Chapter18 Student 2

8/10/2019 Chapter18 Student 2

http://slidepdf.com/reader/full/chapter18-student-2 1/118

PHYSICS CHAPTER 18

1

CHAPTER 18:

Electric current and direct-

current circuits(7 Hours)

w w w . k m s

. m a t r i k . e

d u

. m y / p h y s i c s



PHYSICS CHAPTER 18

2

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

Describe microscopic model of current.

Define and use electric current formulae,

Learning Outcome:

18.1 Electrical conduction (1 hour)

w w w . k m s

. m a t r i k . e

d u

. m y / p h y s i c s



PHYSICS CHAPTER 18

3

18.1.1 Electric current, I

Consider a simple closed circuit consists of wires, a battery anda light bulb as shown in Figure 18.1.

18.1 Electrical conduction

Area, A

e F

E

I

Figure 18.1



PHYSICS CHAPTER 18

4

From the Figure 18.1,

Direction of electric field or electric current :

Positive to negative terminal

Direction of electron flows :

Negative to positive terminal

The electron accelerates because of the electric forceacted on it.

is defined as the total (nett) charge, Q flowing through the

area per unit time, t .

Mathematically,

OR

instantaneous current

average current



PHYSICS CHAPTER 18

5

It is a base and scalar quantities.

The S.I. unit of the electric current is the ampere (A).

1 ampere of current is defined as one coulomb of chargepassing through the surface area in one second.

OR

1

sC1second1

coulomb1

ampere1

Note:

If the charge move around a circuit in the same directionat all times, the current is called direct current (dc), which is

produced by the battery.



PHYSICS CHAPTER 18

6

is defined as the current flowing through a conductor per

unit cross-sectional area.

Mathematically,

It is a vector quantity.

Its unit is ampere per squared metre (A m 2)

The direction of current density, J always in the same

direction of the current I . e.g. in Figure 18.2.

18.1.2 Current density, J

where currentelectric: I conductor theof areasectional-cross: A

I

max J

0 J

Area, A

Figure 18.2





PHYSICS CHAPTER 18

8

Consider a metal rod of length L and cross-sectional area A,

which is applied to the electric field as shown in Figures 18.4.

Suppose there are n free electrons (charge carrier) per unit

volume in the metal rod, thus the number of free electron, N isgiven by

18.1.4 Drift velocity of charges, v d

E

J

I

dv

dv

L

A

Figure 18.4

V

N n ALV and

AL

N n nAL N



PHYSICS CHAPTER 18

9

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

NeQenALQ

t

Lvd

dv

Lt

then the drift velocity vd is given by

t

Q I

d

d

nAev

v L

enAL I

and

OR

where

electrontheof charge:e

Definition

Density of the

free electron



PHYSICS CHAPTER 18

10

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 10 19 C)

Solution :

a. By applying the equation of average current, thus

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

Example 18.1 :

A0.3 I

t

Q I

t N

19

1060.10.3

and NeQ

t Ne I

s55t It Q

550.3Q



PHYSICS CHAPTER 18

11

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 10 19 C)

Solution :

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

b. The current density is given by

Example 18.2 :

32836 m1049.8A;1017m;10900 n I r

nAe

I vd

192628

3

d

1060.1109001049.8

1017

π v

and2πr A

er n

I v 2d

2πr

I

J

3

26

17 10

900 10 J

π



PHYSICS CHAPTER 18

12

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 conductoris 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 10 19 C)

Solution :

By applying the equation of the drift velocity, thus

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

Example 18.3 :

nAe

I vd

192228d1060.11000.31049.8

15004

π v

and4

2πd A

ed n

I

v 2d

4

dv

Lt

3

4

100 10

1.56 10t

A;1500m;10100m;1000.332

I Ld 328 m1049.8n



PHYSICS CHAPTER 18

13

Explain how electrical devices can begin operating almostimmediately after you switch on, even though the individual

electrons in the wire may take hours to reach the device.

Solution :

Example 18.4 :

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

When electrons begin to move out of a battery or sourcetheir motion sets up a propagating influence that movesthrough the wire at nearly the speed of light, causingelectrons everywhere in the wire begin to move.



PHYSICS CHAPTER 18

14


State Ohm’s law.

Define and use resistivity formulae,

Learning Outcome:

18.2 Resistivity and Ohm’s law (½ hour)

w w w . k m s

. m a t r i k . e

d u

. m y / p h y s i c

s



PHYSICS CHAPTER 18

15

States that the potential difference across a metallic

conductor is proportional to the current flowing through it ifits 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 18.5.

