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(i) An a.c. source of voltage V=V_(m) sin omega t is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ? (ii) In a series LR circuit X_(L) =R and power factor of the circuit is P_(1) . When capacitor with capacitance C such that X_(L) = X_( C)is put in series, the power factor becomes P_(2) . Calculate P_(1)//P_(2) |
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Answer» Solution :(i) N/A (ii) In the series LR CIRCUIT, power factor `P_(1) = cos phi = R/Z = R/sqrt(R^(2) + X_(L)^(2))`, Since, `X_(L) =R`, hence power `P_(1) = R/sqrt(R^(2) + R^(2)) = 1/sqrt(2)` When a capacitor is PUT in series then new power factor `P_(2) = R/Z = R/sqrt(R^(2) + (X_(L)-X_(C))^(2))` As `X_(L) =R` and `X_(C) =X_(L)`, hence, we have `P_(2) = R/sqrt(R^(2) + 0) =1` `RARR P_(1)/P_(2) = (1//sqrt(2))/1 = 1/sqrt(2)` |
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