Q#33 (Past Exam Paper – June 2016 Paper 13 Q35)
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
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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