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A `2.5//pi` `muF` capacitor and a 3000 ohm resistance are joined in series to an a.c. source of 200 volt and `50 "sec"^(-1)` frequency. The power factor of the circuit and the power dissipated in it will respectively be-A. 0.6, 0.06WB. 0.06, 0.6WC. 0.6, 4.8WD. 4.8, 0.6W |
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Answer» Correct Answer - C The capacitive reactance is `X_(C) = (1)/(omegaC) = (1)/(2pifC)` `=(1)/(2pi xx 50 xx ((25)/(pi) xx 10^(-6)))=4000 ohm`. The impedance of the circuit is `Z = sqrt((R^(2) + X_(C)^(2)))=sqrt([(3000)^(2) + (4000)^(2)])=500 ohm` Power factor, `cos phi = (R)/(Z) = (3000)/(5000) = 0.6` Power dissipation, `overset(-)p = V_(rms) xx i_(rms) xx cos phi`. `= V_(rms) xx (V_(rms))/(Z) xx cos phi`. `= 200 xx (2000)/(5000) xx 0.6` `= 4.8 "watt"`. |
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