In the network shown in figure, the potential difference $V_{a b}$ when the switch A is open is

3 V

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6 V

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9 V

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12 V

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Solution

The correct option is C $6 V$
When Switch is open, the equivalent resistance $R_{e q} = (6 + 4) | | (3 + 12) = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6 Ω$
The current drawn from supply is $i = \frac{V}{R_{e q}} = \frac{15}{6} = 2.5 A$
The current in paq branch is $i_{1} = \frac{15}{25} \times 2.5 = 1.5 A$ (by current division rule)
The current in paq branch is $i_{2} = \frac{10}{25} \times 2.5 = 1 A$
Thus, $V_{p} - V_{a} = 6 i_{1}$ or $15 - V_{a} = 6 (1.5) = 9$
or $V_{a} = 15 - 9 = 6 V$
$V_{p} - V_{b} = 3 i_{2}$ or $15 - V_{b} = 3 (1) = 2$
or $V_{b} = 15 - 3 = 12 V$
thus, $V_{a b} = V_{b} - V_{a} = 12 - 6 = 6 V$

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