JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Alternating Current

In a series LR circuit, power of 400 W is dissipated from a source of

250 V, 50 Hz

. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as

(\frac{n}{3 π}) μ F

, then value of 'n' is ____.

Answer: 400

Solution

From

P = \frac{V _{r m s}^{2}}{Z} cos ϕ

we get

Z = \frac{( 250 ) ^{2} ( 0.8 )}{400} = 125 Ω

, so

R = Z cos ϕ = 100 Ω

and

X_{L} = Z sin ϕ = 75 Ω

. Unity power factor needs

X_{C} = X_{L} = 75 Ω

. With

L = \frac{X _{L}}{ω} = \frac{75}{2 π ( 50 )} = \frac{3}{4 π} H

, the resonance condition

ω^{2} = \frac{1}{L C}

gives

C = \frac{1}{ω ^{2} L} = \frac{400}{3 π} μ F

. Written as

\frac{n}{3 π} μ F

, this means

n = 400

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Integer answer

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