A potentiometer wire of length 4 m and resistance 5 Ω is connected in series with a resistance of 992 Ω and a cell of e.m.f. 4 V with internal resistance 3 Ω . The length of 0.75 m on potentiometer wi

Question

A potentiometer wire of length 4 m and resistance 5 Ω is connected in series with a resistance of 992 Ω and a cell of e.m.f. 4 V with internal resistance 3 Ω . The length of 0.75 m on potentiometer wire balances the e.m.f. of

TestHub · Accepted Answer

Total resistance of the primary circuit is given by $R_{t o t a l} = R_{w} + R_{s} + r$

Substituting the given values, $R_{t o t a l} = 5 + 992 + 3 = 1000 Ω$

Current in the primary circuit is $I = \frac{E}{R _{t o t a l}} = \frac{4}{1000} = 4 \times 1 0^{- 3} A$

Potential difference across the potentiometer wire is $V_{w} = I \times R_{w} = 4 \times 1 0^{- 3} \times 5 = 20 \times 1 0^{- 3} V$

Potential gradient of the potentiometer wire is $k = \frac{V _{w}}{L} = \frac{20 \times 1 0 ^{- 3}}{4} = 5 \times 1 0^{- 3} V/m$

The e.m.f. balanced by a length $l = 0.75 m$ is $E^{'} = k \times l = 5 \times 1 0^{- 3} \times 0.75 = 3.75 \times 1 0^{- 3} V = 3.75 mV$

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