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In the network shown in figure, reading of the ideal voltmeter V is

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4.5 V

4 V

8 V

6 V

answer is D.

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Detailed Solution

$12 - \frac{Q}{4} - \frac{Q}{2} - \frac{Q}{4} - \frac{Q}{2} = 0 \Rightarrow Q = 8 μC$

$Now, V_{B} - \frac{Q}{4} - \frac{Q}{2} = V_{A} \Rightarrow V_{B} - V_{A} = \frac{3 Q}{4} = \frac{3 x 8}{4} V = 6 V$

Reading = 6 V

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