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A series LCR circuit is connected to an AC source of voltage v and angular frequency Omega.When only the capacitor is removed the current lags behind the voltage by a phase angle'phi' and when only the inductor is removed the current leads the voltage by the same phase angle. Find the current flowing and the average power dissipated in the LCR circuit. |
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Answer» <P> Solution :`tan phi = (X_(L) - X_(C)) = (omegaL - 91)/((omega C)/(R ))`when CAPACITOR is removed `tan phi = (omega L)/(R )` when inductor is removed `tan phi = ((-1)/(omega C))/(R )` -ve SIGN indicates that current leads the voltage `:. omega L = (1)/(omega C)` `omega = (1)/(sqrt(LC))` `rArr LCR` circuit is in resonance, `i_(rms) = (V_(rms))/(R )` `P_(AV) = V_(rms)i_(rms)` `= V_(rms)^(2)//R` |
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