Question 33
A circuit contains a
cell, two resistors of resistances R1 and
R2 and a variable resistor X. The cell has
negligible internal resistance.
V1 is the potential difference across the resistor
of resistance R1.
I2 is the current through the resistor of
resistance R2.
The resistance of X is
reduced.
What is the effect on V1 and I2?
V1 I2
A decreases decreases
B decreases increases
C increases decreases
D increases
increases
Reference: Past Exam Paper – June 2016 Paper 13 Q35
Solution:
Answer:
C.
The variable resistor R is
in parallel with resistor R2, and this combination is in series with
the resistor R1.
When the resistance of X
is reduced, the equivalent resistance of the parallel combination decreases.
This causes the p.d. across the parallel combination to decrease.
The sum of p.d. V1 and the p.d. across the parallel combination is equal to the e.m.f.
of the cell. V = IR. Since the resistance of the combination is reduced, the
p.d. across it is also reduced. So, V1 increases. [A and B incorrect]
As the current reaches the
junction of the parallel combination, it splits.
From Ohm’s law: I = V / R
The greater the
resistance, the small the current. Since the resistance of X has decreased,
more current flow through it. This means that the current I2 decreases.
[C correct]
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