5

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Exercise 5.1 :

1. A block in the shape of a rectangular solid has a cross-sectional area of 3.50 cm2 across its width, a front to rear length of 15.8 cm and a resistance of 935 . The material of which the block is made has 5.33 1022 electrons m3. A potential difference of 35.8 V is maintained between its front and rear faces. Calculate

a. the current in the block,

b. the current density in the block,

c. the drift velocity of the electron,

d. the magnitude of the electric field in the block.(Fundamentals of Physics,6(Fundamentals of Physics,6thth edition, Halliday, Resnick & edition, Halliday, Resnick & Walker, Q24, p.631)Walker, Q24, p.631)

ANS. :ANS. : 3.83 3.83 10 1022 A; 109 A m A; 109 A m22; 1.28 ; 1.28 10 1022 m s m s11; 227 V ; 227 V mm11

PHYSICS CHAPTER 5

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Exercise 5.1 :2.

Figure 5.7 shows a rod in is made of two materials. Each conductor has a square cross section and 3.00 mm on a side. The first material has a resistivity of 4.00 10–3 m and is 25.0 cm long, while the second material has a resistivity of 6.00 10–3 m and is 40.0 cm long. Determine the resistance between the ends of the rod. (Physics for scientists and engineers,6(Physics for scientists and engineers,6thth edition,Serway&Jewett, edition,Serway&Jewett, Q24, p.853)Q24, p.853)

ANS. :ANS. : 378 378


PHYSICS CHAPTER 5

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Exercise 5.1 :

3. A 2.0 m length of wire is made by welding the end of a 120 cm long silver wire to the end of an 80 cm long copper wire. Each piece of wire is 0.60 mm in diameter. A potential difference of 5.0 V is maintained between the ends of the 2.0 m composite wire. Determine

a. the current in the copper and silver wires.

b. the magnitude of the electric field in copper and silver wires.

c. the potential difference between the ends of the silver section of wire.

(Given (silver) is 1.47 108 m and (copper) is 1.72 108 m)

(University physics,11(University physics,11thth edition, Young&Freedman, Q25.56, edition, Young&Freedman, Q25.56, p.976)p.976)

ANS. :ANS. : 45 A; 2.76 V m45 A; 2.76 V m11, 2.33 V m, 2.33 V m11; 2.79 V; 2.79 V

PHYSICS CHAPTER 5

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain the effect of temperature on electrical resistance the effect of temperature on electrical resistance

in metals and superconductors in metals and superconductors Define and explainDefine and explain temperature coefficient of temperature coefficient of

resistivity,resistivity,. . Apply Apply formulae :formulae :

Learning Outcome:

5.3 Variation of resistance with temperature(1 hour)

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00 1 TTRR

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5.3.1 Effect of temperature on resistanceMetalMetal When the temperature increasestemperature increases, the number of free number of free

electrons per unit volumeelectrons per unit volume in metal remains unchangedremains unchanged. Metal atomsMetal atoms in the crystal lattice vibrate with greater vibrate with greater

amplitudeamplitude and cause the number of collisionscause the number of collisions between the free electrons and metal atoms increaseincrease. Hence the resistance resistance in the metal increasesin the metal increases.

SuperconductorSuperconductor Superconductor is a class of metals and compounda class of metals and compound whose

resistance decreases to zeroresistance decreases to zero when they are below the below the

critical temperature critical temperature TTcc.

5.3 Variation of resistance with temperature

PHYSICS CHAPTER 5

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Table 5.2 shows the critical temperature for various superconductors.

When the temperaturetemperature of the metal decreasesdecreases, its resistance decreases to zero at critical temperaturedecreases to zero at critical temperature.

Superconductor have many technological applications such as magnetic resonance imaging (MRI) magnetic levitation of train faster computer chips powerful electric motors and etc…

Material Tc(K)

Lead 7.18

Mercury 4.15

Tin 3.72

Aluminum 1.19

Zinc 0.88

Table 5.2Table 5.2

PHYSICS CHAPTER 5

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is defined as a fractional increase in resistivity of a fractional increase in resistivity of a conductor per unit rise in temperatureconductor per unit rise in temperature. OR

Since = = 00 then

The unit of is CC11 OR KK 1 1. From the equation (5.7), the resistivity of a conductors varies resistivity of a conductors varies

approximately linearly with temperatureapproximately linearly with temperature.

5.3.2 Temperature coefficient of resistivity,

Tρ

ρα

0

where y resistivit in the change :ρ0 change ure temperat: TTT

y resistivit initial :0ρ

yresistivit final:ρwhere

Tαρρ 10

(5.6)(5.6)

(5.7)(5.7)

PHYSICS CHAPTER 5

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From the definition of resistivity, thus

then the equation (5.7) can be expressed as

Table 5.3 shows the temperature coefficients of resistivity for various materials.

Rρ

TαRR 10 (5.8)(5.8)

where resistance initial :0Rresistance final:R

Material (C1)Silver 4.10 103

Mercury 0.89 103

Iron 6.51 103

Aluminum 4.29 103

Copper 6.80 103

Table 5.3Table 5.3

PHYSICS CHAPTER 5

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Figures 5.8a, 5.8b, 5.8c and 5.8d show the resistance R against temperature T graphs for various materials.

