Question

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

• A. $2.5pF$
• B. $5.0pF$
• C. $25pF$
• D. $50pF$

Answer 1

The correct option is A $2.5pF$

Given:
Resistance $R=50\Omega$

Inductance $L=10mH$

Frequency $f=1MHz$

For resonance

$\omega L=\frac{1}{\omega C}$

$C=\frac{1}{{\omega }^{2}L}$

$=\frac{1}{(2\pi f{)}^{2}L}$

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

$=\frac{1}{4{\pi }^2({10}^6{)}^2\times 10\times {10}^{-3}}$

$=2.5\times {10}^{-12}F$

$=2.5pF$