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An inductor of 200 mH, capacitor of 400 uF and a resistor of 10 Omegaare connected in series of a.c. source of 50 V of variable frequency. Calculate the (a) angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of the effective current. (b) value of Q-factor of the circuit. |
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Answer» Solution :Here, L = 200 mH = 0.2 H, C = 400 `muH = 400xx 10^(-6) F = 4 XX 10^(-4) F, R = 10 Omega` and `V_(rms) = 50 V` (a) For maximum power DISSIPATION, the circuit MUST be a resonant circuit having an angular frequency `omega_(0) = 1/sqrt(LC) = 1/sqrt(0.2 xx 4 xx 10^(-4)) = 50sqrt(5) s^(-1)` or `111.8 s^(-1)` `therefore` Corresponding value of effective current `I_(ETA) = V_(rms)/r = 50/10 = 5 A` (b) Q-factor `X_(L)/R = (L omega)/R = (0.2 xx 111.8)/10 = 2.24` |
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