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(D) The circuit consists of two parallel branches connected to a battery of $20 \, V$ with internal resistance $r = 1.5 \, \Omega$.Each branch has a t

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$2.5 \, V \, (V_Q > V_P)$

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(D) The circuit consists of two parallel branches connected to a battery of $20 \, V$ with internal resistance $r = 1.5 \, \Omega$.Each branch has a total resistance of $3 \, \Omega + 2 \, \Omega = 5 \, \Omega$.The equivalent resistance of the two parallel branches is $R_p = \frac{5 	imes 5}{5 + 5} = 2.5 \, \Omega$.The total resistance of the circuit is $R_{eq} = R_p + r = 2.5 \, \Omega + 1.5 \, \Omega = 4 \, \Omega$.The total current in the circuit is $i = \frac{V}{R_{eq}} = \frac{20 \, V}{4 \, \Omega} = 5 \, A$.Since the two branches have equal resistance,the current splits equally,so $i_1 = i_2 = \frac{i}{2} = 2.5 \, A$.Let $X$ be the potential at the left junction. The potential at $P$ is $V_P = V_X - i_1 	imes 3 \, \Omega = V_X - 2.5 	imes 3 = V_X - 7.5 \, V$.The potential at $Q$ is $V_Q = V_X - i_2 	imes 2 \, \Omega = V_X - 2.5 	imes 2 = V_X - 5 \, V$.The potential difference between $P$ and $Q$ is $V_Q - V_P = (V_X - 5) - (V_X - 7.5) = 2.5 \, V$.Since $V_Q - V_P = 2.5 \, V$,it follows that $V_Q > V_P$.

If in the circuit shown below,the internal resistance of the battery is $1.5 \, \Omega$ and $V_P$ and $V_Q$ are the potentials at $P$ and $Q$ respectively,what is the potential difference between the points $P$ and $Q$?

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