Solution  
The, terminal p.d V = 1.0V, R = 5Ω,  
1.0  
퐼 =  
퐴
5
= 0.2 A  
퐸 = 퐼 (푅 + 푟)  
From  
퐸 = 0.25 (5.5 + 푟)  
E = Ir + V  
E = 0.2 × 2 + 1.0 V  
= 1.4V  
E = 1.375 + 0.25r … … … . . (i)  
Case II: Consider the circuit diagram below  
Example 04  
A battery of e.m.f 40V and internal  
resistance of 5Ω, is connected to a resistor of  
15Ω, Calculate the terminal p.d  
Solution  
E = 40V, r = 5Ω, R = 15Ω, V =?  
퐸 = 퐼 (푅 + 푟)  
퐸 = 0.3 (4.5 + 푟)  
From  
E = I ( r + R )  
ꢀ
I =  
E = 1.35 + 0.3r … … . . (ii)  
ꢁ+ꢂ  
40  
Since the e.m.f is the same, the two  
equations can be equated as follows  
I =  
= 2A  
15+5  
V= IR  
= 2 X 15  
= 30V.  
1.375 + 0.25푟 = 1.35 + 0.3푟  
푟 = 0.5Ω  
(b)Determine the e.m.f of a cell  
Example 05  
From equation (i)  
퐸 = 1.375 + 0.25푟  
The terminal potential difference of a battery  
is 12V when an external resistance of 20Ωis  
connected and 13.5V when an external  
resistance of 45Ω is connected. Calculate the  
e.m.f and internal resistance of a battery.  
퐸 = 1.375 + 0.25 × 0.5  
퐸 = 1.5푉  
Example 03  
Soln  
퐸
퐼 =  
… … … . (푖)  
A cell of unknown e.m.f, E and internal;  
resistance 2Ω is connected to a 5Ω,  
resistance. If the terminal P.d V is 1.0V,  
Calculate, E  
푅 + 푟  
푉
( )  
… … . . 푖푖  
I =  
푅
= 5Ω  
푉
푅
퐸
=
푡표푡푎푙 퐸. 푀. 퐹  
푅 + 푟  
푐푢푟푟푒푛푡 =  
푡표푡푎푙 푟푒푠푖푠푡푎푛푐푒  
Case I, when V = 12V, R = 20Ω  
3
5
푐푢푟푟푒푛푡 =  
12  
20  
퐸
=
푐푢푟푟푒푛푡 = 0.6퐴  
20 + 푟  
(
)
E = 12 + 0.6r … … . . 푖푖푖  
(b) Parallel connection  
Case II, when V = 13V, R = 45Ω  
13  
45  
퐸
=
45 + 푟  
( )  
E = 13.5 + 0.3r … … . . 푖푣  
Equate (iii) and (iv)  
Solution  
Total e.m.f = 1.5V  
12 + 0.6r = 13.5 + 0.3r  
r = 5Ω and E = 15 V  
1
1
1
=
=
+
푟푇  
푟
푟2  
1
1
1
1
+
Example 06  
푟푇  
2
2
Two cells each having an e.m.f of 1.5V and  
internal resistances of 2Ω are connected (a)  
In series and (b) in parallel. Find the current  
in each case when the cells are connected to  
an external resistor of resistance 1Ω  
푟푇 = 1Ω  
Resistance in series = 푅1 + 푅2  
= 1Ω + 1Ω  
= 2Ω  
Solution  
(a) Series connection  
푡표푡푎푙 퐸. 푀. 퐹  
푐푢푟푟푒푛푡 =  
푡표푡푎푙 푟푒푠푖푠푡푎푛푐푒  
1.5  
푐푢푟푟푒푛푡 =  
2
푐푢푟푟푒푛푡 = 0.75퐴  
Example 07  
In the figure below, the lamp in the two  
circuits is identical and the cells have the  
same e.m.f. Explain why the lamps in circuit  
B may be brighter than those in circuit A  
Total E.M.F = 1.5 + 1.5  
= 3V  
when  
the  
switches  
are  
closed  
Resistance in series = 푅1 + 푅2 + 푅3  
simultaneously.  
