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Standard X

Physics

Resistance

In the circui...

Question

In the circuit shown in figure, switch $S_{1}$ is initially closed and $S_{2}$ is open. Find $V_{a} - V_{b}$

4 V

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8 V

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12 V

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16 V

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Solution

The correct option is B 8 V
Switch $S_{2}$ is open so capacitor is not in circuit.

Current through 3 $Ω$ registor $= \frac{24}{3 + 3} = 4 A$
Let potential of point 'O' shown in fig. is $V_{o}$
then using ohm's law
$V_{o} - V_{a} = 3 \times 4 = 12 V$ $. . . . (i)$
Now current through $5 Ω$ resistor $= \frac{24}{5 + 1} = 4 A$
So $V_{o} - V_{b} = 4 \times 1 = 4 V$ $. . . . (i i)$

From equation (i) and (ii) $V_{b} - V_{a}$ =12-4=8V.

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

In the circuit shown in figure, switch $S_{1}$ is initially closed and $S_{2}$ is open. Find $V_{a} - V_{b}$