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Question

Physics Question on Current electricity

A 4μF4\, \mu F capacitor is charged to 400 V. If its plates are joined through a resistance of 2kΩ2\,k \Omega, then heat produced in the resistance is

A

0.64 J

B

1.28 J

C

0.16 J

D

0.32 J

Answer

0.32 J

Explanation

Solution

Capacitance (C) =4μF=4×106F;=4\, \mu F = 4\, \times\, 10^{-6}\, F ; Voltage (V) = 400 volts and resistance (R) =2kΩ=2×103Ω.= 2\, k\Omega = 2 \times 10^{3}\, \Omega.
Heat produced = Electrical energy stored =12CV2 = \frac{1}{2} CV^2
=12×(4×106)×(400)2=0.32J.=\frac{1}{2} \times (4 \times10^{-6}) \times (400)^2 = 0.32\, J.