First slide

Capacitance

Question

In given arrangement of the capacitors, one $3 μ F$ capacitor has got of energy $600 μ J$ . Then the potential difference across $2 μ F$ capacitor is

Moderate

A

40 V
B

15 V
C

60 V
D

45 V

Solution

$energy \frac{1}{2} {CV}^{2} = 600 μJ \frac{1}{2} 3 x 10^{- 6} V^{2} = 600 x 10^{- 6} V = 20 V P . D across AB is 60 V P . D across 2 μF is (\frac{60}{8} x 6) V = 45 V [in series V is inversely proportional to C]$

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