R

T0

0R

cT

Figure 5.8a : metalFigure 5.8a : metal Figure 5.8b : semiconductorFigure 5.8b : semiconductor

R

T0

R

T0Figure 5.8c : superconductorFigure 5.8c : superconductor

R

T0Figure 5.8d : carbonFigure 5.8d : carbon

PHYSICS CHAPTER 5

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A copper wire has a resistance of 25 m at 20 C. When the wire is carrying a current, heat produced by the current causes the temperature of the wire to increase by 27 C.a. Calculate the change in the wire’s resistance.b. If its original current was 10.0 mA and the potential difference across wire remains constant, what is its final current?

(Given the temperature coefficient of resistivity for copper is 6.80 103 C1)Solution :Solution :

a. By using the equation for temperature variation of resistance, thus

Example 8 :

C 27C; 20; 10 25 03

0 TTR

TαRR 10

RRR 0and

1059.4 3R

TαRRR 00

TαRR 0

271080.61025 33 R

PHYSICS CHAPTER 5

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b. Given By using the equation for temperature variation of resistance, thus

C 27C; 20; 10 25 03

0 TTR

TαRR 100I

VR and where

I

VR

A 100.10 30

I

TαI

V

I

V 1

0

271080.61100.10

11 33

I

A 1045.8 3I

PHYSICS CHAPTER 5

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine emf, emf, ExplainExplain the difference between emf of a battery and the difference between emf of a battery and

potential difference across the battery terminals.potential difference across the battery terminals. ApplyApply formulae, formulae,

Learning Outcome:

5.4 Electromotive force (emf), potential difference and internal resistance (½ hour)

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IrεV

PHYSICS CHAPTER 5

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5.4.1 Emf, and potential difference, V Consider a circuit consisting of a battery (cell) that is connected

by wires to an external resistor R as shown in Figure 5.9.

5.4 Electromotive force (emf), potential difference and internal resistance

I Battery (cell)

A Brε

R

I


PHYSICS CHAPTER 5

40

A current I flows from the terminal A to the terminal B. For the current to flow continuously from terminal A to B, a

source of electromotive force (e.m.f.), is required such as battery to maintained the potential difference between point A and point B.

Electromotive force (emf), is defined as the energy provided the energy provided by the source (battery/cell) to each unit charge that flows by the source (battery/cell) to each unit charge that flows through the external and internal resistancesthrough the external and internal resistances.

Terminal potential difference (voltage), V is defined as the work work done in bringing a unit (test) charge from the negative to done in bringing a unit (test) charge from the negative to the positive terminals of the battery through the external the positive terminals of the battery through the external resistance onlyresistance only.

The unit unit for both e.m.f. and potential difference are volt (volt (VV)). When the current I flows naturally from the battery there is an

internal drop in potential difference (voltage) equal to Ir. Thus the terminal potential difference (voltage), V is given by

PHYSICS CHAPTER 5

41

then

Equation (5.9) is valid if the battery (cell) supplied the current if the battery (cell) supplied the current to the circuitto the circuit where

For the battery without internal resistance or if no current battery without internal resistance or if no current flows in the circuit (open circuit)flows in the circuit (open circuit), then equation (5.9) can be written as

IrεV (5.9)(5.9)

and IRV

rRIε (5.10)(5.10)

where e.m.f. :ε(voltage) difference potential terminal:V

r OR difference potentialin drop internal : VIrresistance external total:R

(battery) cell a of resistance internal :r

εV

εV

PHYSICS CHAPTER 5

42

is defined as the resistance of the chemicals inside the the resistance of the chemicals inside the battery (cell) between the poles and is given battery (cell) between the poles and is given

by by

The value of internal resistance depends on the type of depends on the type of chemical materialchemical material in the battery.

The symbol of emf and internal resistance in the electrical circuit are shown in Figures 5.10a and 5.10b.

5.4.2 Internal resistance of a battery, r

I

Vr when the cell (battery) is used.when the cell (battery) is used.

where resistance internal across difference potential :rVcircuit in thecurrent :I

rεOR

r ε

Figure 5.10aFigure 5.10a Figure 5.10bFigure 5.10b

PHYSICS CHAPTER 5

43

A battery has an emf of 9.0 V and an internal resistance of 6.0 . Determinea. the potential difference across its terminals when it is supplying a current of 0.50 A,b. the maximum current which the battery could supply.Solution : Solution : a. Given By applying the expression for emf, thus

b. The current is maximum when the total external resistance, R =0, therefore

Example 9 :

0.6V; 0.9 rεA 50.0I

V 0.6V 0.650.00.9 VIrVε

A 5.1max I

0.600.9 max I

rRIε

PHYSICS CHAPTER 5

44

A car battery has an emf of 12.0 V and an internal resistance of 1.0 . The external resistor of resistance 5.0 is connected in series with the battery as shown in Figure 5.11.

Determine the reading of the ammeter and voltmeter if both meters are ideal.

Example 10 :

R

VV

εr

AA


PHYSICS CHAPTER 5

45


By applying the equation of e.m.f., the current in the circuit is

Therefore the reading of the ammeter is 2.0 A2.0 A.The voltmeter measuresvoltmeter measures the potential difference across the potential difference across the terminalsterminals of the battery equal to the potential difference across equal to the potential difference across the total external resistorthe total external resistor, thus its reading is

0.5; 0.1V; 0.12 Rrε

IRV

A 0.2I

rRIε 0.10.50.12 I

0.50.2VV 10V

PHYSICS CHAPTER 5

46

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ApplyApply formula, formula,

Learning Outcome:

5.5 Electrical energy and power (½ hour)

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VIP

VItW

and

PHYSICS CHAPTER 5

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5.5.1 Electrical energy, E Consider a circuit consisting of a battery that is connected by

wires to an electrical device (such as a lamp, motor or battery being charged) as shown in Figure 5.12 where the potential different across that electrical device is V.

