In the circuit shown in figure, switch S 1 is initially closed and S 2 is open. Find V a – V b
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Switch $\mathrm{S}_2$ is open so capacitor is not in circuit.
Current through 3Ω resistor $\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0$ $\frac{24}{3+3} = 4 A$
Let potential of point ‘O’ shown in fig. is $V_0$
then using ohm’s law
$V_0 - V_a = 3 \times 4 = 12V$ ....(i)
Now current through 50 Omega resistor $= \frac{24}{5+1} = 4A$
So $V_0 - V_b = 4 \times 1 = 4V$ .....(ii)
From equation (i) and (ii) $V_{b} - V_{a} = 12 - 4 = 8V.$
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