18.2.1 Ohm’s law

(18.4)

I V

where conductor aof resistance: R

where constantT Then

Figure 18.5 E

I A

l



PHYSICS CHAPTER 18

16

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,

Ed V El V

IRV JA I where

A

ρl JA El

A

ρl Rand

and

OR (18.5)



PHYSICS CHAPTER 18

17

Figures 18.6a, 18.6b, 18.6c and 18.6d show the potential

difference V against current I graphs for various materials.

V

I 0

Gradient, m

= R

Figure 18.6a : metal

V

I 0

Figure 18.6b :

semiconductor



PHYSICS CHAPTER 18

18

V

I 0

Figure 18.6c : carbon

V

I 0

Figure 18.6d : electrolyteNote:

Some conductors have resistances which depend on thecurrents flowing through them are known as Ohmic conductors

and are said to obey Ohm’s law.

Meanwhile, non-ohmic conductors are the conductors where

their resistance depend only of the temperature.



PHYSICS CHAPTER 18

19

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 m 3. Determine

a. 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 10 8 m and charge of

electron, e = 1.60 10 19 C)

Solution :

a. The current density is given by

Example 21.7 :

;m100.8m;100.2A;0.10 3283 na I m50l

A

I J

2a Aand

2a

I

J 23

10.0

2.0 10 J



PHYSICS CHAPTER 18

20

Solution :

d. By using the equation of drift velocity, thus

c. The electric field intensity is

;m100.8m;100.2A;0.10 3283 na I m50l

nAe

I vd

192328d

1060.1100.2100.8

0.10v

and2a A

ena

I v

2d

J E 68 105.21068.1 E



PHYSICS CHAPTER 18

21

Solution :

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

e. From the ohm’s law, therefore

;m100.8m;100.2A;0.10 3283 na I m50l

El V

50042.0V

IRV

R0.101.2



PHYSICS CHAPTER 18

22

18.2.1 Resistance, R

is defined as a ratio of the potential difference across anelectrical component to the current passing through it.

Mathematically,

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

It is a scalar quantity and its unit is ohm ( ) or V A1

In general, the resistance of a metallic conductor increaseswith temperature.

18.2 Resistivity and Ohm’s law

where (voltage)difference potential:V current: I

Note:

If the temperature of the metallic conductor is constant hence its

resistance also constant.

(18.1)



PHYSICS CHAPTER 18

23

Resistivity,

is defined as the resistance of a unit cross-sectional area perunit length of the material.

Mathematically,

It is a scalar quantity and its unit is ohm meter ( m)

It is a measure of a material’s ability to oppose the flow of

an electric current. It also known as specific resistance.

Resistivity depends on the type of the material and on thetemperature.

A good electric conductors have a very low resistivities and

good insulators have very high resistivities.

18.2.2 Resistivity and conductivity

where materialtheof length:l areasectional-cross: A

(18.2)



PHYSICS CHAPTER 18

24

From the eq. (18.2), the resistance of a conductor depends onthe length and cross-sectional area.

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

Conductivity,

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

Mathematically,

It is a scalar quantity and its unit is 1m 1.

Material Resistivity, ( m)

Silver 1.59 10 8

Copper 1.68 10 8

Aluminum 2.82 108

Gold 2.44 10 8

Glass 1010 1014

Table 18.1

(18.3)



PHYSICS CHAPTER 18

25

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 timesgreater than that of wire Q, determine the ratio of their diameters.

Solution :

Given

Example 18.6 :

l l l ρ ρ ρ QPQP ;

QP3 R R and

A

ρl R

Q

P

d

d

Q

QQ

P

PP 3 A

l ρ

A

l ρand

4

2πd A

2

Q

2

P

43

4

πd

ρl

πd

ρl

ORP

Q

d

d



PHYSICS CHAPTER 18

26

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 is7.14 109 A m 2. Calculate

a. the resistivity of the wire,

b. the conductivity of the wire.

Solution :

a. From the definition of resistance, thus

b. The conductivity of the wire is given by

Example 18.7 :

I

V R where

A

ρl R

JA

V

A

ρl 91014.7

240

00.2 ρ

8

1

1.68 10σ

m;1050.0m;00.2V;240 3r l V

29 mA1014.7 J

and JA I

ρσ

1



PHYSICS CHAPTER 18

27

Exercise 18.2 :

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 rearlength of 15.8 cm and a resistance of 935 . The material of

which the block is made has 5.33 1022 electrons m 3. 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,6th edition, Halliday, Resnick &Walker, Q24, p.631)

ANS. : 3.83 10 2 A; 109 A m 2; 1.28 10 2 m s 1; 227 V m 1



PHYSICS CHAPTER 18

28

2.

Figure 5.7 shows a rod in is made of two materials. Eachconductor 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 is25.0 cm long, while the second material has a resistivity of

6.00 10 –3

m and is 40.0 cm long. Determine theresistance between the ends of the rod.

