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A source of alternating emf of 220 V-50 Hz is connected in serieswith a resitanceof 200 Omegaan inductance of 100 mH anda capacitance of 30 mu Fdoesthe current lead or lag the voltageand by what angle ? |
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Answer» Solution :`X_L=2pivL=31.4pi` `X_C=1//2pivC=106.1 Omega` SINCE `X_C GT X_L` currents LEADS the voltage or voltage lags the current `tan PHI=(X_C-X_L)/R=0.3734` `phi=20.5^@` (Since `phi` is positive , current leads the voltage) Given , `V_(rms)`=200 V, f=50 Hz , R=200 W `L=100xx10^(-3)` = 0.1 H, `C=30xx10^(-6) F=3xx10^(-5)` F  Interactive reactance `X_L=2pifL` i.e., `X_L=2xx3.142xx50xx0.1W` i.e., `X_L`=31.42 W Capacitance reactance , `X_C=1/(2pifC)` i.e., `X_C=1/(2xx3.1742xx50xx3xx10^(-5))` i.e., `X_C=1000/9.426=106.08Omega` Reactance of the circuit , `X_L-X_C` = 31.42-106.08 i.e., `X_L-X_C`=-74.6 W `rArr tan phi = (X_L-X_C)/R` i.e., `tan phi =(-74.66)/200`=0.3733 Hence `f=-20^@ 30.` i.e., the current leads the voltage by `20^@30.`. |
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