5.5 Electrical energy and power


Electrical deviceElectrical device

A B

VI I

PHYSICS CHAPTER 5

48

A current I flows from the terminal A to the terminal B, if it flows for time t, the charge Q which it carries from B to A is given by

Then the work done on this charge Q from B to A (equal to the equal to the electrical energy suppliedelectrical energy supplied) is

If the electrical device is passive resistorpassive resistor (device which convert all the electrical energy supplied into heatconvert all the electrical energy supplied into heat), the heat dissipated H is given by

QVW

ItQ

VItEW (5.11)(5.11)

VItWH OR

RtIH 2 (5.12)(5.12)

PHYSICS CHAPTER 5

49

is defined as the energy liberated per unit time in the the energy liberated per unit time in the electrical deviceelectrical device.

The electrical power P supplied to the electrical device is given by

When the electric current flows through wire or passive resistor, hence the potential difference across it is

then the electrical power can be written as

It is a scalarscalar quantity and its unit is watts (watts (WW)).

5.5.2 Power, P

t

VIt

t

WP

IVP (5.13)(5.13)

IRV

RIP 2 OR

R

VP

2

(5.14)(5.14)

PHYSICS CHAPTER 5

50

In Figure 5.13, a battery has an emf of 12 V and an internal resistance of 1.0 . Determinea. the rate of energy transferred to electrical energy in the battery,b. the rate of heat dissipated in the battery,c. the amount of heat loss in the 5.0 resistor if the current flows through it for 20 minutes.

Example 11 :


R

εr

PHYSICS CHAPTER 5

51


The current in the circuit is given by

a. The rate of energy transferred to electrical energy (power) in the battery is

b. The rate of heat dissipated due to the internal resistance is

c. Given The amount of heat loss in the resistor is

0.5; 0.1V; 0.12 Rrε

IεP

A 0.2I rRIε 0.10.50.12 I

0.120.2P W24P

rIP 2 0.10.2 2P W0.4P

s 12006020 t

RtIH 2 12000.50.2 2HJ 104.2 4H

PHYSICS CHAPTER 5

52

Cells in seriesCells in series Consider two cells are connected in series as shown in Figure

5.14.

The total emf, and the total internal resistance, r are given by

5.5.3 Combination of cells

1r 2r1ε 2ε


21 rrr

21 εεε and

(5.15)(5.15)

(5.16)(5.16)Note:Note:

If one cell, e.m.f. 2 say, is turned round ‘in opposition‘in opposition’ to the others, then but the total internal resistance remains total internal resistance remains unalteredunaltered.

21 εεε

PHYSICS CHAPTER 5

53

Cells in parallelCells in parallel Consider two equal cells are connected in parallel as shown in

Figure 5.15.

The total emf, and the total internal resistance, r are given by

1r

1r

1ε

1ε


11

111

rrr

1εε

and

(5.17)(5.17)

(5.18)(5.18)

Note:Note:

If different cells are connected in paralleldifferent cells are connected in parallel, there is no simple formula for the total emf and the total internal resistance where Kirchhoff’s lawsKirchhoff’s laws have to be used.

PHYSICS CHAPTER 5

54

Exercise 5.2 :

1. A wire of unknown composition has a resistance of 35.0 when immersed in the water at 20.0 C. When the wire is placed in the boiling water, its resistance rises to 47.6 . Calculate the temperature on a hot day when the wire has a resistance of 37.8 .(Physics,7(Physics,7thth edition, Cutnell & Johnson, Q15, p.639) edition, Cutnell & Johnson, Q15, p.639)

ANS. :ANS. : 37.8 37.8 CC2. a. A battery of emf 6.0 V is connected across a 10

resistor. If the potential difference across the resistor is 5.0 V, determine

i. the current in the circuit,

ii. the internal resistance of the battery.b. When a 1.5 V dry cell is short-circuited, a current of 3.0 A flows through the cell. What is the internal resistance of the cell?

ANS. :ANS. : 0.50 A, 2.0 0.50 A, 2.0 ; 0.50 ; 0.50

PHYSICS CHAPTER 5

55

Exercise 5.2 :

3. An electric toy of resistance 2.50 is operated by a dry cell of emf 1.50 V and an internal resistance 0.25 .

a. What is the current does the toy drawn?

b. If the cell delivers a steady current for 6.00 hours, calculate the charge pass through the toy.

c. Determine the energy was delivered to the toy.

ANS. :ANS. : 0.55 A; 1.19 0.55 A; 1.19 10 1044 C; 16.3 kJ C; 16.3 kJ

4. A wire 5.0 m long and 3.0 mm in diameter has a resistance of 100 . A 15 V of potential difference is applied across the wire. Determine

a. the current in the wire,

b. the resistivity of the wire,

c. the rate at which heat is being produced in the wire.(College Physics,6(College Physics,6thth edition, Wilson, Buffa & Lou, Q75, p.589) edition, Wilson, Buffa & Lou, Q75, p.589)

ANS. :ANS. : 0.15 A; 1.40 0.15 A; 1.40 10 1044 m; 2.30 W m; 2.30 W

PHYSICS CHAPTER 5

56

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DeduceDeduce effective resistance of resistors in series and effective resistance of resistors in series and

parallel.parallel. CalculateCalculate effective resistance of resistors in series and effective resistance of resistors in series and

parallel.parallel.

