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Q.

If in the circuit shown below. The internal resistance of the battery is 1.5 and V_P and V_Q are the potentials at P and Q respectively, the potential difference between the points P and Q is

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a

zero

b

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

c

$2 .5 V (V_{Q} > V_{P})$

d

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

answer is D.

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$i = \frac{V}{R_{net}} = \frac{20}{1 .5 + 2 .5} = 5 A$
Current through APB = CQD =i₁ = i₂ = 2.5A
$\begin{array}{l} V_{A} - V_{P} = 7 .5 V \\ V_{C} - V_{Q} = 5 V \\ (V_{A} - V_{P}) - (V_{C} - V_{Q}) = V_{Q} - V_{P} = 7 .5 - 5 = 2.5 V (\sin c e V_{A} = V_{C}) \\ V_{Q} - V_{P} = 2 .5 V ∴ V_{Q} = 2 .5 + V_{P} ∴ V_{Q} > V_{P} \end{array}$

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