When a 60 text{ mH} inductor and a resistor a | vedclass

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(A) Given, inductance $L = 60 	ext{ mH} = 60 	imes 10^{-3} 	ext{ H}$.Phase difference in $L-R$ circuit, $	heta_{1} = 60^{\circ}$.Capacitance $C =

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$\frac{1}{2 \pi} 	imes 10^{4} 	ext{ Hz}$

Vedclass · Answer

(A) Given, inductance $L = 60 	ext{ mH} = 60 	imes 10^{-3} 	ext{ H}$.Phase difference in $L-R$ circuit, $	heta_{1} = 60^{\circ}$.Capacitance $C = 0.5 	ext{ } \mu	ext{F} = 0.5 	imes 10^{-6} 	ext{ F}$.Phase difference in $R-C$ circuit, $	heta_{2} = 30^{\circ}$.For $L-R$ circuit, $	an 	heta_{1} = \frac{X_{L}}{R} = \frac{\omega L}{R} \quad \dots(i)$.For $R-C$ circuit, $	an 	heta_{2} = \frac{X_{C}}{R} = \frac{1}{\omega CR} \quad \dots(ii)$.Dividing $(i)$ by (ii): $\frac{	an 	heta_{1}}{	an 	heta_{2}} = \frac{\omega L / R}{1 / (\omega CR)} = \omega^{2} LC$.Substituting values: $\frac{	an 60^{\circ}}{	an 30^{\circ}} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3 = \omega^{2} LC$.$\omega^{2} = \frac{3}{LC} = \frac{3}{60 	imes 10^{-3} 	imes 0.5 	imes 10^{-6}} = \frac{3}{30 	imes 10^{-9}} = 10^{8}$.$\omega = \sqrt{10^{8}} = 10^{4} 	ext{ rad/s}$.Since $\omega = 2 \pi f$, then $f = \frac{\omega}{2 \pi} = \frac{10^{4}}{2 \pi} 	ext{ Hz}$.

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