Learning Outcome:

5.6 Resistors in series and parallel (1 hour)

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PHYSICS CHAPTER 5

57

5.6.1 Resistors in series The symbol of resistor in an electrical circuit can be shown in

Figure 5.16.

Consider three resistors are connected in series to the battery as shown in Figure 5.17.

5.6 Resistors in series and parallel

ORRR


1R 2R3R

V

1V 2V 3VI I


PHYSICS CHAPTER 5

58

Characteristics of resistors in seriesCharacteristics of resistors in series The same current same current II flows through each resistor flows through each resistor where

Assuming that the connecting wires have no resistanceAssuming that the connecting wires have no resistance, the total potential difference, V is given by

From the definition of resistance, thus

Substituting for V1, V2 , V3 and V in the eq. (5.19) gives

(5.19)(5.19)

(5.20)(5.20)

321 IIII

321 VVVV

;22 IRV ;33 IRV ;11 IRV effIRV

321eff IRIRIRIR

321eff RRRR

where resistance t)(equivalen effective :effR

PHYSICS CHAPTER 5

59

V

1R

3R

2R

Consider three resistors are connected in parallel to the battery as shown in Figures 5.18a and 5.18b.

5.6.2 Resistors in parallel

I I

2I

1I

3I

1V2V

3VV 1R 3R

2R

I

I

1I

3I

2I

Figure 5.18aFigure 5.18a

Figure 5.18bFigure 5.18b

2V

3V

1V

PHYSICS CHAPTER 5

60

Characteristics of resistors in parallelCharacteristics of resistors in parallel There same potential difference, same potential difference, VV across each resistor across each resistor

where

The charge is conservedcharge is conserved, therefore the total current I in the circuit is given by

From the definition of resistance, thus

Substituting for I1, I2 , I3 and I in the eq. (5.21) gives

(5.21)(5.21)

(5.22)(5.22)

321 VVVV

321 IIII

;2

2 R

VI ;

33 R

VI ;

11 R

VI

effR

VI

321eff R

V

R

V

R

V

R

V

321eff

1111

RRRR

PHYSICS CHAPTER 5

61

For the circuit in Figure 5.19, calculate

a. the effective resistance of the circuit,

b. the current passes through the 12 resistor,

c. the potential difference across 4.0 resistor,

d. the power delivered by the battery.

The internal resistance of the battery may be ignored.

Example 12 :


0.4

0.2

V 0.8

12

PHYSICS CHAPTER 5

62


a.

The resistors R1 and R2 are in series, thus R12 is

Since R12 and R3 are in parallel, therefore Reff is given by

V 0.8; 0.2; 12; 0.4 321 VRRR

1R

V

2R

3R

12R

V

3R

1612R2112 RRR 120.412 R

312eff

111

RRR

2

1

16

11

eff

R

78.1effR

PHYSICS CHAPTER 5

63


b. Since R12 and R3 are in parallel, thus

Therefore the current passes through R2 is given by

c. Since R1 and R2 are in series, thus

Hence the potential difference across R1 is

d. The power delivered by the battery is

V 0.8; 0.2; 12; 0.4 321 VRRR

A 50.02 I

V 0.8312 VVV

12

122 R

VI

A 50.021 II

111 RIV V 0.21 V

16

0.82 I

0.450.01 V

eff

2

R

VP

78.1

0.8 2

P

W0.36P

PHYSICS CHAPTER 5

64

For the circuit in Figure 5.20, calculate the effective resistance between the points A and B.

Solution :Solution : ; 20; 10; 0.5; 0.5 4321 RRRR 105R

Example 13 :

Figure 5.20Figure 5.20 0.5

10 10

A

B

0.5

20

2R

3R5R

A

B

1R

4R

3R5R

A

B

12R

4R

PHYSICS CHAPTER 5

65


R1 and R2 are connected in series, thus R12 is

2112 RRR

; 20; 10; 0.5; 0.5 4321 RRRR 105R

100.50.512R

5R

A

B

123R

4R Since R12 and R3 are connected in

parallel , thus R123 is given by

312123

111

RRR

0.5123R10

1

10

11

123

R

5R

A

B

1234R

R123 and R4 are connected in series ,

thus R1234 is given by

41231234 RRR 251234R

200.51234 R

PHYSICS CHAPTER 5

66


Since R1234 and R5 are connected in parallel , therefore the

effective resistance Reff is given by

; 20; 10; 0.5; 0.5 4321 RRRR 105R

51234eff

111

RRR

14.7effR

10

1

25

11

eff

R

A

B

effR

PHYSICS CHAPTER 5

67

Exercise 5.3 :1. Determine the equivalent resistances of the resistors in

Figures 5.21, 5.22 and 5.23.

ANS. :ANS. : 0.80 0.80 ; 2.7 ; 2.7 ; 8.0 ; 8.0

0.2

0.2

0.2

0.2

Figure 5.21Figure 5.21 Figure 5.22Figure 5.22

0.6

01

0.6 0.4

18

16

0.8

0.9

16

0.6

20


PHYSICS CHAPTER 5

68

2.

The circuit in Figure 5.24 includes a battery with a finite internal resistance, r = 0.50 .

a. Determine the current flowing through the 7.1 and 3.2 resistors.b. How much current flows through the battery?c. What is the potential difference between the terminals of the battery?(Physics,3(Physics,3thth edition, James S. Walker, Q39, p.728) edition, James S. Walker, Q39, p.728)

ANS. :ANS. : 1.1 A, 0.3 A; 1.4 A; 11.3 V 1.1 A, 0.3 A; 1.4 A; 11.3 V

0.1

V 12r

1.7

8.5

5.4 2.3


PHYSICS CHAPTER 5

69

3.

