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Q.

Figure shows a simple potentiometer circuit for measuring a small e.m.f. produced by a thermocouple. The meter wire PQ has a resistance 5 Ω and the driver cell has an e.m.f. of 2 V. If a balance point is obtained 0.600 m along PQ when measuring an e.m.f. of 6.00 mV, what is the value of resistance R

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a

995 Ω

b

1995 Ω

c

2995 Ω

d

none of these

answer is A.

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Detailed Solution

The voltage per unit light of the metre wire PQ is .6.00mV0.600m i.e. 10 mV/m. Length of PQ wire is 1 meter. Hence potential difference across the meter wire is 10 mV/m x 1m = 10 mV. The current drawn from the driver cell is i=10mV5Ω=2mA. The resistance R=(2V−10mV)2mA=(2000mV−10mV)2mA=1990mV2mA=995Ω

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