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If in the circuit shown below, the internal resistance of the battery is $1.5 Ω$ 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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$4 v o l t (V_{P} > V_{Q})$

$4 v o l t (V_{Q} > V_{P})$

$2.5 v o l t (V_{Q} > V_{P})$

answer is D.

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Detailed Solution

$R_{e q} = \frac{5}{2} Ω$

$i = \frac{20}{\frac{5}{2} + 1.5} = 5 A$

Potential difference between X and P,

$V_{x} - V_{P} = (\frac{5}{2}) \times 3 = 7.5 V (i)$

$V_{x} - V_{Q} = \frac{5}{2} \times 2 = 5 V (i i)$

On solving $(i) & (i i)$ $V_{P} - V_{Q} = - 2.5 v o l t; V_{Q} > V_{P} .$

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