Q. A $LCR$ series $A.C.$ circuit is tuned to resonance. The impedence of the circuit is now
A) $R$
B) $\left\lbrack R^{2} + \left( \frac{1}{\omega C} - \omega L \right)^{2} \right\rbrack^{1/2}$
C) $\left\lbrack R^{2} + (\omega L)^{2} + \left( \frac{1}{\omega C} \right)^{2} \right\rbrack^{1/2}$
D) $\left\lbrack R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack^{1/2}$
Explanation
(a) In $LCR$ series circuit, impedance $Z$ of the circuit is given by
$Z = \sqrt{(R)^{2} + \left( X_{L} - X_{C} \right)^{2}}$ where $X_{L} = \omega L,\ X_{C} = 1/\omega C$
At resonance $X_{L} = X_{C}\ \therefore Z = R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.