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Questions  

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 

 

a
40 V
b
15 V
c
60 V
d
45 V

detailed solution

Correct option is D

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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Two capacitors C1 and C2 having capacitance 2μF and 4μF respectively are connected as shown in the figure. Initially C1 has charge 4μC and C2 is uncharged. After long time closing the switch S :


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