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CBSE Class 12-science Answered

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
Expert Answer

Capacitance of a charged capacitor, C1 = 4 Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.F = 4 × 10-6 F

Supply voltage, V1= 200 V

Electrostatic energy stored in C1is given by,

begin mathsize 11px style straight E subscript 1 equals 1 half straight C subscript 1 straight V subscript 1 squared
space space space space space equals space 1 half space straight x space 4 space straight x space 10 to the power of negative 6 end exponent space straight x space left parenthesis 200 right parenthesis squared
space space space space space equals space 8 space straight x space 10 to the power of negative 2 end exponent space straight J end style

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 = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. V

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

begin mathsize 11px style straight E subscript 2 equals 1 half open parentheses straight C subscript 1 plus straight C subscript 2 close parentheses space straight V subscript 2 squared
space space space space space equals 1 half space left parenthesis 2 space plus space 4 right parenthesis space straight x space 10 to the power of negative 6 end exponent space straight x space open parentheses 400 over 3 close parentheses squared
space space space space space equals space 5.33 space straight x space 10 to the power of negative 2 end exponent straight J end style

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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