1. 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
2. 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 , r
I
V
21.3 Electromotive force (emf), internal resistance
and potential difference . (1/2 hour)
3. 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
4. Internal Resistance (r)
In reality, when a battery is supplying current, its
terminal voltage is less than its e.m.f, ξ
5. This reduces voltage is due to energy dissipation in the
battery. In effect, the battery has internal resistance (r).
Mathematically,
r
I
R
I
r
I
V
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
6. • 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.
7. 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 : r
I
R
I
)
( r
R
I
05
.
0
3
12
A
I 93
.
3
8. The terminal voltage, IR
V
)
3
(
93
.
3
V
V 8
.
11
Another alternative :
Using : r
I
V
r
I
V
)
05
.
0
(
93
.
3
12
V
V 8
.
11
9. LEARNING OUTCOMES
a) Apply power, P = IV and
electrical energy,W = VIt
21.4 Electrical energy and power (1/2 hour)
10. 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.
V
Q
W
But Q = It , so : t
I
V
W
Applying V = IR to above equation gives :
t
R
I
W 2
R
t
V
W
2
)
1
(
)
2
(
11. As Power, P =
t
W
; it follows from (1) & (2) that :
I
V
P
R
I
P 2
R
V
P
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)
12. 21.5 Resistors in series and parallel (1 hour)
LEARNING OUTCOMES
a) Deduce and calculate effective resistance
of resistors in series and parallel
13. The current in resistors are same
The PD applied across the series combination of
resistors will divide between the resistors.
3
2
1 V
V
V
V
)
1
(
3
2
1
IR
IR
IR
V
Resistors in Series
21.5 : RESISTORS IN SERIES AND PARALLEL
14. )
2
(
E
IR
V
Substitute (2) into (1) :
3
2
1 IR
IR
IR
IRE
Canceling the common I’s , we get :
3
2
1 R
R
R
RE
Extending this result to the case of n resistors connected
in series combination:
n
E R
R
R
R
R
3
2
1
To replace the resistors by one resistor which has an
equivalent resistance & maintain the same current, we
have :
15. Resistors in Parallel
Potential difference across the resistors are same
Current divides into different path at the junction.
3
2
1 I
I
I
I
)
1
(
3
2
1
R
V
R
V
R
V
I
16. 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
(
E
R
V
I
Substitute (2) into (1) :
3
2
1 R
V
R
V
R
V
R
V
E
Canceling the common V’s , we get :
3
2
1
1
1
1
1
R
R
R
RE
17. Extending this result to the case of n resistors connected
in parallel combination:
n
E R
R
R
R
R
1
1
1
1
1
3
2
1
Example 6
What is the equivalent resistance of the resistors in
figure below ?
A
B
R1
R2
R3
R4
18. Another view of the circuit :
A
B
R1
R2
R3
R4
4
3
34
1
1
1
R
R
R
2
1
1
1
1
2
1
34
R
Circuit has been reduced to :
A
B
R1
R2
R34
R34 & R2 in series, thus:
34
2
234 R
R
R
2
3
2
1
1
Solution
19. Circuit has been reduced to :
A
B
R1
R234
234
1
1
1
1
R
R
RE 3
5
3
2
1
1
5
3
E
R
Example 7
Find the current in & voltage of the 10 Ω resistor shown
in Figure.
3
R
1
R
2
R
20. Solution
1st, find the RE for the circuit & get the current, I flow in
the circuit
2
1
1
1
1
R
R
Rp
2
1
10
1
10
6
67
.
1
p
R
p
E R
R
R 3
0
.
5
67
.
1
67
.
6
E
R
21. The current flow in the circuit :
E
R
V
I 67
.
6
18
A
7
.
2
As R3 & R1 in series combination
1
3 V
V
V
3
1 IR
V
V )
5
(
7
.
2
18
V
V 5
.
4
1
1
1
1
R
V
I
10
5
.
4
A
45
.
0
