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. $\frac{1}{2\pi }\times 10^{4}Hz$
• B. $\frac{1}{\pi }\times 10^{4}Hz$
• C. $\frac{3}{2\pi }\times 10^{4}Hz$
• D. $\frac{1}{2\pi }\times 10^{8}Hz$

Answer 1

Answer: (A)

$from\left(i\right)and\left(ii\right)$
$\frac{tan60^{\circ }}{tan30^{\circ }}=\frac{\omega L}{R\times \frac{1}{\omega CR}}={\omega }^{2}LC$
$\frac{\sqrt{3}}{1\sqrt{3}}={\omega }^2\times 60\times 10^{-3}\times \frac{0.5}{10}\times 10^{-6}\Rightarrow 3={\omega }^2\times 3\times 10^{-8}or,{\omega }^2=10^{8}or,\omega =10^4$
$2\pi f=10^4\therefore f=\frac{10^4}{2\pi }$