= 2Ω + 2Ω + 1Ω  
4.25Ω resistor. Calculate the current flowing  
in 4.25Ω resistor.  
Solution  
Total internal resistance  
1
1
1
=
+
푟
1.5  
푟 = 0.75Ω  
1.5  
Answer  
Total resistance in the circuit = r + R  
푅 = (0.75 + 4.25)Ω  
푅 = 5Ω  
In A, the resistance offered by the circuit is  
higher since the cells are in series in B, since  
the cells are I parallel the effective of the  
cell is less, therefore more current is drawn.  
Now  
퐸
퐼 =  
푅 + 푟  
Example 08  
2
퐼 =  
5
The  
potential  
difference  
between  
the  
terminals of a cell in an open circuit was  
2.2V and when it was connected through a  
resistance of 5Ω, the potential difference fell  
to 1.8V. Find the internal resistance of the  
cell.  
퐼 = 0.4퐴  
Example 10  
Four cells each of e.m.f 2V and internal  
resistance 0.1Ω are connected in series. The  
combination is connected in series to an  
ammeter of negligible resistance, a 1.6Ω  
resistor and unknown resistance푅1. The  
current in the circuit is 2A.  
Solution  
Given: E = 2.2V; R = 5Ω; V = 1.8V; r= ?  
(퐸 − 푉)  
Draw a labeled circuit diagram for the above  
arrangement and calculate:  
푟 =  
× 푅  
푉
(a) The total resistance in the circuit  
(b) The total e.m.f  
(2.2 − 1.8)  
1.8  
푟 =  
× 5Ω  
(c) The value of 푅1.  
푟 = 1.11Ω  
(d) The p.d across 푅1.  
Example 09  
Solution  
Two cells having an e.m.f of 2V and internal  
resistance of 1.5Ω are connected in parallel.  
The arrangement is then connected to a  
Solution  
V = Current × resistance  
V = 2A × 2Ω  
V = 4V  
Example 13  
Why does the voltage across the terminals of  
a cell fall when it is delivering current?  
Answers  
(a) The total e.m.f  
E.m.f = (2+2+2+2+2) V  
= 8V  
The voltage across the terminals of a cell  
falls when delivering current due to the  
presence of internal resistance of a cell. The  
e.m.f of a cell not only provides the p.dto  
drive the current through the resistor but  
also through itself. The lost p.d is known as  
the lost voltage.  
(b) The value of 푅1.  
Solution  
Internal resistance = (0.1 + 0.1 + 0.1 +  
Example 14  
0.1)Ω  
Would you expect two identical cells in  
parallel to drive more current through a  
resistor than one cell does?  
푟 = 0.4Ω  
Total resistance in series = 1.6 + 0. 4 + 푅1  
= 2 + 푅1  
Answers  
From  
NO, because;  
푡표푡푎푙 푒. 푚. 푓  
퐼 =  
The p.d across the resistor is the same with  
two cells in parallel as it is with only one  
cell. The current through the resistor is thus  
the same in each case. Each cell supplies  
about half the current.  
푡표푡푎푙 푟푒푠푖푠푡푎푛푐푒  
8
2 =  
푅1 + 2  
푅1 = 2Ω  
Example 15  
Why do two identical cells in series drive  
more current through a resistor than one  
does and why do they not double the  
current?  
(c) The total resistance in the circuit  
Total internal resistance = 0.4 × 0.1Ω =  
0.4Ω  
Total resistance in series = 1.6 + 04 +  
푅1  
Answers  
Two identical cells in series provide twice  
the e.m.f as one cell does.  
= 2 + 푅1  
= 2 + 2  
= 4Ω  
The current in the circuit does not double  
because the circuit now includes the internal  
resistance of two cells rather than one cell  
(d) The p.d across 푹ퟏ.