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If in the circuit shown below. The internal resistance of the battery is 1.5 $Ω$ and $V_{P} a n d V_{Q}$ are the potentials at P and Q respectively, the potential difference between the points P and Q is

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$4 V (V_{Q} > V_{P})$

$4 V (V_{P} > V_{Q})$

$2.5 V (V_{Q} > V_{P})$

answer is D.

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Detailed Solution

If in the circuit shown below, the internal resistance of the battery is ..

$i = \frac{V}{R_{e f f}} = \frac{20}{\frac{5}{2} + 1.5} = \frac{20}{4} = 5 A$

$i_{1} = i_{2} = \frac{5}{2} = 2.5 A$

$V_{A} - V_{P} = 2.5 \times 3 = 7.5 V ... (1)$

$V_{A} - V_{Q} = 2.5 \times 2 = 5 V$ $.... (2)$

$(1) - (2) \to V_{Q} - V_{P} = 2.5 V$

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