Question

A series LCR circuit is connected across a source of emf $E=\sin (100\pi t-\frac{\pi }{6})$. The current from the supply is $I=\sin (100\pi t-\frac{\pi }{12})$. Draw the impedence triangle for the circuit.

Answer 1

$I_0=1$

$E_0=1$

$I_0=\frac{E_0}{\sqrt{R^2+({X_L-X_C)}^2}}$

Then, $\sqrt{R^2+({X_L-X_C)}^2}=1$

$\varphi =\frac{-\pi }{12}-(-\frac{\pi }{6})=\frac{\pi }{12}$

Current Leading Therefore, $cos\varphi =0.965$

$X_C-X_L=0.965$

$R=\sqrt{1-({X_C-X_L)}^2}=0.25$