In a series LR circuit |Alternating Current |JEE-Main 2020

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JEE MAIN 2020

06-09-2020 S2

Question

In a series LR circuit, power of 400 W is dissipated from a source of $250 \mathrm{~V}, 50 \mathrm{~Hz}$. 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 $\_\_\_\_$ .

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Solution

$\begin{aligned} & P=\frac{(250)^2}{Z} \cos \phi \\ & 500=\frac{(250)^2}{Z} \\ & Z=125 \Omega \\ & R=100 \Omega \\ & X_L=75 \Omega \\ & 75=\frac{1}{2 \pi f C} \\ & C=\frac{1}{2 \pi \times 75 \times 50}=\frac{1}{7500 \pi} \\ & C=\left(\frac{10^6}{2500} \times \frac{1}{3 \pi}\right) \mu F \\ & C=\frac{400}{3 \pi} \mu F\end{aligned}$

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