A 50 text{ Hz} AC circuit has a 10 text{ mH} | vedclass

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(B) For the power factor to be unity $(\cos \phi = 1)$,the circuit must be in resonance.In an $LCR$ series circuit at resonance,the inductive reactanc

Vedclass · Accepted Answer

$1.014 	imes 10^{-3} 	ext{ F}$

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(B) For the power factor to be unity $(\cos \phi = 1)$,the circuit must be in resonance.In an $LCR$ series circuit at resonance,the inductive reactance equals the capacitive reactance,i.e.,$X_L = X_C$.This implies $\omega L = \frac{1}{\omega C}$,where $\omega = 2 \pi f$.Substituting the values: $2 \pi f = \frac{1}{\sqrt{LC}}$.Given $f = 50 	ext{ Hz}$,$L = 10 	ext{ mH} = 10 	imes 10^{-3} 	ext{ H}$.$2 \pi 	imes 50 = \frac{1}{\sqrt{10 	imes 10^{-3} 	imes C}}$.$100 \pi = \frac{1}{\sqrt{0.01 	imes C}}$.Squaring both sides: $(100 \pi)^2 = \frac{1}{0.01 	imes C}$.$10000 	imes \pi^2 = \frac{1}{0.01 	imes C}$.$C = \frac{1}{10000 	imes \pi^2 	imes 0.01} = \frac{1}{100 	imes \pi^2} \approx \frac{1}{100 	imes 9.8696} \approx \frac{1}{986.96} \approx 1.0132 	imes 10^{-3} 	ext{ F}$.Thus,the required capacitance is approximately $1.014 	imes 10^{-3} 	ext{ F}$.

$A$ $50 \text{ Hz}$ $AC$ circuit has a $10 \text{ mH}$ inductor and a $2 \text{ } \Omega$ resistor in series. The value of capacitance to be placed in series in the circuit to make the circuit power factor as unity is

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