What is the potential difference $V_{c d}$ in $V$ ?

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Solution

For upper branch $3 μ F$ and $6 μ F$ are in series. So voltage across $3 μ F$ is $V_{3} = \frac{6}{3 + 6} V_{a b} = \frac{6}{9} \times 360 = 240 V$
Also $V_{3} = V_{a} - V_{d} = 240 V . . . . (1)$
Similarly for lower branch voltage across $6 μ F$ is
$V_{6} = V_{a} - V_{c} = \frac{3}{9} \times 360 = 120 V . . . (2)$
$(1) - (2), V_{c} - V_{d} = 240 - 120 = 120 V$

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