Question

An $L-C-R$ series circuit with $L=0.120H,R=240\Omega$, and $C=7.30\mu F$ carries an rms current of $0.450A$ with a frequency of $400Hz$. What are the phase angle and power factor for this circuit?

Answer 1

$X_L=\omega L=2\pi fL=2\times 3.14\times 400\times 0.120=301.44\Omega$

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\times 3.14\times 400\times 7.30\times {10}^{-6}}=54.53\Omega$
Impedance in LCR series circuit is $Z=\sqrt{R^2+(X_L-X_C{)}^2}=\sqrt{(240{)}^2+(301.44-54.53{)}^2}=344.33\Omega$

$\therefore$ The phase angle is $\varphi ={\cos }^{-1}\left(R/Z\right)={\cos }^{-1}\left(240/344.33\right)={45.8}^o$

$\therefore$ Power factor $=\cos \varphi =\frac{R}{Z}=\frac{240}{344.33}=0.697$