JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Alternating Current

In a series LR circuit with

X_{L} = R

, power factor is

P_{1}

. 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}

P_{2}

will be :

Answer: (B)

Solution

The power factor is

cos ϕ = \frac{R}{Z}

. With only

R

and

X_{L} = R

Z = R^{2} + R^{2} = R 2

, so

P_{1} = \frac{1}{2}

. Adding the capacitor with

X_{C} = X_{L}

cancels the reactance, the impedance becomes

R

, and

P_{2} = 1

. The ratio is

P_{1} : P_{2} = \frac{1}{2} : 1 = 1 : 2

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option