Q. In AC circuit in which inductance and capacitance are joined in series. Current is found to be maximum when the value of inductance is 0.5 H and the value of capacitance is $8\ \mu F$. The angular frequency of applied alternating voltage will be
A) 4000 Hz
B) 5000 Hz
C) $2 \times 10^{5}\ Hz$
D) 500 Hz
Explanation
(d) For current to be maximum , $X_{L} = X_{C}$
Hence, resonant frequency
$f = \frac{1}{2\pi} = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 8 \times 10^{- 6}}} = \frac{10^{3}}{4\pi}$
But angular frequency
$\ \ \ \ \ \ \ \ \ \ \ \omega = 2\pi f$
$\omega = \frac{2\pi \times 1000}{4\pi} = 500\ Hz$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.