An inductance of frac{200}{pi} mH,a capacita | vedclass

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Question

(B) The phase angle $	heta$ in an $LCR$ series circuit is given by $	an 	heta = \frac{X_L - X_C}{R}$.First,calculate the inductive reactance: $X_L

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) The phase angle $	heta$ in an $LCR$ series circuit is given by $	an 	heta = \frac{X_L - X_C}{R}$.First,calculate the inductive reactance: $X_L = 2 \pi f L = 2 \pi 	imes 50 	imes \left(\frac{200}{\pi} 	imes 10^{-3}ight) = 20 \ \Omega$.Next,calculate the capacitive reactance: $X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi 	imes 50 	imes \left(\frac{10^{-3}}{\pi}ight)} = \frac{1}{100 	imes 10^{-3}} = 10 \ \Omega$.Given resistance $R = 10 \ \Omega$.Substituting these values into the formula: $	an 	heta = \frac{20 - 10}{10} = \frac{10}{10} = 1$.Therefore,$	heta = 	an^{-1}(1) = \frac{\pi}{4}$.

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

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