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A 2 uF capacitor, 100 Omegaresistor and 8 H inductor are connected in series with an a.c. source. Find the frequency of the a.c. source for which the current drawn in the circuit is maximum. If the peak value of emf of the source is 200 V, calculate the (i) maximum current, and (ii) inductive and capacitive reactance of the circuit at resonance. |
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Answer» Solution :Here `C = 2muF = 2 xx 10^(-6) F, R = 100 Omega, L = 8H` and `V_(m) = 200 V` `therefore` Frequency of a.c source for current in the circuit to be maximum = resonant frequency `omega_(r) = 1/sqrt(LC) = 1/sqrt(8 xx 2 xx 10^(-6)) rArr omega_(r) = 250 s^(-1)` (i) The maximum current `I_(m) = V_(m)/R = 200/100 = 2A` (ii) Inductive reactance `X_(L) = Lomega = 8 xx 250 = 2000 Omega` and capacitive reactance `X_(C ) =1/(C omega) = 1/((2 xx 10^(-6)) xx 250) = 2000 Omega` |
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