Q. In an L – C – R circuit, capacitance is changed from C to 2C. For the resonant frequency to remain unchanged, the inductance should be changed from L to
A) 4L
B) 2L
C) $L/2$
D) $L/4$
Explanation
(c) In the condition of resonance
$X_{L} = X_{C}$
or $\omega L = \frac{1}{\omega C}$ …(i)
Since, resonant frequency remains unchanged,
So, $\sqrt{LC} =$constant
or LC = constant
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L_{1}C_{1} = L_{2}C_{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ L \times C = L_{2} \times 2\ C$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L_{2} = \frac{L}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.