(Physics for scientists and engineers,6th edition,Serway&Jewett,Q24, p.853)

ANS. : 378

Figure 18.7



PHYSICS CHAPTER 18

29

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. Eachpiece 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 silverwires.

c. the potential difference between the ends of the silver

section of wire.

(Given (silver) is 1.47 108

m and (copper) is 1.7210 8 m)

(University physics,11th edition, Young&Freedman, Q25.56,

p.976)

ANS. : 45 A; 2.76 V m 1, 2.33 V m 1; 2.79 V



PHYSICS CHAPTER 18

30


Explain the effect of temperature on electrical resistance

in metals and superconductors

Define and explain temperature coefficient of resistivity,.

Apply resistance :

Learning Outcome:

18.3 Variation of resistance with temperature(1 hour)

w w w . k m s

. m a t r i k . e

d u

. m y / p h y s i c

s



PHYSICS CHAPTER 18

31

18.3.1 Effect of temperature on resistance

Metal

When the temperature increases, the number of free

electrons per unit volume in metal remains unchanged.

Metal atoms in the crystal lattice vibrate with greateramplitude and cause the number of collisions between thefree electrons and metal atoms increase. Hence the resistancein the metal increases.

Superconductor Superconductor is a class of metals and compound whose

resistance decreases to zero when they are below the

critical temperature T c.

18.3 Variation of resistance with

temperature



PHYSICS CHAPTER 18

32

Table 18.2 shows the critical temperature for varioussuperconductors.

When the temperature of the metal decreases, its resistance

decreases 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 T c( K)

Lead 7.18

Mercury 4.15

Tin 3.72

Aluminum 1.19

Zinc 0.88

Table 18.2

Video 18.1

Video 18.2

http://localhost/var/www/apps/conversion/tmp/scratch_4/Super%20Conductor.avi

http://localhost/var/www/apps/conversion/tmp/scratch_4/Magnetic%20Train%20with%20superconductors.avi

http://localhost/var/www/apps/conversion/tmp/scratch_4/Magnetic%20Train%20with%20superconductors.avi

http://localhost/var/www/apps/conversion/tmp/scratch_4/Super%20Conductor.avi



PHYSICS CHAPTER 18

33

is defined as a fractional increase in resistivity of a

conductor per unit rise in temperature.OR

Since =0

then

The unit of is C 1 OR K 1.

From the equation (18.7), the resistivity of a conductors

varies approximately linearly with temperature.

18.3.2 Temperature coefficient of resistivity,

where yresistivitin thechange: ρ0changeure temperat: T T T

yresistivitinitial:0 ρ

yresistivitfinal: ρwhere

(18.6)

(18.7)



PHYSICS CHAPTER 18

34

From the definition of resistivity, thus

then the equation (18.7) can be expressed as

Table 18.3 shows the temperature coefficients of resistivity forvarious materials.

R ρ

(18.8)

where resistanceinitial:0 Rresistancefinal: R

Material ( C 1)

Silver 4.10 10 3

Mercury 0.89 10 3

Iron 6.51 10 3

Aluminum 4.29 10 3

Copper 6.80 10 3

Table 18.3



PHYSICS CHAPTER 18

35

Figures 18.7a, 18.7b, 18.7c and 18.7d show the resistance Ragainst temperature T graphs for various materials.

R

T 0

0 R

cT

Figure 18.7a : metal Figure 18.7b : semiconductor

R

T 0

R

T 0Figure 18.7c : superconductor

R

T 0Figure 18.7d : carbon



PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

37

Solution :

b. Given

By using the equation for temperature variation of resistance,thus

C27C;20;1025 03

0 T T R

T α R R 100 I

V Randwhere

I

V R

A100.10 30 I

T α

I

V

I

V 1

0

271080.61100.10

11 3

3 I

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

38


Define emf,

Explain the difference between emf of a battery and

potential difference across the battery terminals.

Apply formulae,

Learning Outcome:

18.4 Electromotive force (emf), potential

difference and internal resistance (½ hour)

w w w . k m s

. m a t r i k . e

d u

. m y / p h y s i c

s

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

39

18.4.1 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 18.8.

18.4 Electromotive force (emf), potential

difference and internal resistance

I Battery (cell)

A Br ε

R

I

Figure 18.8

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

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 asbattery to maintained the potential difference between point Aand point B.

Electromotive force (emf), is defined as the energy providedby the source (battery/cell) to each unit charge that flows

through the external and internal resistances.

Terminal potential difference (voltage), V is defined as the work

done in bringing a unit (test) charge from the negative tothe positive terminals of the battery through the externalresistance only.

