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An inductance of (200/pi) mH, a capacitance of (10^(-3)/pi)F and a resistance of 10Omega are connected in series with an a.c. source 220 V, 50Hz. Then what is the phase angle of the circuit ? |
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Answer» `pi/4` rad `L=200/pi` mH `C=10^(-3)/pi`F `R=10 Omega` `RARR X_L =2pifL=2pixx50xx200/pixx10^(-3)` `=20Omega` `rArr X_C =1/(2pifC)=1/(2pixx50xx10^(-3)/pi)` `therefore X_C=10Omega` `rArr tan theta =(X_L-X_C)/R` `=(20-10)/10` `=10/10` `tan theta =1` `therefore theta = tan^(-1) (1)` `therefore theta =pi/4` Here, `X_L gt X_C` so, current lags behind by VOLTAGE by `pi/4` rad . |
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