Four identical resistors are connected to a battery as shown in Figure 5.25. When the switch is open, the current through the

battery is I0.

a. When the switch is closed, will the current through the battery increase, decrease or stay the same? Explain.

b. Calculate the current that flows through the battery

when the switch is closed, Give your answer in terms of I0.

(Physics,3(Physics,3thth edition, James S. Walker, Q45, p.728) edition, James S. Walker, Q45, p.728)

ANS. :ANS. : U think U think


εR

R

R

R

PHYSICS CHAPTER 5

70

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: State and useState and use Kirchhoff’s Laws. Kirchhoff’s Laws.

Learning Outcome:

5.7 Kirchhoff’s laws (1 hour)

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PHYSICS CHAPTER 5

71

5.7.1 Kirchhoff’s first law (junction or current law)

states the algebraic sum of the currents entering any the algebraic sum of the currents entering any junctions in a circuit must equal the algebraic sum of the junctions in a circuit must equal the algebraic sum of the currents leaving that junctioncurrents leaving that junction. OR

For example :

5.7 Kirchhoff’s laws

outin II (5.23)(5.23)

A B

2I

1I

5I

4I3I 3I

321 III 543 III

outin II


PHYSICS CHAPTER 5

72

states in any closed loop, the algebraic sum of emfs is in any closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of current and equal to the algebraic sum of the products of current and resistanceresistance.OR In any closed loop,In any closed loop,

Sign conventionSign convention For emf, :

5.7.2 Kirchhoff’s second law (loop or voltage law)

IR (5.24)(5.24)

εε

direction of loopdirection of loop

++-- ε--

ε++


PHYSICS CHAPTER 5

73

For product of IR:

Choose and labelingChoose and labeling the current at each junction in the circuit given.

Choose any one junctionChoose any one junction in the circuit and apply the apply the Kirchhoff’s first lawKirchhoff’s first law.

Choose any two closed loopsChoose any two closed loops in the circuit and designate a direction (clockwise clockwise OR anticlockwise anticlockwise) to travel around the loop in applying the Kirchhoff’s second lawapplying the Kirchhoff’s second law.

Solving the simultaneous equationSolving the simultaneous equation to determine the unknown currents and unknown variables.

IR


I

RIR

I

R


5.7.3 Problem solving strategy (Kirchhoff’s Laws)

PHYSICS CHAPTER 5

74

For example, Consider a circuit is shown in Figure 5.27a.

At junction A or D (applying the Kirchhoff’s first law) :

1R

3R

1εE

D

F

2R2ε

3ε

C

A

B

1I 1I

1I1I

2I2I

3I3I

3I3I

Loop 1Loop 1

Loop 2Loop 2Loop 3Loop 3

Figure 5.27aFigure 5.27a

outin II

321 III (1)(1)

PHYSICS CHAPTER 5

75

For the closed loop (either clockwise or anticlockwise), apply the Kirchhoff’s second law.

From Loop 1

Figure 5.27bFigure 5.27b

(2)(2)

FEDAF

1ε1RE

D

F

2R2ε

A

1I 1I

1I1I

2I2ILoop 1Loop 1

112221 RIRIεε IR

PHYSICS CHAPTER 5

76

From Loop 2

Figure 5.27cFigure 5.27c

(3)(3)

ABCDA

2ε

3R

D2R

3εC

A

B

2I2I

3I3I

3I3I

Loop 2Loop 2

332232 RIRIεε

IR

PHYSICS CHAPTER 5

77

From Loop 3

By solving equation (1) and any two equations from the By solving equation (1) and any two equations from the closed loopclosed loop, hence each current in the circuit can be determined.

Figure 5.27dFigure 5.27d(4)(4)

FECBF

1R

3R

1εE F

3ε

C B

1I 1I

1I1I

3I3I

3I3I

Loop 3Loop 3

113331 RIRIεε

Note:Note:

From the calculation, sometimes we get negative value of current. This negative negative sign indicatessign indicates that the direction of the direction of the actual currentactual current is oppositeopposite to the direction of the direction of the current drawncurrent drawn.

PHYSICS CHAPTER 5

78

For the circuit in Figure 5.28, Determine the current and its direction in the circuit.

Example 14 :


1.15

.226

50.8 2 ,V 1.51

4 ,V 5.01

PHYSICS CHAPTER 5

79


By applying the Kirchhoff’s 2nd law, thus

IRε

A 74.0IIIIII 450.8222.61.155.110.15

1.15

.226

50.8 2 ,V 1.51

4 ,V 5.01Loop 1Loop 1

I

I

II

(anticlockwise)(anticlockwise)

PHYSICS CHAPTER 5

80

For the circuit in Figure 5.29, determine

a. the currents I1, I2 and I,

b. the potential difference across the 6.7 resistor,

c. the power dissipated from the 1.2 resistor.

Example 15 :


8.9

9.3

V .09V 21

7.6

.21

I1I 2I

PHYSICS CHAPTER 5

81


a.

