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Fig. shows series LCR circuit with `L = 5.0 H, C = 80 mu F, R = 40 Omega` connected to a variable frequency 240 V source. Calculate (i) the angular freuquency of the source which drives the circuit at resonance. (ii) Current at the resonanting frequency. (iii) the rms pot. drop across the capacitor at resonance. |
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Answer» Here, `L = 5.0 H`, `C = 80 mu F = 80 xx 10^(-6) F, R = 40 ohm` `E_(v) = 240 V, omega = ?, I = ?, V_(C ) = ?` At resonance, `X_(L) = X_(C )` `omega = (1)/(omega C)` or `omega = (1)/(sqrt(LC))` `omega = (1)/(sqrt(5.0 xx 80 xx 10^(-6))) = (1000)/(20) = 50 rad//s` (ii) At resonance. `Z = R =40 Omega` `I_(v) = (E_(v))/(Z) = (240)/(40) = 6 A` (iii) `V_(C ) = I_(v) X_(C )` `= (I_(v))/(omega C) = (6)/(50 xx 80 xx 10^(-6)) = (6000)/(4) = 1500 V` |
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