The unit for both e.m.f. and potential difference are volt (V).

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 18



PHYSICS CHAPTER 18

41

then

Equation (18.9) is valid if the battery (cell) supplied the

current to the circuit where

For the battery without internal resistance or if no currentflows in the circuit (open circuit), then equation (18.9) can bewritten as

(18.9)

and IRV

(18.10)

where e.m.f.:ε

(voltage)difference potentialterminal:V r OR difference potentialindropinternal: V Ir

resistanceexternaltotal: R(battery)cellaof resistanceinternal:r

V ε

V ε

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

42

is defined as the resistance of the chemicals inside the

battery (cell) between the poles and is given by

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

The symbol of emf and internal resistance in the electrical circuit

are shown in Figures 18.9a and 18.9b.

18.4.2 Internal resistance of a battery, r

when the cell (battery) is used.

where resistanceinternalacrossdifference potential:r V circuitin thecurrent: I

r εOR

Figure 18.9a Figure 18.9b

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

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 :

a. GivenBy applying the expression for emf, thus

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

Example 18.9 :

0.6V;0.9 r ε

A50.0 I

0.650.00.9 V Ir V ε

0.600.9 max I

r R I ε

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

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 serieswith the battery as shown in Figure 18.10.

Determine the reading of the ammeter and voltmeter if both meters

are ideal.

Example 18.10 :

R

V

r

A

Figure 18.10

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

45

Solution :

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

Therefore the reading of the ammeter is .

The voltmeter measures the potential difference across the

terminals of the battery equal to the potential difference across

the total external resistor , thus its reading is

0.5;0.1V;0.12 Rr ε

IRV

r R I ε

0.10.50.12 I

0.50.2V



PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

47

is defined as the energy liberated per unit time in the

electrical 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 scalar quantity and its unit is watts (W).

18.5.2 Power, P

t

VIt

t

W P

(18.11)

IRV

OR (18.12)

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

48

In Figure 18.11, 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 18.11 :

Figure 18.11

R

r

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

49

Solution :

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. GivenThe amount of heat loss in the resistor is

0.5;0.1V;0.12 Rr ε

I ε P

A0.2 I r R I ε 0.10.50.12 I

0.120.2 P

r I P 2 0.10.2

2 P

s12006020t

Rt I H 2 12000.50.22

H

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

50

18.5.1 Electrical energy, E

Consider a circuit consisting of a battery that is connected bywires to an electrical device (such as a lamp, motor or batterybeing charged) as shown in Figure 18.12 where the potential

different across that electrical device is V .

18.5 Electrical energy and power

Figure 18.12

Electrical device

A B

V I I

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

51

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

electrical energy supplied) is

If the electrical device is passive resistor (device whichconvert all the electrical energy supplied into heat), the heat

dissipated H is given by

QV W

It Q

(18.13)

OR

(18.14)

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

52

Cells in series

Consider two cells are connected in series as shown in Figure18.13.

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

18.5.3 Combination of cells

1r 2r 1ε 2ε

Figure 18.13

and

(18.15)

(18.16)Note:

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

21 εεε

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

53

Cells in parallel

Consider two equal cells are connected in parallel as shown in

Figure 18.14.

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

1r

1r

1ε

1ε

Figure 18.14

and

(18.17)

(18.18)

Note:

If different cells are connected in parallel, there is no simpleformula for the total emf and the total internal resistance where

Kirchhoff’s laws have to be used.

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

54

Exercise 18.3 :

1. A wire of unknown composition has a resistance of 35.0

when immersed in the water at 20.0 C. When the wire isplaced 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,7th edition, Cutnell & Johnson, Q15, p.639)

ANS. : 37.8 C2. 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 Aflows through the cell. What is the internal resistance of thecell?

ANS. : 0.50 A, 2.0 ; 0.50

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

55

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. : 0.55 A; 1.19 104 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,6th edition, Wilson, Buffa & Lou, Q75, p.589)

ANS. : 0.15 A; 1.40 10 4 m; 2.30 W

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

56


Deduce and calculate effective resistance of resistors

in series and parallel.

Learning Outcome:

18.6 Resistors in series and parallel (1 hour)

w w w . k m s . m a t r i k . e

d u . m y / p h y s i c

s

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

57

18.6.1 Resistors in series

The symbol of resistor in an electrical circuit can be shown inFigure 18.15.

Consider three resistors are connected in series to the batteryas shown in Figure 18.16.

