Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_{p} =$ $1 Ω$ as shown in the figure. An external resistance of $R_{0} = 2 Ω$ is connected via the sliding contact. The electric current in the circuit is

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0.9 A

0.3 A

1.0 A

1.35 A

answer is C.

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Detailed Solution

The circuit can be modified as

$\begin{array}{l} ∴ R_{eq} = 0.5 + \frac{0.5 \times 2}{2 + 0.5} = (\frac{5}{10} + \frac{10}{25}) Ω \\ = \frac{45}{50} = \frac{9}{10} = 0.9 \\ ∴ i = \frac{0.9}{0.9} = 1 A \end{array}$

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