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 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
Text Solution
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The voltage per unit light of the meter wire PQ is
$\left(\frac{6.00\,mV}{0.600\,m} \right)$
i.e. $10 \text{ mV} / \text{m}$ .
Hence potential difference across the meter wire PQ is $10\ mV /m \times 1m = 10\ mV$ .
The current drawn from the driver cell is
$i = \frac{10 \ mV}{5 \ \Omega} = 2 \ mA$ .
The resistance R is
$R = \frac{(2V - 10 mV)}{2 mA} = \frac{1990 \ mV}{2 mA} = 995 \ \Omega$ .
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