Question

A 4uf capacitor is charged by a 200v supply. it is then disconnected from the supply and it connected to another charge 2uf capacitor how much electroststic energy of the first capicator is lost in the form of heat and electrostatic radition ?

Answer 1

Answer:

2.67×10−2J

Explanation:

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

Supply voltage, V1=200V

Electrostatic energy stored in C1 is given by,

E1=21C1V12

      =21×4×10−6×(200)2

      =8×10−2J

Capacitance of an uncharged capacitor, C2=2μF=2×10−6F 

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

∴V2(C1+C2)=C1V1

V2×(4+2)×10−6=4×10−6×200

V2 = 400 / 3

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

E2=21(C1+C2)V22

     =21(2+4)×10−6×(3400)2

     =5.33×10−2J

Hence, amount of electrostatic energy lost by capacitor C1

=E1−E2

=0.08−0.0533=0.0267

=2.67×10−2J

i hope that it helps you my dear frnd

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