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asked in JEE by (30 points)
A 5uf capacitor having a charge of 20u C is discharged through a wire of resistance 5 ohm. Find the heat dissipated in the wire between 25-50 u secafter the connections are made.
commented by (328 points)
Your question seems incomplete as the answer is in terms of e

1 Answer

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answered by (2k points)

The potential difference across the capacitor

V0 = Q/C = 20/5 = 4V

The initial current,

i0 = V/R = 4/5 = 0.8A

τ = CR = 20*5 = 100 µs

The current in the resistor at any time t

i = i0e-t/τ

The heat dissipated,

After substituting and simplifying, we get

H = 4.7 µJ

commented by (328 points)
You answer could be probably wrong as time constant is not equal to 100 µs, it is 25 µs

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