In the circuit of figure given below, the magnitudes of V_{ L } \text { and } V_{ C } are twice that of V_{ R } . The inductance of the coil is

(a) 2.14 mH (b) 5.30 H

(c) 31.8 mH (d) 1.32 H

Step-by-Step

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Given that for resonance

\left|V_{ L }\right|=\left|V_{ C }\right|=2 V_{ R }

Voltage across inductor V_{ L }=I_{ R } \times j \omega L .

At resonance, I_{ R }=5 \angle 0^{\circ}=\frac{5}{5}=1 A .

\begin{aligned} & \left|V_{ L }\right|=\omega L=2 V_{ R } \text { where } \omega L=2 \times 5=10 \\ & 2 \pi f \cdot L=10 \quad \text { (where } \omega=2 \pi f) \\ & 2 \times \pi \times 50 \times L=10 \Rightarrow L=31.8 mH \end{aligned}

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