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(C) The circuit consists of two parts connected in series.First part: Three resistors of $6\, \Omega$ each are connected in parallel. The equivalent r

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$3$

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(C) The circuit consists of two parts connected in series.First part: Three resistors of $6\, \Omega$ each are connected in parallel. The equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$, so $R_1 = 2\, \Omega$.Second part: Two resistors of $6\, \Omega$ are in series, giving $6 + 6 = 12\, \Omega$. This is in parallel with another $6\, \Omega$ resistor. The equivalent resistance $R_2$ is given by $R_2 = \frac{12 	imes 6}{12 + 6} = \frac{72}{18} = 4\, \Omega$.The total resistance between $P$ and $Q$ is $R_{eq} = R_1 + R_2 = 2\, \Omega + 4\, \Omega = 6\, \Omega$.The potential difference $V_P - V_Q = I 	imes R_{eq} = 0.5\, A 	imes 6\, \Omega = 3\, V$.

Resistances of $6\, \Omega$ each are connected in the manner shown in the adjoining figure. With the current $0.5\, A$ as shown in the figure, the potential difference $V_P - V_Q$ is .............. $V$.

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