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Q.

In given arrangement of the capacitors, one  3 μF capacitor has got  of energy 600 μJ. Then the potential difference across 2 μF capacitor is

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a

40 V

b

15 V

c

60 V

d

45 V

answer is D.

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Detailed Solution

energy 12CV2=600 μJ 123 x10-6V2= 600x 10-6 V= 20 V P.D across AB is 60 V  P.D across 2 μF is 608 x6 V =45 V   in series V is inversely proportional to C
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