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Question: After charging a capacitor of capacitance \(4 \mu F\) upto a potential 400 V, its plates are connec...

After charging a capacitor of capacitance 4μF4 \mu F upto a potential 400 V, its plates are connected with a resistance of 1kΩ1 \mathrm { k } \Omega. The heat produced in the resistance will be

A

0.16 J

B

1.28 J

C

0.64 J

D

0.32 J

Answer

0.32 J

Explanation

Solution

This is the discharging condition of capacitor and in this condition energy released

U=12CV2U = \frac { 1 } { 2 } C V ^ { 2 } =12×4×10−6×(400)2=0.32 J= \frac { 1 } { 2 } \times 4 \times 10 ^ { - 6 } \times ( 400 ) ^ { 2 } = 0.32 \mathrm {~J} =0.32 J= 0.32 \mathrm {~J}