Question

The tuning circuit of a radio receiver has a resistance of $50\Omega$, an inductor of $10\text{mH}$ and a variable capacitor. A $1\text{MHz}$ radio wave produces a potential difference of $0.1\text{mV}.$ The value of the capacitor to produce the resonance is $($take${\pi }^2=10)$

• A. $50\text{pF}$
• B. $25\text{pF}$
• C. $5.0\text{pF}$
• D. $2.5\text{pF}$

Answer 1

The correct option is D $2.5\text{pF}$

The resonant frequency of the series $LCR$ circuit is,

$f=\frac{1}{2\pi }\frac{1}{\sqrt{LC}}$

$\Rightarrow LC=\frac{1}{4{\pi }^2{f}^2}$

$C=\frac{1}{4{\pi }^2{f}^2}\times \frac{1}{L}=\frac{1}{4\times 10\times {10}^{12}}\times \frac{1}{10\times {10}^{-3}}$

$C=\frac{1}{4}\times {10}^{-11}=2.5\times {10}^{-12}\text{F}\text{(or)}$

$C=2.5\text{pF}$