Question

When a 60 mH inductor and resistor are connected in series with an AC voltage source, the voltage leads the current by 60°. If the inductor is replaced by a 0.5 mF capacitor, the voltage lags behind the current by 30°. What is the frequency of the AC supply?

• A. $\frac{1}{2\pi}\times10^{4}Hz$
• B. $\frac{1}{\pi}\times10^{4}Hz$
• C. $\frac{3}{2\pi}\times10^{4}Hz$
• D. $\frac{1}{2\pi}\times10^{8}Hz$

Answer 1

Answer: (A)

$from \left(i\right) and \left(ii\right)$
$\frac{tan 60^{\circ}}{tan 30^{\circ}} =\frac{\omega L}{R\times\frac{1}{\omega CR}} =\omega^{2}LC$
$\frac{\sqrt{3}}{1\sqrt{3}}=\omega^{2}\times60\times10^{-3}\times\frac{0.5}{10}\times10^{-6} \Rightarrow 3=\omega^{2}\times3\times10^{-8} or, \omega^{2}=10^{8} or,\omega=10^{4}$
$2\pi f =10^{4} \therefore f =\frac{10^{4}}{2\pi}$