Resistances of $6\, ohm$ each are connected in the manner shown in adjoining figure. With the current $0.5\,ampere$ as shown in figure, the potential difference ${V_P} - {V_Q}$ is .............. $V$
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${V_p} - {V_q} = \left( {\frac{6}{3} + \frac{{12 \times 6}}{{12 + 6}}} \right)\,(0.5) = (2 + 4)\,(0.5) = 3\,V$
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