18.6 Resistors in series and parallel

OR

R R

Figure 18.15

1 R 2 R3 R

V

1V 2V 3V

I I

Figure 18.16

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

58

Characteristics of resistors in series

The same current I flows through each resistor where

Assuming that the connecting wires have no resistance, the

total potential difference, V is given by

From the definition of resistance, thus

Substituting for V 1, V 2 , V 3 and V in the eq. (18.19) gives

(18.19)

(18.20)

where resistancet)(equivaleneffective:eff R

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

59

V

1 R

3 R

2 R

Consider three resistors are connected in parallel to the battery

as shown in Figures 18.17a and 18.17b.

18.6.2 Resistors in parallel

I I

2 I

1 I

3 I

1V 2V

3V V 1 R

3 R2 R

I

I

1 I

3 I

2 I

Figure 18.17a

Figure 18.17b

2V

3V

1V

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

60

Characteristics of resistors in parallel

There same potential difference, V across each resistor

where

The charge is conserved, therefore the total current I in the

circuit is given by

From the definition of resistance, thus

Substituting for I 1, I

2, I

3and I in the eq. (18.21) gives

(18.21)

(18.22)

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

61

For the circuit in Figure 18.18, 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 18.15 :

Figure 18.18

0.4

0.2

V0.8

12

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

62

Solution :

a.

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

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

V0.8;0.2;12;0.4 321 V R R R

1 R

V

2 R

3 R

12 R

V

3 R

1612 R

2112 R R R 120.412 R

312eff

111

R R R 2

1

16

11

eff R

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

63

Solution :

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

V0.8;0.2;12;0.4 321 V R R R

V0.8312 V V V

12

122

R

V I

A50.021 I I

111 R I V

16

0.82 I

0.450.01V

eff

2

R

V P

78.1

0.82

P

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

64

For the circuit in Figure 18.19, calculate the effective resistancebetween the points A and B.

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

Example 18.16 :

Figure 18.19 0.5

10 10

A

B

0.5

20

2 R

3 R5 R

A

B

1 R

4 R

3 R5 R

A

B

12 R

4 R



PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

66

Solution :

Since R1234 and R5 are connected in parallel , therefore the

effective resistance Reff is given by

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

51234eff

111

R R R

10

1

25

11

eff R

A

B

eff R

PHYSICS CHAPTER 18



PHYSICS CHAPTER 18

67

Exercise 18.4 :1. Determine the equivalent resistances of the resistors in

Figures 18.20, 18.21 and 18.22.

ANS. : 0.80 ; 2.7 ; 8.0

0.2

0.2

0.2

0.2

Figure 18.20 Figure 18.21

0.6

01

0.6

0.4

18

16

0.8

0.9

16

0.6

20

Figure 18.22

PHYSICS CHAPTER 18



68

2.

The circuit in Figure 18.23 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 ofthe battery?

(Physics,3th edition, James S. Walker, Q39, p.728)

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

0.1

r

1.7

8.5

5.4 2.3

Figure 18.23

PHYSICS CHAPTER 18



69

3.

Four identical resistors are connected to a battery as shown inFigure 18.24. When the switch is open, the current through

the battery is I 0.

a. When the switch is closed, will the current through thebattery 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 I 0.


ANS. : U think

Figure 18.24

ε R

R

R

R

PHYSICS CHAPTER 18



70


State and use Kirchhoff’s Laws.

Learning Outcome:

18.7 Kirchhoff’s laws (1 hour)

w w w . k m s . m a t r i k . e

d u . m y / p h y s i c s

PHYSICS CHAPTER 18



71

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

states the sum of the currents entering any junctions in acircuit must equal the sum of the currents leaving that

junction.

OR

For example :

18.7 Kirchhoff’s laws

(18.23)

A B

2 I

1 I

5 I

4 I 3 I 3 I

321 I I I 543 I I I

outin I I

Figure 18.26

PHYSICS CHAPTER 18



72

states in any loop, the sum of emfs is equal to the sum of

the products of current and resistance.OR In any loop,

Sign convention

For emf, :

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

(18.24)

εε

direction of loop

+- ε-ε

+

direction of loop

PHYSICS CHAPTER 18



73

For product of IR:

Choose and labeling the current at each junction in the circuitgiven.

Choose any one junction in the circuit and apply theKirchhoff’s first law.

Choose any two closed loops in the circuit and designate a

direction (clockwise OR anticlockwise) to travel around theloop in applying the Kirchhoff’s second law.

Solving the simultaneous equation to determine the unknowncurrents and unknown variables.

IR

direction of loop

I

R IR

I

R

direction of loop

18.7.3 Problem solving strategy (Kirchhoff’s Laws)

PHYSICS CHAPTER 18



74

For example, Consider a circuit is shown in Figure 18.25a.

