# 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 $\left ( \frac{n}{3\pi } \right )\mu F,$ then value of n is _______ Option: 1 400 Option: 2 200 Option: 3 100 Option: 4 50

$\begin{array}{l} P=\frac{(250)^{2}}{Z} \cos \phi \\ \\ 500=\frac{(250)^{2}}{Z} \\\\ Z=125 \Omega \\\\ R=100 \Omega \\\\ X_{L}=75 \Omega \end{array}$

$\begin{array}{l} 75=\frac{1}{2 \pi f C} \\ \\ C=\frac{1}{2 \pi \times 75 \times 50}=\frac{1}{7500 \pi} \\\\ C=\left(\frac{10^{6}}{2500} \times \frac{1}{3 \pi}\right) \mu F \\\\ C=\frac{400}{3 \pi} \mu F \end{array}$

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