A resistor of 100 Omega,an inductor of self-i | vedclass

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(D) When the current and voltage are in phase,the circuit is in resonance. At resonance,the inductive reactance $X_L$ equals the capacitive reactance

Vedclass · Accepted Answer

$2.5 	imes 10^{-5} 	ext{ F}, 400 	ext{ W}$

Vedclass · Answer

(D) When the current and voltage are in phase,the circuit is in resonance. At resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$. Given $L = \frac{4}{\pi^2} 	ext{ H}$ and $f = 50 	ext{ Hz}$. $X_L = 2\pi f L = 2 	imes \pi 	imes 50 	imes \frac{4}{\pi^2} = \frac{400}{\pi} \Omega$. Since $X_L = X_C$,we have $\frac{1}{2\pi f C} = \frac{400}{\pi}$. Substituting $f = 50 	ext{ Hz}$: $\frac{1}{2 	imes \pi 	imes 50 	imes C} = \frac{400}{\pi} \implies \frac{1}{100 \pi C} = \frac{400}{\pi} \implies C = \frac{1}{40000} = 2.5 	imes 10^{-5} 	ext{ F}$. At resonance,the impedance $Z = R = 100 \Omega$. The power dissipated is $P = \frac{V^2}{R} = \frac{200^2}{100} = \frac{40000}{100} = 400 	ext{ W}$. Thus,the capacitance is $2.5 	imes 10^{-5} 	ext{ F}$ and the power dissipated is $400 	ext{ W}$.

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