Question 1.SGPYQ.4: In the circuit of figure given below, the magnitudes of VL a......

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}