In the circuit shown in figure, switch $S_1$ is initially closed and $S_2$ is open. Find $V_a -V_b$ .............. $V$
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Switch ${S_2}$ is open so capacitor is not in circuit.
Current through $3\,\Omega $ resistor $ = \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\,\Omega $ resistor $ = \frac{{24}}{{5 + 1}} = 4\,A$
So ${V_0} - {V_b} = 4 \times 1 = 4\,V$ ..... $(ii)$
From equation $(i)$ and $(ii)$ ${V_b} - {V_a} = 12 - 4 = 8\,V.$
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