In a series LR circuit, power of 400W is dissipated from a source of… | JEE Main 2020 (6 Sep, Shift 2)

Q62JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Power, Power Factor & Wattless Current

In a series LR circuit, power of 400W 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 $\left(\frac{n}{3\pi}\right)\mu$ F, then value of n is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 400

$P = V I\cos\phi \Rightarrow I = \frac{400}{250\times 0.8} = 2$ A
$Z = 250/2 = 125\ \Omega$ , $R = Z\cos\phi = 100\ \Omega$ , $X_L = \sqrt{125^2-100^2} = 75\ \Omega$
Unity power factor needs $X_C = 75\ \Omega$ : $C = \frac{1}{100\pi\times 75} = \frac{1}{7500\pi}$ F
$C = \frac{10^6}{7500\pi}\ \mu$ F $= \frac{400}{3\pi}\ \mu$ F, so $n = 400$ .

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