Question

When a \(60\) mH inductor and resistor are connected in series with an AC source, the voltage leads the current by \(60^{\circ }\). If the inductor is replaced by a \(0.5\ \mu\)F capacitor, the voltage lags behind the current by \(30^{\circ }\). The frequency (in Hz) of AC supply is \(\left(\tan 30^{\circ }=\frac{1}{\sqrt{3}},\,\tan 60^{\circ }=\sqrt{3}\right)\)

• A. \(\frac{10^{4}}{\pi }\)
• B. \(\frac{10^{4}}{4\pi }\)
• C. \(\frac{10^{4}}{2\pi }\)
• D. \(\frac{3\times 10^{4}}{2\pi }\)

Hint:

Write the tangent of the phase angle for each circuit and divide the two relations to eliminate the resistance.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For the L-R circuit, \(\tan60^{\circ}=\dfrac{\omega L}{R}=\sqrt{3}\). For the C-R circuit, \(\tan30^{\circ}=\dfrac{1}{\omega CR}=\dfrac{1}{\sqrt{3}}\).

Step 2: Divide the relations
\(\dfrac{\omega L/R}{1/(\omega CR)}=\omega^{2}LC=\dfrac{\sqrt{3}}{1/\sqrt{3}}=3\).

Step 3: Calculation
\(\omega^{2}=\dfrac{3}{LC}=\dfrac{3}{60\times10^{-3}\times0.5\times10^{-6}}=10^{8}\), so \(\omega=10^{4}\ \text{rad/s}\).
\(f=\dfrac{\omega}{2\pi}=\dfrac{10^{4}}{2\pi}\ \text{Hz}\).

Final Answer:
The frequency is 10 to the power 4 over 2 pi hertz. \[ \boxed{\text{(C) }\dfrac{10^{4}}{2\pi}} \]