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In a series AC circuit consists of an inductor of magnitude 2 mH, resistance of magnitude 0.1 ohm  and having a Capacitance C. Here the current leads the voltage by 45 °. Find the value of capacitance if angular frequency (\omega) of AC is 50 rad/s. 

 

Option: 1

100 F


Option: 2

10 mF


Option: 3

100  \mu F


Option: 4

100 mF


Answers (1)

best_answer

\\In\ series\ A\ C\ circuit. \\Since\ current\ leads\ the\ voltage\ by\ \phi=45^{\circ}.

{\text { So, }} \\ \tan \phi=\frac{X_C-X_L}{R}

X_c=\frac{1}{\omega c} \quad, \quad X_L=\omega L

\\where \\ \begin{aligned} & \omega=50 \mathrm{rad} / \mathrm{s}, L=2 \mathrm{mH} . \\ & R=0.1 \Omega, \phi=45^{\circ} . \end{aligned}

\tan 45^{\circ}=\frac{\frac{1}{50 C}-50 \times 2 \times 10^{-3}}{0.1}

\Rightarrow 1=\frac{\frac{1}{50 c}-100 \times 10^{-3}}{0.1}

\Rightarrow 0.1 \times 1=\frac{1}{50 C}-10^{-1}

\Rightarrow 0.1+0.1=\frac{1}{50 c} \quad \Rightarrow c=\frac{1}{50 \times 0.2}

\begin{aligned} \Rightarrow C=\frac{1}{10} F \Rightarrow C & =0.1 \mathrm{~F} \\ & =100 \mathrm{mF} \end{aligned}

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SANGALDEEP SINGH

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