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A voltage V= V_(m) sin omegatis applied to a series L-C-R circuit. Derive the expression for the average power dissipated over a cycle. Under what condition (i) no power is dissipated even though the current flows through the circuit, (ii) maximum power is dissipated in the circuit ? |
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Answer» SOLUTION :Let, in general APPLIED voltage and current flowing in a LCR series a.c. circuit be given by `V = V_(m) sin omega t` and `I = I_(m) sin (omega t-phi)` `therefore` INSTANTANEOUS power `P = VI = V_(m) sin omegat. I_(m) sin (omega t-phi)` `=V_(m)I_(m)[sin^(2)omegat cos phi - 1/2 sin 2 omegat sin phi]` `therefore` Average power for one complete cycle of a.c. `VECP = V_(m)I_(m) [vec(sin^(2)omega) cos phi -1/2 vec(sin2 omegat) sin phi]` But we know that average VALUE of `sin^(2)omegat`for a complete cycle is `1/2`, whereas average value of `sin 2omega t` for one complete cycle is zero. Hence, `vecP` or `P_(av) = V_(m)I_(m) [1/2 cos phi] = 1/2 V_(m) I_(m) cos phi = V_(rms) I_(rms) cos phi` (i) If phase difference `phi` between V and I be `pi/2`or a pure inductor or a pure capacitor or a series LC circuit) then `cos phi = cos pi/2 =0`and hence average power dissipated is zero even though the current flows through the circuit. (ii) If phase difference `phi = 0^(@)`(as for a pure resistor or for a resonant circuit) then `cos phi = cos 0^(@)=1`and hence average power `P_(av) = V_(rms) I_(rms)`which is maximum power dissipated in the circuit. |
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