Q. A resistor $30\ \Omega$, inductor of reactance $10\ \Omega$ and capacitor of reactance $10\ \Omega$ are connected in series to an AC voltage source $e = 300\sqrt{2}\sin{(\omega t)}$. The current in the circuit is
A) $10\sqrt{2}$ A
B) 10 A
C) $30\sqrt{11}$ A
D) $30/\sqrt{11}$ A
Explanation
(b) $e = 300\sqrt{2}\sin{\omega t}$
$I_{0} = \frac{e_{0}\ }{Z} = \frac{300\sqrt{2}}{\sqrt{(30)^{2} + (10 - 10)^{2}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \{\because Z = \sqrt{R^{2} + \left( X_{L} - X_{C} \right)^{2}\}}$
$= \frac{300\sqrt{2}}{30} = 10\sqrt{2}\ A$
$\therefore\ \ \ $Current$\ I = \frac{I_{0}}{\sqrt{2}} = 10\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.