Q#42 (Past Exam Paper – November 2016 Paper 12 Q38)
In the circuit below, P is a potentiometer of total resistance 10 Ω and Q is a fixed resistor of resistance 10 Ω. The battery has an e.m.f. of 4.0 V and negligible internal resistance. The voltmeter has a very high resistance.
The slider on the potentiometer is moved from X to Y and a graph of voltmeter reading V is plotted against slider position.
Which graph is obtained?
Solution:
Answer: A.
The circuit consists of a potentiometer and a fixed resistor connected in series.
The voltmeter is connected to the lower terminal of the fixed resistor at one side and the other side is connected to the jockey of the potentiometer.
The voltmeter measures the p.d. across these components, depending on the position of the jockey.
The resistance of the fixed resistor Q is always included in the measurement of the voltmeter while the resistance of the potentiometer depends on the position of the jockey.
From Kirchhoff’s law, the sum of the p.d. across the components is equal to the e.m.f. in the circuit.
When the jockey positioned at X, all of the resistance of the potentiometer is accounted for. So, the sum of p.d. is equal to the e.m.f. (= 4.0 V). [B and D incorrect]
As the jockey is moved towards Y, the resistance of the potentiometer being measured by the voltmeter decreases. So, the voltmeter reading decreases.
At Y, none of the resistance of the potentiometer is included on the voltmeter reading – only the resistance of Q.
Since the resistance of P and Q are equal, the e.m.f. of the source is divided equally between them. At position Y, only the p.d. across Q is being measured and this is half the value of the e.m.f. [A is correct]
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