Question
Physics Question on Current electricity
A 4μF capacitor is charged to 400 V. If its plates are joined through a resistance of 2kΩ, 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=4×10−6F; Voltage (V) = 400 volts and resistance (R) =2kΩ=2×103Ω.
Heat produced = Electrical energy stored =21CV2
=21×(4×10−6)×(400)2=0.32J.