First slide

Electrical measuring instruments

Question

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

Difficult

A

995 $Ω$
B

1995 $Ω$
C

2995 $Ω$
D

none of these

Solution

The voltage per unit light of the metre wire PQ is . $(\frac{6.00 mV}{0.600 m})$ 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 $i = \frac{10 mV}{5 Ω} = 2 mA$ .
The resistance $R = \frac{(2 V - 10 mV)}{2 mA} = \frac{(2000 m V - 10 mV)}{2 mA} = \frac{1990 mV}{2 mA} = 995 Ω$

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