Question

(a) Determine the value of phase difference between the current and the voltage in the given series LCR circuit.



(b) Calculate the value of additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity.

Answer 1

(a) $V=V_{0}sin(1000t+\varphi )$

Comparing this with $V=V_{0}sin(\omega t+\varphi )$

We have,

$\omega =1000$

Given: $L=100\times {10}^{-3}H,R=400\Omega ,C=2\mu F$

$X_L=\omega L$

$X_L=1000\times 100\times {10}^{-3}={10}^2=100\Omega$

$X_C=\frac{1}{\omega C}=\frac{1}{1000\times 2\times {10}^{-6}}=\frac{1}{2}\times {10}^3$

$=500\Omega$

Phase difference,

$tan\varphi =\frac{X_L-X_C}{R}$

$tan\varphi =\frac{X_C-X_L}{R},X_C>X_L$

we have, $R=400\Omega$

$tan\varphi =\frac{500-100}{400}=\frac{400}{400}=1$

$tan\varphi =1$

$\therefore tan\varphi =tan{45}^{\circ }$

$\varphi ={45}^{\circ }$

(b) The power factor of the circuit is unity. It means that the given circuit is in resonance. It is possible, if another capacitor C' is used in the circuit.

$X_C^{\prime }=X_L$

$\frac{1}{\omega C^{\prime }}=\omega L$

$C^{\prime }=\frac{1}{{\omega }^{2}L}=\frac{1}{(1000{)}^2\times 100\times {10}^{-3}}$

$C^{\prime }=10\mu F$

Since, $C^{\prime }>C$, so on additional capacitor of $8\mu F$ would be connected in parallel to the capacitor of $C=2\mu F$.