## a battery of 10 volt is connected to a capacitor of capacitance 0.1 if the battery is now removed and this capacitor is connected

Question

a battery of 10 volt is connected to a capacitor of capacitance 0.1 if the battery is now removed and this capacitor is connected to a parallel with another uncharged capacitor of capacitance 0.2 how much of electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation​

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2 years 2021-06-22T03:16:34+00:00 1 Answers 0 views 0

Explanation:

Capacitance of a charged capacitor, C

1

​

=4μF=4×10

−6

F

Supply voltage, V

1

​

=200V

Electrostatic energy stored in C1 is given by,

E

1

​

=

2

1

​

C

1

​

V

1

2

​

=

2

1

​

×4×10

−6

×(200)

2

=8×10

−2

J

Capacitance of an uncharged capacitor, C

2

​

=2μF=2×10

−6

F

When C

2

​

is connected to the circuit, the potential acquired by it is V

2

​

.

According to the conservation of charge, initial charge on capacitor C

1

​

is equal to the final charge on capacitors, C

1

​

and C

2

​

∴V

2

​

(C

1

​

+C

2

​

)=C

1

​

V

1

​

V

2

​

×(4+2)×10

−6

=4×10

−6

×200

V

2

​

=

3

400

​

V

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

E

2

​

=

2

1

​

(C

1

​

+C

2

​

)V

2

2

​

=

2

1

​

(2+4)×10

−6

×(

3

400

​

)

2

=5.33×10

−2

J

Hence, amount of electrostatic energy lost by capacitor C

1

​

=E

1

​

−E

2

​

=0.08−0.0533=0.0267

=2.67×10

−2

J