Then, the resistor colour code will be as  
shown below  
Solution  
Given  
퐸1 = 1.63푉 , 푙1 = 85푐푚, 퐸2 = 1.07푉  
Then, the new length 푙1 is obtained from the  
formula:  
THE POTENTIAL MEETRE  
퐸1  
퐿1  
=
Potentiometer is a device used to compare  
the e.m.f. (electromotive force) of two cells,  
to measure the internal resistance of a cell,  
and potential difference across a resistor  
The potential metre is a variable resistor  
with three terminals.  
퐸2  
퐿1  
퐸1  
푙2 =  
푙1  
퐸2  
1.07푉  
푙2 =  
× 85푐푚  
1.63푉  
푙2 = 55.79푐푚  
The new balancing length will be 55.79푐푚  
In potentiometer, a contact is moved to  
select desired proportions of the total  
voltage across them.  
Example 02  
A Daniel cell of e.m.f. 1.08V gives a  
balance point 0.52m long a potentiometer  
wire. The potential difference across its ends  
of a resistor of value 5Ω gives a balance  
point 0.78m long the potentiometer wire.  
What is the current through the 5Ω resistor?  
퐸1  
퐸2  
퐿1  
퐿1  
=
Solution  
It is clear that from the experiment, as the  
e.m.f increases, the length of the  
Consider the diagram below  
potentiometer wire increases as well.  
Example 01  
The balance length of a potentiometer wire  
for a cell of e.m.f, 퐸1 = 1.63푉 is 85cm. If  
the cell is replaced by another one of 퐸2 =  
1.07푉, Calculate the new balance length.  
By using voltmeter – ammeter method  
The Battery, a switch, voltmeter, ammeter  
variable  
resistor  
and  
resistor,  
R
are  
connected as shown below. The voltmeter is  
connected across (parallel) to the resistor.  
퐸1  
퐸2  
퐿1  
퐿1  
=
=
1.08  
0.52  
푉
0.78  
푉 = 1.62푉  
With the switch open the voltmeter reading  
is recorded, when the switch is closed the  
reading V, and ammeter reading I are  
recorded and the graph of v against I is  
plotted, from the graph, the slope represents  
the resistance R.  
푝. 푑  
푐푢푟푟푒푛푡 =  
퐼 =  
푟푒푠푖푠푡푎푛푐푒  
1.62푉  
5Ω  
푐푢푟푟푒푛푡 = 0.324퐴  
Observations  
When the switch is open, no current flows  
through the resistor and therefore both the  
ammeter and the voltmeter reading is zero.  
When the current through the resistor  
Applications  
Potentiometers  
with  
circular  
resistance  
tracks are used as balance controls in audios  
amplifier. This is because when the e.m.f  
increases the balancing point also increases.  
increases,  
the voltage  
across it  
also  
increases. The graph of V against I is a  
straight line whose gradient gives resistance.  
MEASUREMENT OF RESISTANCE  
∆푉  
푆푙표푝푒 =  
∆퐼  
There are two methods used to measure  
resistance of a conductor in the laboratory,  
namely;  
푆푙표푝푒 = 푅  
(a) By using voltmeter – ammeter method  
(b) By using a whetstone (metre) bridge  
Note: The resistance obtained cannot be  
accurate since the voltmeter takes some little  
current, thus allowing not all of it flows  
through the resistor.  
By using the Wheatstone bridge  
The Wheatstone bride consists of four  
resistors and a galvanometer, as shown in  
figure below. The operation of the bridge  
involves making adjustments to one or two  
of the resistors until there is no deflection in  
the galvanometer.  
Therefore, when the bridge is balanced  
P
R
X
=
Q
METRE BRIDGE  
The figure below shows a practical form of  
Wheatstone bridge known as metre –  
bridge  
Therefore, when the bridge is balanced  
K
L
=
M
N
The Wheatstone bridge is more accurate in  
measuring resistance than the voltmeter –  
ammeter  
method  
because  
the  
value  
obtained by using Wheatstone bridge  
method depend on the accuracy of the  
current  
–
measuring  
instrument  
(galvanometer) used.  
The resistance can be obtained as follows  
R
RX  
=
L1  
L2  
Taking the average of the values eliminates  
errors due to non – uniformity of the wire  
and end corrections.  
Example 01  
In an experiment to determine the resistance  
a nichrome wire using the metre bridge, the  
balance point was found to be at 38cm mark.  
If the resistance in the right gap needed to  
balance the bridge was 25Ω, calculate the  
value of the unknown resistor.  
Solution  
LB  
LA  
RX =  
× R  
RX = X , R = 10Ω, LB = 52cm and  
LA = 48cm  
Therefore  
LB  
RX =  
× R  
LA  
48cm  
Solution  
RX =  
× 10Ω  
52cm  
Since AB = 10 cm and AC = 38 cm  
length BC = 100 cm − 38 cm  
RX = 9.23Ω  
Therefore, the resistance X is 9.23Ω  
length BC = 68 cm  
EQUIVALENT RESISTANCE  
R
25 Ω  
=
38 cm  
62 cm  
Equivalent resistor is a single device which  
can be used to replace a number of resistors  
in a circuit.  
38 cm × 25 Ω  
R =  
62 cm  
Equivalent resistance is the value of the  
single resistance which can replace a  
number of resistances in a circuit.  
R = 15.32Ω  
This is the resultant (total) resistance in the  
circuit. It is also called the effective  
resistance.  
Example 02  
A metre bridge is set up as shown in the  
figure below using a standard 10Ω resistor.  
The galvanometer shows zero deflection  
when the jockey contact is at 48cm from end  
A. Determine the resistance of resistor, X.  
RESISOTR CONNECTIONS  
In the circuit, the resistors can be connected  
in major two ways such as:  
(i) Series connections