CBSE Class 12 · 2004 · Current ElectricityShort Answer

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The circuit shows a potentiometer used to measure a small EMF produced by a thermocouple connected between X and Y. The driver cell C has EMF

2 V

and negligible internal resistance; the potentiometer wire PQ is

1.00 m

long with resistance

5 Ω

. The balance point S is

400 mm

from P. Calculate the EMF

V

generated by the thermocouple.

Answer: $V \approx 4 mV$ .

Solution

The thermocouple EMF equals the potential drop along PS. With the series resistance of the circuit (shown in the figure) the current makes the drop across the whole

1 m

wire small; the potential gradient is

k = \frac{(drop across PQ)}{1 m}

, and

V = k \times (balancing length) = k \times 0.40 m

. For the archive circuit the wire drop is

10 mV

, giving

k = 10 mV m^{- 1}

and

V = 10 \times 0.40 = 4 mV .

CBSE Class 12 Board · 2004 · Subjective

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