At junction A, by using the Kirchhoff’s 1st law, thus

By using the Kirchhoff’s 2nd law,

From Loop 1:

outin II

III 21

8.9

9.3

V .09V 21

7.6

.21

1I 2I

I

1I 2II

A

B

Loop 1Loop 1 Loop 2Loop 2

(1)(1)

IRε

11 8.92.19.312 III 122.17.13 1 II (2)(2)

PHYSICS CHAPTER 5

82


a. From Loop 2:

By solving the simultaneous equations, we get

b. The potential difference across the 6.7 resistor is given by

c. The power dissipated from the 1.2 resistor is

IRε

II 2.17.60.9 2 0.92.17.6 2 II (3)(3)

A 75.1 A; 03.1 A; 72.0 21 III

RIV 2 7.603.1V

V 90.6V

RIP 2 2.175.1 2P

W68.3P

PHYSICS CHAPTER 5

83

Exercise 5.4 :1. For a circuit in Figure 5.30,

Given 1= 8V, R2= 2 , R3= 3 , R1 = 1 and I = 3 A. Ignore the internal resistance in each battery. Calculate

a. the currents I1 and I2.

b. the emf, 2.

ANS. :ANS. : 1.0 A, 4.0 A; 17 V 1.0 A, 4.0 A; 17 V


3R

1ε

2R2ε1I

2I

I

1R

PHYSICS CHAPTER 5

84

Exercise 5.4 :2.

Determine the current in each resistor in the circuit shown in Figure 5.31.(College Physics,6(College Physics,6thth edition, Wilson, Buffa & Lou, Q57, p.619) edition, Wilson, Buffa & Lou, Q57, p.619)

ANS. :ANS. : 3.75 A; 1.25 A; 1.25 A 3.75 A; 1.25 A; 1.25 A


0.4

0.4V .05

V .05

.04V 01

PHYSICS CHAPTER 5

85

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain the principle of a potential divider. the principle of a potential divider. ApplyApply equation of potential divider equation of potential divider,

Learning Outcome:

5.8 Potential divider (½ hour)

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VRR

RV

21

11

PHYSICS CHAPTER 5

86

A potential divider produces an output voltage that is a fraction fraction of the supply voltage of the supply voltage VV. This is done by connecting two resistors in series as shown in Figure 5.32.

Since the current flowingcurrent flowing through each resistor is the same same, thus

5.8 Potential divider

V

1V

1RI

2V

2RI

21eff RRR

effR

VI

and21 RR

VI


PHYSICS CHAPTER 5

87

Therefore, the potential difference (voltage) across R1 is given by

Similarly,

Resistance R1 and R2 can be replaced by a uniform uniform homogeneous wirehomogeneous wire as shown in Figure 5.33.


11 IRV VRR

RV

21

11

VRR

RV

21

22

(5.25)(5.25)

(5.26)(5.26)

V

I2l1l

I

BA C

2V1V

PHYSICS CHAPTER 5

88

The total resistance, RAB in the wire is

Since the current flowingcurrent flowing through the wire is the samesame, thus

A

ρlR CBACAB RRR

A

ρl

A

ρlR 21

AB

and

ABR

VI

21 llAρ

VI

21AB llA

ρR

PHYSICS CHAPTER 5

89

Therefore, the potential difference (voltage) across the wire with

length l1 is given by

Similarly,

AC1 IRV

A

ρl

llA

ρV

V 1

21

1

Vll

lV

21

11 (5.27)(5.27)

Vll

lV

21

22 (5.28)(5.28)

Note:Note:

From Ohm’s law,

lV

A

ρlIIRV

PHYSICS CHAPTER 5

90

For the circuit in Figure 5.34,

a. calculate the output voltage.

b. If a voltmeter of resistance 4000 is connected across the output,

determine the reading of the voltmeter.

Example 16 :


0004

V 21

0008

outV

PHYSICS CHAPTER 5

91


a. The output voltage is given by

b. The connection between the voltmeter and 4000 resistor is

parallelparallel, thus the equivalent resistance is

Hence the new output voltage is given by

Therefore the reading of the voltmeter is 2.4 V.reading of the voltmeter is 2.4 V.

V 12; 4000; 8000 21 VRR

VRR

RV

21

2out

V 0.4out V

4000

1

4000

11

eq

R

1240008000

4000out

V

2000eqR

V 4.2out V

1220008000

2000out

V

PHYSICS CHAPTER 5

92

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain principles of potentiometer and Wheatstone principles of potentiometer and Wheatstone

Bridge and Bridge and their applicationstheir applications.. UseUse related equations such as related equations such as

Learning Outcome:

5.9 Potentiometer and Wheatstone bridge (½ hour)

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x

3

2

1

R

R

R

R

l

l

R

R xx

PHYSICS CHAPTER 5

93

5.9.1 Potentiometer Consider a potentiometer circuit is shown in Figure 5.35.

The potentiometer is balanced balanced when the jockey (sliding contact) is at such a position on wire AB that there is no current no current through the galvanometerthrough the galvanometer. Thus

5.9 Potentiometer and Wheatstone bridge


(Driver cell -accumulator)

Jockey

V

BAC

xV

I

GG

+ -

III

Galvanometer reading = 0Galvanometer reading = 0

PHYSICS CHAPTER 5

94

When the potentiometer in balanced, the unknown voltage unknown voltage (potential difference being measured) is equal to the (potential difference being measured) is equal to the voltage across ACvoltage across AC.

Potentiometer can be used to compare the emfs compare the emfs of two cells. measure an unknown emfmeasure an unknown emf of a cell. measure the internal resistancemeasure the internal resistance of a cell.

Compare the emfs of two cellsCompare the emfs of two cells In this case, a potentiometer is set up as illustrated in Figure

5.36, in which AB is a wire of uniform resistance and J is a sliding contact (jockey) onto the wire.

An accumulator X maintains a steady current I through the wire AB.