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

1 R

3 R

1εE

D

F

2 R2ε

3ε

C

A

B

1 I 1 I

1 I 1 I

2 I 2 I

3 I 3 I

3 I 3 I

Loop 1

Loop 2Loop 3

Figure 18.25a

outin I I

321 I I I (1)

PHYSICS CHAPTER 18



75

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

From Loop 1

Figure 18.25b

(2)

FEDAF

1ε 1 RE

D

F

2 R2ε A

1 I 1 I

1 I

1

I

2 I 2 I Loop 1

112221 R I R I εε

IR

PHYSICS CHAPTER 18



76

From Loop 2

Figure 18.25c

(3)

ABCDA

2ε

3 R

D2 R

3ε

C

A

B

2

I 2

I

3 I 3 I

3 I 3 I

Loop 2

332232 R I R I εε

IR

PHYSICS CHAPTER 18



77

From Loop 3

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

determined.

Figure 18.25d

(4)

FECBF

1 R

3 R

1εE F

3ε

C B

1 I 1 I

1 I 1 I

3 I 3 I

3 I 3 I

Loop 3

113331 R I R I εε

Note:From the calculation,

sometimes we get

negative value of

current. This negative

sign indicates thatthe direction of the

actual current is

opposite to the

direction of the

current drawn.

PHYSICS CHAPTER 18



78

For the circuit in Figure 18.26, Determine the current and its direction

in the circuit.

Example 18.17 :

Figure 18.26

1.15

.226

50.8 2,V1.51

4,V5.01

PHYSICS CHAPTER 18



79

Solution :

By applying the Kirchhoff’s 2nd law, thus

IRε

I I I I I 450.8222.61.155.110.15

1.15

.226

50.8 2,V1.51

4,V5.01Loop 1

I

I

I I

PHYSICS CHAPTER 18



80

For the circuit in Figure 18.27, determine

a. the currents I 1, I 2 and I ,

b. the potential difference across the 6.7 resistor,

c. the power dissipated from the 1.2 resistor.

Example 18.18 :

Figure 18.27

8.9

9.3

V.09V21

7.6

.21

I 1 I 2 I

PHYSICS CHAPTER 18



81

Solution :

a.

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

By using the Kirchhoff’s 2nd law,

From Loop 1:

outin I I

I I I 21

8.9

9.3

V.09V21

7.6

.21

1 I 2 I

I

1 I 2 I I

A

B

Loop 1 Loop 2

(1)

IRε

11 8.92.19.312 I I I

122.17.13 1 I I (2)

PHYSICS CHAPTER 18



82

Solution :

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

I I 2.17.60.9 2

0.92.17.6 2 I I (3)

A75.1A;03.1A;72.021 I I I

R I V 2

7.603.1V

R I P 2

2.175.12

P

PHYSICS CHAPTER 18



83

Exercise 18.5 :

1. For a circuit in Figure 18.28,

Given 1= 8V,

R2= 2,

R3= 3,

R1 = 1 and I = 3 A.Ignore the internal resistance in each battery. Calculate

a. the currents I 1 and I 2.

b. the emf, 2.

ANS. : 1.0 A, 4.0 A; 17 V

Figure 18.28

3 R

1ε

2 R2ε

1 I

2 I

I

1 R

PHYSICS CHAPTER 18



84

2.

Determine the current in each resistor in the circuit shown in

Figure 18.29.(College Physics,6th edition, Wilson, Buffa & Lou, Q57, p.619)

ANS. : 3.75 A; 1.25 A; 1.25 A

Figure 18.29

0.4

0.4V.05

V.05

.04V01

PHYSICS CHAPTER 18



85


Explain the principle of a potential divider.

Apply equation of potential divider ,

Learning Outcome:

18.8 Potential divider (½ hour)

w w w . k m

s . m a t r i k . e d

u . m y / p h y s i c s

PHYSICS CHAPTER 18



86

A potential divider produces an output voltage that is a fraction

of the supply voltage V . This is done by connecting tworesistors in series as shown in Figure 18.30.

Since the current flowing through each resistor is the same,

thus

18.8 Potential divider

V

1V

1 R I

2V

2 R I

21eff R R R

eff R

V I

and21 R R

V I

Figure 18.30

PHYSICS CHAPTER 18



87

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

by

Similarly,

Resistance R1 and R2 can be replaced by a uniform

homogeneous wire as shown in Figure 18.31.

Figure 18.31

11 IRV (18.25)

(18.26)

V

I

2l 1l I

BA C

2V 1V

PHYSICS CHAPTER 18



88

The total resistance, RAB in the wire is

Since the current flowing through the wire is the same, thus

A

ρl RCBACAB R R R

A

ρl

A

ρl R 21

AB

and

AB R

V I

21 l l A

ρ

V I

21AB l l A

ρ R

PHYSICS CHAPTER 18



89

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

length l 1 is given by

Similarly,

AC1 IRV A ρl

l l A

ρV V 1

21

1

(18.27)

(18.28)

Note:

From Ohm’s law,

l V

A

ρl I IRV

PHYSICS CHAPTER 18



90

For the circuit in Figure 18.32,

a. calculate the output voltage.

