A 5.0 µF capacitor having a charge of 20 µC is discharged through a wire of resistance 5.0 Ω . Find the heat dissipated in the wire between 25 to 50 µs after the connections are made. (Given: e – 2 = 0.135)
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(40 (1 – e – 2 )e – 2 µJ = 4.7 µJ.)
Sol. i 0 = 
= 
amp.
i = i 0 e – t/RC H = 
= 
= 
= 40 × 10
– 6 
= 4.7 µJ
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