Question

A transistor-oscillator using a resonant circuit with an inductor L (of negligible resistance) and a capacitor C in series produce oscillations of frequency f. If L is doubled and C is changed to 4C, the frequency will be :

• A. $\frac{\text{f}}{4}$
• B. 8f
• C. $\frac{\text{f}}{2\sqrt 2}$
• D. 2f

Answer 1

Answer: (C)

$\frac{\text{f}}{2\sqrt 2}$

Answer 2

In a series $LC$ circuit, frequency of $LC$ oscillations is given by $f=\frac{1}{2 \pi \sqrt{L C}}$ or $f \propto \frac{1}{\sqrt{L C}}$ $\Rightarrow \frac{f_{1}}{f_{2}}=\sqrt{\frac{L_{2} C_{2}}{L_{1} C_{1}}}$

Given $L_{1}=L, C_{1}=C, L_{2}=2 L, C_{2}=4 C, f_{1}=f$

$\therefore \frac{f}{f_{2}}=\sqrt{\frac{2 L \times 4 C}{L C}}=\sqrt{8}$ $\Rightarrow f_{2}=\frac{f}{2 \sqrt{2}}$

Therefore, the correct option is (C): $\frac{\text{f}}{2\sqrt 2}$