First slide

Electrical measuring instruments

Question

A potentiometer wire PQ of 1 m length is connected to a standard cell $E_{1}$ . Another cell $E_{2}$ of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is:

Moderate

A

0.02 V/cm
B

0.01 V/cm
C

0.04 V/cm
D

0.03 V/cm

Solution

As switch S is open
$\to$ Under balancing condition, no current flows through secondary circuit.
So $V_{P J} = E_{2}$
(Pot. gradient) $= \frac{E_{2}}{l_{P J}}$
(Pot. gradient) $= \frac{1.02}{0.51} (\frac{V}{c m}) = 0.02 \frac{V}{c m}$

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