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Q.

In the circuit show in figure. The internal resistance of battery is $1.5 Ω$ and $V_{p}$ and $V_{q}$ are the potentials at P and Q respectively, The potencial difference between the point P and Q is

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a

Zero

b

$4 v (V_{p} > V_{q})$

c

$4 v (V_{q} > V_{p})$

d

$2.5 v (V_{q} > V_{p})$

answer is D.

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Detailed Solution

$\begin{aligned} R_{1} = 3 + 2 = 5 Ω R_{2} = 2 + 3 = 5 Ω \\ R_{1} = \frac{5 \times 5}{5 + 5} = 2.5 Ω \\ R^{1} = 2.5 + 1.5 = 4 Ω i = \frac{20}{4} = 5 A \\ i_{1} = i_{2} = 2.5 A \\ V_{x} - V_{p} = 2.5 (3) = 7.5 V \\ V_{x} - V_{Q} = 2.5 (2) = 5 V \end{aligned}$
$V_{Q} - V_{P} = 2.5 V, V_{Q} > V_{P}$

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