The length of a potentiometer wire is 2.5 m | vedclass

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(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing

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$6.4$

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(C) The total resistance in the primary circuit is $R_{total} = R_{wire} + R_{series} = 8 \ \Omega + 242 \ \Omega = 250 \ \Omega$. The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{2.5 \ V}{250 \ \Omega} = 0.01 \ A$. The potential drop across the entire wire is $V_{wire} = I 	imes R_{wire} = 0.01 \ A 	imes 8 \ \Omega = 0.08 \ V$. The potential gradient $k$ is the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{0.08 \ V}{2.5 \ m} = 0.032 \ V/m$. The potential difference across a length $l = 20 \ cm = 0.2 \ m$ is $V' = k 	imes l = 0.032 \ V/m 	imes 0.2 \ m = 0.0064 \ V = 6.4 \ mV$.

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