A potentiometer wire of length 100 cm and re | vedclass

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(C) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r_{internal} = 3 \ \Omega + 8 \ \Omega + 1 \ \Omega = 12 \ \Omega$.The

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$0.50$

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(C) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r_{internal} = 3 \ \Omega + 8 \ \Omega + 1 \ \Omega = 12 \ \Omega$.The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{4 \ V}{12 \ \Omega} = \frac{1}{3} \ A$.The potential difference across the potentiometer wire is $V_{wire} = I 	imes R_{wire} = \frac{1}{3} \ A 	imes 3 \ \Omega = 1 \ V$.The potential gradient $K$ of the wire is $K = \frac{V_{wire}}{L} = \frac{1 \ V}{100 \ cm} = 0.01 \ V/cm$.The e.m.f. $E$ of the cell balanced at length $\ell = 50 \ cm$ is $E = K 	imes \ell = 0.01 \ V/cm 	imes 50 \ cm = 0.50 \ V$.

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