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Question

In the circuit shown in figure, switch S1 is initially closed and S2 is open. Find VaVb

A
4 V
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B
8 V
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C
12 V
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D
16 V
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Solution

The correct option is B 8 V

Switch S2 is open so capacitor is not in circuit.



Current through 3 Ω registor =243+3=4 A
Let potential of point 'O' shown in fig. is Vo
then using ohm's law
VoVa=3× 4=12V ....(i)
Now current through 5Ω resistor =245+1=4A
So VoVb=4× 1=4V ....(ii)

From equation (i) and (ii) VbVa=12-4=8V.


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