A 5 micro Farad capacitor is charged to 12 Volt. The positive plate of this capacitor is now connected to the negative terminal of a 12 Volt battery and vice versa. Calculate the heat developed in the connecting wires.
Asked by dipak kumar Saikia | 11th Jul, 2013, 06:28: PM
Q = CV
Initially, Q = 5*10^-6 * 12 = 60*10^-6C
Now, when the positive plate of the capacitor is connected to the negative terminal of the 12 V battery and vice versa, then, redistribution of charge on capacitor will take place and 2Q charge will flow through the circuit.
Energy generated = 1/2*Q*V
Energy generated = 1/2*2Q*V = 60*10^-6 * 12 = 720 microjoules.
Answered by | 19th Jul, 2013, 09:53: AM
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