Question

In a series circuit with R = 10Ω and C = 25 μF, at what frequency will the current lead the voltage by 30°?

(a) 11.4 kHz

(b) 4.5 kHz

(c) 1.1 kHz

(d) 24.74 kHz

The question was posed to me in examination.

Query is from Advanced Problems on Resonance topic in section Resonance & Magnetically Coupled Circuit of Network Theory

Answer 1

Right option is (a) 11.4 kHz

To elaborate: From the impedance diagram, 10 -jXC = Z∠-30°

∴ – XC = 10 tan (-30°) = -5.77 Ω

∴ XC = 5.77 Ω

Then, XC = \(\frac{1}{2πfC}\) or, f = \(\frac{1}{2πX_C C} \)

 = \(\frac{1}{2π×5.77×25×10^{-6}}\)

 = \(\frac{10^6}{906.18}\)

= 1103 HZ = 1.1 kHz.