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 _____

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answer is 400.

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Detailed Solution

$\begin{array}{l} \cos θ = 0.8 \\ \Rightarrow \tan θ = \frac{3}{4} = \frac{X_{L}}{R} \\ \Rightarrow X_{L} = \frac{3 R}{4} \\ And Z= \sqrt{{X_{L}}^{2} + R^{2}} \\ \Rightarrow Z = \frac{5 R}{4} \\ P o w e r P = \frac{V^{2}}{Z^{2}} \times R \\ \Rightarrow 400 = \frac{{(250)}^{2} \times R}{\frac{25 R^{2}}{16}} \\ \Rightarrow R = 100 Ω \\ So, X_{L} = 75 Ω \\ For power fractor to be 1, X_{C} = X_{L} \\ So, \frac{1}{2 π f C} = 75 \\ \Rightarrow C = \frac{400}{3 π} μ F \end{array}$