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An L-C-R series circuit consists of a resistance of 10 \Omega a capacitor of reactance 60 \Omega and an inductor coil. The circuit is found to resonate when put across a 300 V, 100 Hz supply. The inductance of coil is (take, \pi= 3)

Option: 1

0.1 H


Option: 2

0.01 H


Option: 3

0.2 H


Option: 4

0.02 H


Answers (1)

best_answer

Angular velocity,

\mathrm{\begin{aligned} & \omega_0=2 \pi \mathrm{v}=2 \pi \times 100 \\ & \omega_0=2 \times 3 \times 100 \\ & =600 \mathrm{rads}^{-1} \end{aligned} \quad[\because \pi=3]}

\mathrm{Further, \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} Also, \mathrm{X}_{\mathrm{C}}=\frac{1}{\mathrm{C} \omega_0}=60 \Omega}
\mathrm{\Rightarrow \mathrm{C}=\frac{1}{\omega_0 \times 60}=\frac{1}{600 \times 60} \Rightarrow \mathrm{C}=\frac{1}{36 \times 10^3} \mathrm{~F} }
So, put values in eq. (i), we get

\mathrm{\begin{aligned} & 600=\frac{1}{\sqrt{L\left(\frac{1}{36 \times 10^3}\right)}} \\ & \Rightarrow 36 \times 10^4=\frac{36 \times 10^3}{L} \Rightarrow \mathrm{L}=\frac{36 \times 10^3}{36 \times 10^4}=\frac{1}{10}=0.1 \mathrm{H} \end{aligned} }

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Divya Prakash Singh

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