Question

In an $L C R$ circuit, the resonating frequency is $500\, kHz$. 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)

$f_{1}=500 \times 10^{3} Hz$
$L' \rightarrow 2 L, C'\rightarrow \frac{1}{8} C$
$f=\frac{1}{2 \pi \sqrt{L C}} $
$\Rightarrow f \propto \frac{1}{\sqrt{C L}}$
$\frac{f_{1}}{f_{2}}=\frac{\sqrt{L_{2} C_{2}}}{\sqrt{L_{1} C_{1}}}=\sqrt{\frac{(2 L)\left(\frac{1}{8} C\right)}{L C}}$
$\frac{f_{1}}{f_{2}}=\frac{1}{2} $
$\Rightarrow f_{2}=1000 \,kHz$