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A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2μF capacitor. How much electrostatic energy of the first capacitor is lost in the heat and electromagnetic radiation?
Asked by Topperlearning User | 22 Apr, 2015, 08:40: AM

Capacitance of a charged capacitor, C1 = 4 F = 4 × 10-6 F

Supply voltage, V1= 200 V

Electrostatic energy stored in C1is given by,

Capacitance of an uncharged capacitor, C = 2 μF = 2 × 10-6 F

When C2 is connected to the circuit, the potential acquired by it is V2.

According to the conservation of charge, initial charge on capacitor C1 is equal to the  final charge on capacitors, C1 and C2.

therefore, V2 (C1 + C2) = C1V1

V2 × (4 + 2)  × 10-6 × 200

V2 = V

Electrostatic energy for the combination of two capacitors is given by,

Hence, amount of electrostatic energy lost by capacitor C1

= E1 - E2

= 0.08 - 0.0533 = 0.0267

= 2.67 x 10-2 J

Answered by | 22 Apr, 2015, 10:40: AM

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