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

determine the reading of the voltmeter.

Example 18.19 :

Figure 18.32

0004

V21

0008

outV

PHYSICS CHAPTER 18



91

Solution :

a. The output voltage is given by

b. The connection between the voltmeter and 4000 resistor is

parallel, thus the equivalent resistance is

Hence the new output voltage is given by

Therefore the reading of the voltmeter is 2.4 V.

V12;4000;8000 21 V R R

V R R

RV 21

2out

V0.4outV

4000

1

4000

11

eq R

1240008000

4000outV

2000eq R

V4.2outV

1220008000

2000outV

PHYSICS CHAPTER 18



92


Explain principles of potentiometer and Wheatstone

Bridge and their applications.

Use related equations,

and

Learning Outcome:

18.9 Potentiometer and Wheatstone bridge (½ hour)

w w w . k m


u . m y / p h y s i

c s

PHYSICS CHAPTER 18



93

18.9.1 Potentiometer

Consider a potentiometer circuit is shown in Figure 18.33.

The potentiometer is balanced when the jockey (sliding contact)

is at such a position on wire AB that there is no current

through the galvanometer . Thus

18.9 Potentiometer and Wheatstone bridge

Figure 18.33

(Driver cell -accumulator)

Jockey

V

BAC

xV

I

G

+ -

I

I I

Galvanometer reading = 0

PHYSICS CHAPTER 18



94

When the potentiometer in balanced, the unknown voltage

(potential difference being measured) is equal to the

voltage across AC.

Potentiometer can be used to

compare the emfs of two cells.

measure an unknown emf of a cell. measure the internal resistance of a cell.

Compare the emfs of two cells

In this case, a potentiometer is set up as illustrated in Figure18.34, 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.

PHYSICS CHAPTER 18X



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

Figure 18.34

X

BA I

G

I

(2)

(1)

2εS

I I

1ε

CJ

D1l 2l

PHYSICS CHAPTER 18



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 V ε

ACAC IRV where

A

ρl R 1

ACand

11 l A

ρI ε (1)

then

AD2 V ε

ADAD IRV where

A

ρl R 2

ADand

22 l A

ρI ε (2)

then

PHYSICS CHAPTER 18



97

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

Measure an unknown emf of a cell By using the same circuit shown in Figure 18.34, the value of

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

A standard cell is one in which provides a constant and

accurately known emf . Thus the emf 2 can be calculated byusing the equation (18.29).

(18.29)

PHYSICS CHAPTER 18



98

Measure the internal resistance of a cell

Consider a potentiometer circuit as shown in Figure 18.35.

Figure 18.35

ε

BA I

G

I

1ε

0l C

J

S R

r

I I

PHYSICS CHAPTER 18



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 theemf 1 exactly balances the emf from the accumulator

(galvanometer reading is zero) at point C. Hence

After the switch S is closed, the current I 1 flows through theresistance box R and the jockey J moved until the galvanometer

reading is zero (balanced condition) at point D as shown in

Figure 18.36.

AC1 V ε

ACAC IRV where

A

ρl

R

0

ACand

01 l A

ρI ε (1)

then

PHYSICS CHAPTER 18



100

Figure 18.36

ε

BA I

G

I

1ε

J

S

R

r

I I

1 I

Dl

1 I

1 I

1 I

1 I

PHYSICS CHAPTER 18

H



101

Hence

From the equation of emf,

ADV V

ADAD IRV where

A

ρl

RADand

l A

ρI V (2)

then

r I V ε 11

1

1

I

V εr

R

V I 1and

RV

V εr 1 (3)

PHYSICS CHAPTER 18

B b tit ti (1) d (2) i t th (3) t



102

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

The value of internal resistance, r is determined by plotting

the graph of 1 /l against 1 /R . Rearranging eq. (4) :

R

l

l l r 0

Rl

l r 10 (4)

c xm y Then compare with

00

111

l Rl

r

l

PHYSICS CHAPTER 18



103

Therefore the graph is straight line as shown in Figure 18.37.

0

,Gradientl

r m

0

1

l

R

1

l

1

0

Figure 18.37

PHYSICS CHAPTER 18Example 18 20 :



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 18.38.

