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(C) The potentiometer wire has a total resistance of $1 \Omega$. Since the sliding contact is in the middle,the wire is divided into two parts,each ha

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$1.0$

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(C) The potentiometer wire has a total resistance of $1 \Omega$. Since the sliding contact is in the middle,the wire is divided into two parts,each having a resistance of $0.5 \Omega$.One part of the wire $(0.5 \Omega)$ is in parallel with the external resistance $R_e = 2 \Omega$.The equivalent resistance of this parallel combination is $R_p' = \frac{0.5 	imes 2}{0.5 + 2} = \frac{1}{2.5} = 0.4 \Omega$.This parallel combination is in series with the other part of the potentiometer wire $(0.5 \Omega)$.Therefore,the total equivalent resistance of the circuit is $R_{	ext{eq}} = 0.5 \Omega + 0.4 \Omega = 0.9 \Omega$.The total current drawn from the battery is $i = \frac{V}{R_{	ext{eq}}} = \frac{0.9 	ext{ V}}{0.9 \Omega} = 1.0 	ext{ A}$.

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