In a series LR circuit with X_L = R. power factor is pi. If a capacitor of capacitance C with X_c = X_L is added to the circuit the power factor becomes P_2. The ratio of P_1 to P_2 will be:

Question

Accepted Answer

Correct answer: B. (b) P= RZ ⇒P_1= R R^2+X_L^2= RR 2( asX_L=R) P_1= 1 2 p_2= R R^2+(X_L-X_L)^2=P_2=1

$In a series LR circuit with X_{L} = R. power factor is pi. If a capacitor of capacitance C with X_{c} = X_{L} is added to the circuit the power factor becomes P_{2} . The ratio of P_{1} to P_{2} will be:$

Solution

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