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Question

(B) The phase angle $	heta$ between current $I$ and voltage $V$ in an $LCR$ series circuit is given by the formula:$	an 	heta = \frac{X_L - X_C}{R}

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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(B) The phase angle $	heta$ between current $I$ and voltage $V$ in an $LCR$ series circuit is given by the formula:$	an 	heta = \frac{X_L - X_C}{R}$First,calculate the inductive reactance $X_L$:$X_L = 2 \pi f L = 2 \pi 	imes 50 	imes \left( \frac{200}{\pi} 	imes 10^{-3} ight) = 20 \, \Omega$Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi 	imes 50 	imes (10^{-3} / \pi)} = \frac{1}{0.1} = 10 \, \Omega$Given resistance $R = 10 \, \Omega$.Substituting these values into the phase angle formula:$	an 	heta = \frac{20 - 10}{10} = \frac{10}{10} = 1$Since $	an 	heta = 1$,we have $	heta = 	an^{-1}(1) = \frac{\pi}{4}$ radians.Thus,the phase angle of the circuit is $\frac{\pi}{4}$.

An inductance of $\left(\frac{200}{\pi}\right) \text{mH}$,a capacitance of $\left(\frac{10^{-3}}{\pi}\right) \text{F}$,and a resistance of $10 \, \Omega$ are connected in series with an $AC$ source of $220 \, \text{V}, 50 \, \text{Hz}$. The phase angle of the circuit is:

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