Q. In the circuit shown below, the ac source has voltage $V = 20\cos{(\omega t)}$ volts with $\omega = 2000\ rad/sec$. The amplitude of the current will be nearest to
A) $2A$
B) 3.3$A$
C) $2/\sqrt{5}\ A$
D) $\sqrt{5}\ A$
Explanation
(a) $R = 6 + 4 = 10\Omega$
$X_{L} = \omega L = 2000 \times 5 \times 10^{- 3} = 10\Omega$
$X_{C} = \frac{1}{\omega C} = \frac{1}{2000 \times 50 \times 10^{- 6}} = 10\Omega$
$\therefore Z = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}} = 10\Omega$
Amplitude of current $= i_{0} = \frac{V_{0}}{Z} = \frac{20}{10} = 2A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.