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A voltage V = V_(0) sin omega t is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, maximum power dissipated in the circuit ? |
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Answer» Solution :APPLIED voltage `= V_(0) sin omega t` Current in the circuit `= I_(0) sin (omega t - phi)` where `phi` is the phase lag of the current with respect to the voltage applied, Hence,instantaneous power dissipation `= V_(0) sin omega t xx I_(0) sin(omega t - phi)` `= (V_(0)I_(0))/(2)[2 sin omega t. sin (omega t - phi)]` `= (V_(0)I_(0))/(2)[cos phi - cos (2 omega t - phi)]` Therefore average power for ONE complete cycle = average of `[(V_(0)I_(0))/(2) {cos phi - cos (2 omega - phi)}]` The average of the SECOND TERM over a complete cycle is zero. Hence, average power dissipated over one complete cycle `= (V_(0)I_(0))/(2) cos phi` Conditions : (i) No power is dissipated when R = 0 (or `phi = 90^(@))` [NOte : We can also WRITE, this condition cannot be satisfied LCR circuit..] (ii)Maximum power is dissipated when `X_(L) = X_(C)` or `omega L = (1)/(omega C)` (or `phi = 0`) |
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