The frequency of AC at which a 16 mu F capaci | vedclass

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(B) Given: Capacitance $C = 16 \ \mu F = 16 	imes 10^{-6} \ F$. Inductance $L = \frac{10}{\pi^2} \ mH = \frac{10}{\pi^2} 	imes 10^{-3} \ H$. For the

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$1.25$

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(B) Given: Capacitance $C = 16 \ \mu F = 16 	imes 10^{-6} \ F$. Inductance $L = \frac{10}{\pi^2} \ mH = \frac{10}{\pi^2} 	imes 10^{-3} \ H$. For the reactances to be equal,$X_L = X_C$. Substituting the formulas for inductive and capacitive reactance: $L \omega = \frac{1}{C \omega}$. Since $\omega = 2 \pi f$,we have $L(2 \pi f) = \frac{1}{C(2 \pi f)}$. Rearranging for frequency $f$: $f^2 = \frac{1}{4 \pi^2 LC}$,which gives $f = \frac{1}{2 \pi \sqrt{LC}}$. Substituting the values: $f = \frac{1}{2 \pi} \sqrt{\frac{1}{(\frac{10}{\pi^2} 	imes 10^{-3}) 	imes (16 	imes 10^{-6})}}$. $f = \frac{1}{2 \pi} \sqrt{\frac{\pi^2}{160 	imes 10^{-9}}} = \frac{1}{2 \pi} \sqrt{\frac{\pi^2}{1.6 	imes 10^{-7}}} = \frac{1}{2 \pi} 	imes \frac{\pi}{\sqrt{1.6 	imes 10^{-7}}}$. Calculating the value: $f = \frac{1}{2} 	imes \frac{1}{\sqrt{16 	imes 10^{-8}}} = \frac{1}{2 	imes 4 	imes 10^{-4}} = \frac{10^4}{8} = 1250 \ Hz = 1.25 \ kHz$.

The frequency of $AC$ at which a $16 \mu F$ capacitor and a $\frac{10}{\pi^2} \ mH$ inductor will have the same reactance is: (in $kHz$)

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