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 18.20 :

Figure 18.38

A

SO

GB

X

Y

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 Ytouches 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 18

S V0112001

Rl



105

Solution :

a. Given

When G = 0 (balance condition), thus

V0.1;12m;00.1 BOSOS ε Rl m50.0m;75.0 OX1OY1 l l

Aε

SO

GBε

X Y

0

OY1l

OX1l

Since wire OS is uniform thus

OS

OS

1OXOX1 R

l

l R

and

0.61200.1

50.0OX1 R

0.91200.1

75.0OY1 R

BOX1 εV OX11OX1 R I V and

1 I 1 I

1 I

1 I 1 I

BOX11 ε R I 0.10.61 I

A17.01 I

PHYSICS CHAPTER 18

S l ti V0112001

Rl



106

Solution :

a. Therefore the potential difference across OY is given by

b. Given

V0.1;12m;00.1 BOSOS ε Rl

OY11OY1 R I V 0.917.0OY1V

m625.0m;00.1 OX2OY2 l l

Aε

SO

GBε

X

Y

0

OY2l

OX2l

2 I 2 I

2 I

2 I 2 I Since wire OS is uniform thus

OS

OS

2OXOX2 R

l

l R

and

5.712

00.1

625.0OX2 R

121200.1

00.1OY2 R

PHYSICS CHAPTER 18

S l ti V0112001

Rl



107

Solution :

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) :


BOX2 εV OX22OX2 R I V and

BOX22 ε R I 0.15.72 I

A13.02 I

OY22OY2 R I V 1213.0OY2V

r R I εA

)( 1OY1A r R I ε

r ε 0.917.0A(1)

PHYSICS CHAPTER 18

S l ti V0112001

Rl



108

Solution :

c. For case in the question (b) :

(1) = (2):

d. The emf of cell A is


)( 2OY2A r R I εr ε 1213.0A

(2)

r r 1213.00.917.0

r ε 0.917.0A

65.00.917.0Aε

PHYSICS CHAPTER 18

18 9 2 Wh t t b id



109

It is used to measured the unknown resistance of the

resistor . Figure 18.39 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 balanced when no currentflows through the galvanometer .

18.9.2 Wheatstone bridge

ε

BA G

C

D

1 R2 R

3 R x R

0

I I

2 I

1 I

2 I

1 I

Figure 18.39



PHYSICS CHAPTER 18

The application of the Wheatstone bridge is Metre Bridge



111

The application of the Wheatstone bridge is Metre Bridge.

Figure 18.40 shows a Metre bridge circuit.

The metre bridge is balanced when the jockey J is at such a

position on wire AB that there is no current through the

galvanometer . Thus the current I 1 flows through the resistance

Rx and R but current I 2 flows in the wire AB.

0

Accumulator

Jockey

Thick copper

strip

(Unknown

resistance) (resistance box)

Wire of uniform

resistance

x R

A

ε

G

B

R

J

2l 1l

Figure 18.40

I I

1 I

2 I

1 I

PHYSICS CHAPTER 18

Let V : p d across R and V : p d across R



112

Let V x : p.d. across Rx and V : p.d. across R,

At balance condition,

By applying Ohm’s law, thus

Dividing gives

AJx V V JBV V and

AJ2x1 R I R I JB21 R I R I and

A

ρl R 1

AJJB2

AJ2

1

x1

R I

R I

R I

R I where and

A

ρl R 2

JB

A

ρl

A

ρl

R

R

2

1

x

(18.31)

PHYSICS CHAPTER 18



113

An unknown length of platinum wire 0.920 mm in diameter is placed

as the unknown resistance in a Wheatstone bridge as shown inFigure 18.41.

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 10 8 m.

Example 18.21 :

Figure 18.41

PHYSICS CHAPTER 18

Solution : ;046;038m;109200 3

RRd



114

Solution :

At balance condition, the ammeter reading is zero thus theresistance of the platinum wire is given by

From the definition of resistivity, thus

;0.46;0.38m;10920.0 21 R Rd ;mΩ106.10;48.3 8

3 ρ R

1

2

3

x

R

R

R

R

21.4x R

0.38

0.46

48.3

x R

l

A R ρ x

4

2d Aand

l d R ρ

4

2x

l 410920.021.4106.10

238

PHYSICS CHAPTER 18

Exercise 18 6 :



115

Exercise 18.6 :1. In Figure 18.42, PQ is a uniform wire of length 1.0 m and

resistance 10.0 .

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

2S

1ε

PQ

G

2ε

T

1 R

2 R

1S

Figure 18.42

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 balancelength 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 18



116

R

2. The circuit shown in Figure 18.43 is known as a Wheatstone

bridge.

Determine the value of the resistor R such that the currentthrough the 85.0 resistor is zero.


ANS. : 7.50

Figure 18.43



PHYSICS CHAPTER 18



Next Chapter…CHAPTER 19:

Magnetic Field

w w . k m


u . m y / p h y s i c s

Documents

Chapter18 Student 2