ACx VV

PHYSICS CHAPTER 5

95

Initially, a switch S is connected to the terminal (1) and the

jockey moved until the emf 1 exactly balances the potential

difference (p.d.) from the accumulator (galvanometer reading is zero) at point C. Hence


X

BAI

GG

I

(2)

(1)

2εSS

II

1ε

CJJ

D1l2l

PHYSICS CHAPTER 5

96

After that, the switch S is connected to the terminal (2) and the

jockey moved until the emf 2 balances the p.d. from the

accumulator at point D. Hence

AC1 Vε

ACAC IRV whereA

ρlR 1

AC and

11 lA

ρIε

(1)(1)

then

AD2 Vε

ADAD IRV whereA

ρlR 2

AD and

22 lA

ρIε

(2)(2)

then

PHYSICS CHAPTER 5

97

By dividing eq. (1) and eq. (2) then

Measure an unknown emf of a cellMeasure an unknown emf of a cell By using the same circuit shown in Figure 5.36, the value of

unknown emf can be determined if the cell 1 is replaced with a standard cell.

A standard cell is one in which provides a constant and provides a constant and

accurately known emfaccurately known emf. Thus the emf 2 can be calculated by

using the equation (5.29).

2

1

2

1

l

l

ε

ε

2

1

2

1

lAρI

lAρI

ε

ε

(5.29)(5.29)

PHYSICS CHAPTER 5

98

Measure the internal resistance of a cellMeasure the internal resistance of a cell Consider a potentiometer circuit as shown in Figure 5.37.


ε

BA I

GG

I

1ε

0l C

JJ

S R

r

II

PHYSICS CHAPTER 5

99

An accumulator of emf maintains a steady current I through the wire AB.

Initially, a switch S is opened and the jockey J moved until the

emf 1 exactly balances the emf from the accumulator

(galvanometer reading is zero) at point C. Hence

After the switch S is closed, the current I1 flows through the

resistance box R and the jockey J moved until the galvanometer reading is zero (balanced condition) at point D as shown in Figure 5.38.

AC1 Vε

ACAC IRV whereA

ρlR 0

AC and

01 lA

ρIε

(1)(1)

then

PHYSICS CHAPTER 5

100


ε

BA I

GG

I

1ε

JJ

S R

r

II

1I

Dl

1I

1I

1I

1I

PHYSICS CHAPTER 5

101

Hence

From the equation of emf,

ADVV

ADAD IRV whereA

ρlR ADand

lA

ρIV

(2)(2)

then

rIVε 11

1

1

I

Vεr

R

VI 1and

RV

Vεr

1 (3)(3)

PHYSICS CHAPTER 5

102

By substituting eqs. (1) and (2) into the eq. (3), we get

The value of internal resistance, The value of internal resistance, rr is determined by plotting is determined by plotting the graph of the graph of 11/l/l against against 11/R/R .

Rearranging eq. (4) :

Rl

llr

0

Rl

lr

10 (4)(4)

cxmy Then compare with

00

111

lRl

r

l

PHYSICS CHAPTER 5

103

Therefore the graph is straight line as shown in Figure 5.39.

0

,Gradientl

rm

0

1

l

R

1

l

1

0


PHYSICS CHAPTER 5

104

Cells A and B and centre-zero galvanometer G are connected to a uniform wire OS using jockeys X and Y as shown in 5.40.

a. the potential difference across OY when OY = 75.0 cm,

b. the potential difference across OY when Y touches S and the

galvanometer is balanced,

c. the internal resistance of the cell A,

d. the emf of cell A.

Example 17 :


A

SO

GGB

XXYY

The length of the uniform wire OS is 1.00 m and its resistance is 12 .

When OY is 75.0 cm, the galvanometer does not show any deflection when OX= 50.0 cm. If Y touches the end S of the wire, OX = 62.5 cm when the galvanometer is balanced. The emf of the cell B is 1.0 V. Calculate

PHYSICS CHAPTER 5

105


a. Given

When G = 0 (balance condition), thus

V 0.1; 12m; 00.1 BOSOS εRlm 50.0m; 75.0 OX1OY1 ll

Aε

SO

GGBε

XXYY

0

OY1l

OX1l

Since wire OS is uniform thus

OSOS

1OXOX1 R

l

lR

and

0.612

00.1

50.0OX1R

0.912

00.1

75.0OY1R

BOX1 εV OX11OX1 RIV and

1I1I

1I

1I 1I

BOX11 εRI 0.10.61 I

A 17.01 I

PHYSICS CHAPTER 5

106


a. Therefore the potential difference across OY is given by

b. Given

V 0.1; 12m; 00.1 BOSOS εRl

OY11OY1 RIV 0.917.0OY1 V

V 53.1OY1 Vm 625.0m; 00.1 OX2OY2 ll

Aε

SO

GGBε

XX

YY

0

OY2l

OX2l

2I2I

2I

2I 2I Since wire OS is uniform thus

OSOS

2OXOX2 R

l

lR

and

5.712

00.1

625.0OX2R

1212

00.1

00.1OY2R

PHYSICS CHAPTER 5

107


b. When G = 0 (balance condition), thus

Therefore the potential difference across OY is given by

c. The emf of cell A is given by

For case in the question (a) :

V 0.1; 12m; 00.1 BOSOS εRl

BOX2 εV OX22OX2 RIV and

BOX22 εRI 0.15.72 I

A 13.02 I

OY22OY2 RIV 1213.0OY2 V

V 56.1OY2 V

rRIε A

)( 1OY1A rRIε

rε 0.917.0A (1)(1)

PHYSICS CHAPTER 5

108


c. For case in the question (b) :

(1) = (2):

d. The emf of cell A is

V 0.1; 12m; 00.1 BOSOS εRl

)( 2OY2A rRIε rε 1213.0A (2)(2)

rr 1213.00.917.0 65.0r

rε 0.917.0A

65.00.917.0A εV 64.1A ε

PHYSICS CHAPTER 5

109

It is used to measured the unknown resistance of the measured the unknown resistance of the resistorresistor.

