An inductance of frac{300}{pi} text{ mH},a ca | vedclass

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(C) The phase angle $\phi$ in an $LCR$ series circuit is given by $	an \phi = \frac{X_L - X_C}{R}$.Given: $L = \frac{300}{\pi} 	ext{ mH} = \frac{0.3

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$	an^{-1}(1)$

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(C) The phase angle $\phi$ in an $LCR$ series circuit is given by $	an \phi = \frac{X_L - X_C}{R}$.Given: $L = \frac{300}{\pi} 	ext{ mH} = \frac{0.3}{\pi} 	ext{ H}$,$C = \frac{1}{\pi} 	ext{ mF} = \frac{1}{\pi} 	imes 10^{-3} 	ext{ F}$,$R = 20 \ \Omega$,$f = 50 	ext{ Hz}$.Angular frequency $\omega = 2 \pi f = 2 	imes \pi 	imes 50 = 100 \pi 	ext{ rad/s}$.Inductive reactance $X_L = \omega L = 100 \pi 	imes \frac{0.3}{\pi} = 30 \ \Omega$.Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100 \pi 	imes \frac{1}{\pi} 	imes 10^{-3}} = \frac{1}{0.1} = 10 \ \Omega$.Substituting these values into the formula:$	an \phi = \frac{30 - 10}{20} = \frac{20}{20} = 1$.Therefore,$\phi = 	an^{-1}(1)$.

An inductance of $\frac{300}{\pi} \text{ mH}$,a capacitance of $\frac{1}{\pi} \text{ mF}$ and a resistance of $20 \ \Omega$ are connected in series with an a.c. source of $240 \text{ V}, 50 \text{ Hz}$. The phase angle of the circuit is

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