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(D) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 20 \ \Omega + 10 \ \Omega = 30 \ \Omega$. The current flowing through

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(D) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 20 \ \Omega + 10 \ \Omega = 30 \ \Omega$. The current flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{3 \ V}{30 \ \Omega} = 0.1 \ A$. The potential difference across the wire is $V_{wire} = I 	imes R_{wire} = 0.1 \ A 	imes 20 \ \Omega = 2 \ V$. The potential gradient along the wire is defined as the potential drop per unit length: $	ext{Potential gradient} = \frac{V_{wire}}{L} = \frac{2 \ V}{10 \ m} = 0.2 \ V/m$.

$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is

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