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Innova Junior College
Tutorial 11: DC Circuits
Assignment
1
(a) The resistance per unit length of a potentiometer wire A, of length 1.0 m, is found to vary
linearly from 9.0  m-1 at one end to 10.0  m-1 at the other, as shown in Fig. 1.1. The
p.d. applied across its ends is 2.0 V.
[4]
2.0 V
A
x
Resistance per unit length ( m-1)
10.0
9.0
0
0
1.0
x/m
Fig. 1.1
Using the above information, find the p.d. between x = 0 and x = 0.5 m.
Suggested Solution
From Fig 1.1., the area under the graph from 0 to x is the resistance of the wire from 0 to
x:
Resistance between x = 0 to 0.5 m = Area under graph from x = 0 to 0.5 m
(9.0  9.5)(0.5)
=
 4.63 
2
Similarly, the resistance between x = 0 and x= 1.0 m is the area under the graph from
x = 0 to x = 1.0 m:
Resistance between x =0 to 1.0 m = Area under graph from x =0 to 1.0 m
=
(9.0  10.0)(1.0)
 9.5 
2
The same current flows through the whole wire. Hence potential divider rule applies:
V0 to 0.5
E

R0 to 0.5 RT
V0 to 0.5 2.0

4.625 9.5
V0 to 0.5 = 0.973 V
(b) The potentiometer wire A is now connected at mid point to a second potentiometer wire
B, of constant resistance per unit length and of length 1.0 m, as shown in Fig. 1.2.
Assume all cells have negligible internal resistance.
2.0 V
0.5 m
P
Q
A
galvanometer
R
B
S
L
1.5 V
Fig. 1.2
What is the length L when there is no current in the galvanometer?
When no current flow through the galvanometer, VPQ = VRS
And since VPQ = V0 to 0.5
= 0.973 V
VRS = 0.973 V
Since the same current flows through wire B,
l
V = IR and R 
A
I l
Therefore V 
A
Vαl
VRS lRS

E
ltotal
0.973 l RS

1.5
1.0
lRS = 0.65 m
[3]
2
A student sets up a potentiometer circuit as shown in Fig. 2. The e.m.f. of the cells and
resistance of the resistors are as indicated, and the galvanometer G shows null deflection.
2.0 V, r = 0 
R
150 cm, 50 
A
B
Y
R1
Fig. 2
(a) If R is such that the potential difference per centimetre along the 150-cm long wire AB is
0.010 V cm–1, find
(i) the required value of R,
(ii) the power provided by the driver cell, and
(iii) the power dissipated in the potentiometer wire.
(b) Discuss whether the value of R1 affects the balance point.
Suggested solution
(a) (i) Given that the p.d. per cm of wire AB is 0.010 V cm-1,
VAB = 150 x 0.010 =1.5 V
The resistor R and the wire AB is connected in series:
VAB
E

RAB RT
1.5
2.0

50 50  R
R = 17 Ω
(ii) Power by driver cell
1.5
= IT E 
 2.0
50
= 60 mW
(iii) Power dissipated by wire
V 2 1.52
 R 
R
50
= 45 mW
[3]
[2]
[2]
[2]
(b) For balance point, there is no current flowing through the galvanometer, i.e. the circuitry
can be seen as the one below instead:
R
2.0 V, r = 0 
150 cm, 50 
A
Y
C
B
E
R1
D
For the lower circuitry, it behaves like an open circuit. This implies no current flow
through R1. Thus, its resistance does not affect the balance point.