Note bab 5_tf025_dc-part_b

CHAPTER 22 : ELECTRIC CURRENT AND DIRECT-CURRENT CIRCUITS

21.1 Ohm’s Law and Resistivity

21.2 Variation of resistance with temperature

21.3 Electromotive force (emf), internal resistance and

potential difference

21.4 Electrical energy and power

21.5 Resistors in series and parallel

21.6 Kirchhoff’s Laws

LEARNING OUTCOMES

a) Define emf, ξ ξ

b) Explain the different between emf of a

battery and potential difference across

the battery terminal

c) Apply voltage , rIV

21.3 Electromotive force (emf), internal resistance

and potential difference . (1/2 hour)

Electromotive Force e.m.f (ξ)

e.m.f (ξ) of a battery is the maximum PD across its terminals

when it is not connected to a circuits. ( Refer Figure a )

Terminal voltage, V is the voltage across the terminals of a battery when there is a current flowing through it. ( Refer Figure b )

SI unit ξ & V : volt ( V )

ξ

Figure a

V

Figure b

21.3 : Electromotive Force (emf), Internal Resistance &

Potential Difference

Internal Resistance (r)

In reality, when a battery is supplying current, its terminal voltage is less than its e.m.f, ξ

This reduces voltage is due to energy dissipation in the

battery. In effect, the battery has internal resistance (r).

Mathematically,

rIRI

rIV

But V = IR , so :

where ξ : e.m.f. of the battery

I : current flows through the circuit

R : external resistance

r : internal resistance

V : terminal voltage

• The internal resistance of a cell is the resistance due to

the chemicals in the cell.

• The internal resistance of an electrical generator is due

to the resistance of the coils in the generator.

• The internal resistance of a cell constitutes part of the

total resistance in a circuit.

• The maximum current that can flow out from a cell is

determined by the internal resistance of the cell.

Example 5

A battery has an e.m.f. of 12 V and an internal

resistance of 0.05 Ω. Its terminals are connected to a

load resistance of 3 Ω.

Find the current in the circuit & the terminal voltage of

the battery.

Solution

Using : rIRI

)( rRI

05.03

12

AI 93.3

The terminal voltage, IRV

)3(93.3

VV 8.11

Another alternative :

Using : rIV

rIV

)05.0(93.312

VV 8.11

LEARNING OUTCOMES

a) Apply power, P = IV and

electrical energy,W = VIt

21.4 Electrical energy and power (1/2 hour)

21.4 Electrical Energy (W) & Power (P)

The electric energy, W is the amount of energy given up

by a charge Q in passing through an electric device.

VQW

But Q = It , so : tIVW

Applying V = IR to above equation gives :

tRIW 2

R

tVW

2

)1(

)2(

As Power, P =t

W; it follows from (1) & (2) that :

IVP

RIP 2

R

VP

2

Exercise

1. Find the power of a heater if 2A of current flows

inside it & generates 3Ω of resistance. (12 W)

2. A radio operating on 12V draws 0.30A. How much

energy does it use in 4 hours ? (51840 J)

21.5 Resistors in series and parallel (1 hour)

LEARNING OUTCOMES

a) Deduce and calculate effective resistance

of resistors in series and parallel

The current in resistors are same

The PD applied across the series combination of

resistors will divide between the resistors.

321 VVVV)1(321 IRIRIRV

Resistors in Series

21.5 : RESISTORS IN SERIES AND PARALLEL

)2(EIRV

Substitute (2) into (1) :

321 IRIRIRIRE

Canceling the common I’s , we get :

321 RRRRE

Extending this result to the case of n resistors connected

in series combination:

nE RRRRR 321

To replace the resistors by one resistor which has an

equivalent resistance & maintain the same current, we

have :

Resistors in Parallel

Potential difference across the resistors are same

Current divides into different path at the junction.

321 IIII

)1(321

R

V

R

V

R

VI

Resistors connected in parallel can be replaced with an

equivalent resistor, RE that has the same potential difference V & the same total current I as the actual

resistors.

)2(ER

VI

Substitute (2) into (1) :

321 R

V

R

V

R

V

R

V

E

Canceling the common V’s , we get :

321

1111

RRRRE

Extending this result to the case of n resistors connected

in parallel combination:

nE RRRRR

11111

321

Example 6

What is the equivalent resistance of the resistors in

figure below ?

A

BR1

R2

R3

R4

Another view of the circuit :

A

B

R1

R2

R3R4

4334

111

RRR2

1

1

1

1

2

134R

Circuit has been reduced to :

A

B

R1

R2

R34

R34 & R2 in series, thus:

342234 RRR

2

3

2

11

Solution

Circuit has been reduced to :

A

B

R1R234

2341

111

RRRE 3

5

3

2

1

1

5

3ER

Example 7

Find the current in & voltage of the 10 Ω resistor shown

in Figure.

3R

1R

2R

Solution

1st, find the RE for the circuit & get the current, I flow in

the circuit

21

111

RRRp2

1

10

1

10

6

67.1pR

pE RRR 3

0.567.1

67.6ER

The current flow in the circuit :

ER

VI

67.6

18A7.2

As R3 & R1 in series combination

13 VVV

31 IRVV )5(7.218

VV 5.41

1

1

1

R

VI

10

5.4A45.0