Question

The turning 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 values of the capacitor to produce resonance is (Take ${\pi }^2=10$):

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

Answer 1

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

$L=10\text{mH}={10}^{-2}\text{H}$
$f=1\text{MHz}={10}^6\text{Hz}$
$f=\frac{1}{2\pi \sqrt{LC}}$
$f^2=\frac{1}{4{\pi }^{2}LC}$
$\Rightarrow C=\frac{1}{4{\pi }^2{f}^{2}L}=\frac{1}{4\times 10\times {10}^{12}\times {10}^{-2}}=\frac{{10}^{-11}}{4}=2.5\text{pF}$