Question

In an $LCR$ circuit, the resonating frequency is $500kHz$. If the value of $L$ is doubled and value of $C$ is decreased to $\frac{1}{8}$ times of its initial values, then the new resonating frequency in $kHz$ will be

• A. 250
• B. 500
• C. 1000
• D. 2000

Answer 1

Answer: (C)

$f_1=500\times 10^{3}Hz$
$L^{\prime }\to 2L,C^{\prime }\to \frac{1}{8}C$
$f=\frac{1}{2\pi \sqrt{LC}}$
$\Rightarrow f\propto \frac{1}{\sqrt{CL}}$
$\frac{f_1}{f_2}=\frac{\sqrt{L_2{C}_2}}{\sqrt{L_1{C}_1}}=\sqrt{\frac{(2L)\left(\frac{1}{8}C\right)}{LC}}$
$\frac{f_1}{f_2}=\frac{1}{2}$
$\Rightarrow f_2=1000kHz$