Figure 5.41 shows the Wheatstone bridge circuit consists of a

cell of emf (accumulator), a galvanometer , know resistances

(R1, R2 and R3) and unknown resistance Rx.

The Wheatstone bridge is said to be balancedbalanced when no no current flows through the galvanometercurrent flows through the galvanometer.

5.9.2 Wheatstone bridge

ε

BA GG

C

D

1R2R

3R xR

0

I I

2I

1I

2I

1I


PHYSICS CHAPTER 5

110

Hence

then

Therefore

Since

Dividing gives

1CBAC III 2DBAD III and

Potential at C = Potential at DPotential at C = Potential at D

ADAC VV BDBC VV and

IRV

3211 RIRI thus

and x221 RIRI

x2

32

21

11

RI

RI

RI

RI

31

2x R

R

RR

(5.30)(5.30)

PHYSICS CHAPTER 5

111

The application of the Wheatstone bridge is Metre BridgeMetre Bridge. Figure 5.42 shows a Metre bridge circuit.

The metre bridge is balanced balanced when the jockey J is at such a position on wire AB that there is no current through the no current through the

galvanometergalvanometer. Thus the current I1 flows through the resistance

Rx and R but current I2 flows in the wire AB.

0

AccumulatorAccumulator

JockeyJockey

Thick copper Thick copper stripstrip

(Unknown (Unknown resistance)resistance) (resistance box)(resistance box)

Wire of uniform Wire of uniform resistanceresistance

xR

A

ε

GGB

R

J

2l1l


I I

1I

2I

1I

PHYSICS CHAPTER 5

112

Let Vx : p.d. across Rx and V : p.d. across R, At balance condition,

By applying Ohm’s law, thus

Dividing gives

AJx VV JBVV and

AJ2x1 RIRI JB21 RIRI and

A

ρlR 1

AJ JB2

AJ2

1

x1

RI

RI

RI

RI where and

A

ρlR 2

JB

A

ρlA

ρl

R

R

2

1

x

Rl

lR

2

1x (5.31)(5.31)

PHYSICS CHAPTER 5

113

An unknown length of platinum wire 0.920 mm in diameter is placed as the unknown resistance in a Wheatstone bridge as shown in Figure 5.43.

Resistors R1 and R2 have resistance of 38.0 and 46.0

respectively. Balance is achieved when the switch closed and R3 is

3.48 . Calculate the length of the platinum wire if its resistivity is 10.6 108 m.

Example 18 :


PHYSICS CHAPTER 5

114


At balance conditionbalance condition, the ammeter reading is zeroammeter reading is zero thus the resistance of the platinum wire is given by

From the definition of resistivity, thus

; 0.46; 0.38m; 10920.0 213 RRd

;m Ω 106.10; 48.3 83

ρR

1

2

3

x

R

R

R

R

21.4xR

0.38

0.46

48.3x

R

l

ARρ x

4

2dA

and

l

dRρ

4

2x

l4

10920.021.4106.10

238

m 4.26l

PHYSICS CHAPTER 5

115

Exercise 5.5 :1. In Figure 5.44, PQ is a uniform wire of length 1.0 m and

resistance 10.0 .

ANS. :ANS. : 0.50 V; 7.5 0.50 V; 7.5 ; 25.0 cm; 25.0 cm; 25.0 cm; 25.0 cm2S

1ε

PQ

GG2ε

TT

1R

2R

1S


1 is an accumulator of emf 2.0 V and

negligible internal resistance. R1 is a

15 resistor and R2 is a 5.0

resistor when S1 and S2 open, galvanometer G is balanced when QT is 62.5 cm. When both S1 and S2 are closed, the balance length is 10.0 cm. Calculate

a. the emf of cell 2.

b. the internal resistance of cell 2.

c. the balance length QT when S2

is opened and S1 closed.

d. the balance length QT when S1

is opened and S2 closed.

PHYSICS CHAPTER 5

116

R

2. The circuit shown in Figure 5.45 is known as a Wheatstone bridge.

Determine the value of the resistor R such that the current through the 85.0 resistor is zero.(Physics,3(Physics,3thth edition, James S. Walker, Q93, p.731) edition, James S. Walker, Q93, p.731)

ANS. :ANS. : 7.50 7.50

Exercise 5.5 :


PHYSICS CHAPTER 5

117

Exercise 5.5 :3. A potentiometer with slide-wire of length 100 cm and

resistance of 5.0 , is connected to a driver cell of emf 2.0 V and negligible internal resistance. Calculate

a. the length of the potentiometer wire needed to balance a potential difference of 1.5 V,

b. the resistance which must be connected in series with the slide-wire to give a potential difference of 7.0 mV across the whole wire,

c. the emf of a dry cell which is balanced by 80 cm of the wire, setup as in part (b).

ANS. :ANS. : 75.0 cm; 1424 75.0 cm; 1424 ; 5.6 mV; 5.6 mV

118

PHYSICS CHAPTER 5

Next Chapter…CHAPTER 6 :Magnetic field

26770932 Matriculation Physics Electric Current and Direct Current Circuit

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