Question 416: AC fundamentals

Question

The frequency of the output signal of an LC oscillator circuit is '$$\mathrm{F}$$' Hz with a capacitance of 0.1 $$\mu \mathrm{F}$$. If the value of the capacitor is increased to $$0.2~ \mu \mathrm{F}$$, then the frequency of the output signal will be

Accepted Answer

In an LC oscillator circuit, the frequency of the output signal is determined by the inductance $$L$$ and the capacitance $$C$$. The expression for the frequency $$F$$ of an LC oscillator is given by:
$$ F = \frac{1}{2\pi\sqrt{LC}} $$
Given that the initial frequency is $$\mathrm{F}$$ with a capacitance of $$0.1 \mu \mathrm{F}$$, we can write:
$$ \mathrm{F} = \frac{1}{2\pi\sqrt{L \cdot 0.1 	imes 10^{-6}}} $$
Now, if the capacitance is increased to $$0.2 \mu \mathrm{F}$$, the new frequency $$F_{	ext{new}}$$ can be expressed as:
$$ F_{	ext{new}} = \frac{1}{2\pi\sqrt{L \cdot 0.2 	imes 10^{-6}}} $$
To find the ratio of the new frequency to the original frequency, we can divide the expression for $$F_{	ext{new}}$$ by the expression for $$F$$:
$$ \frac{F_{	ext{new}}}{F} = \frac{\frac{1}{2\pi\sqrt{L \cdot 0.2 	imes 10^{-6}}}}{\frac{1}{2\pi\sqrt{L \cdot 0.1 	imes 10^{-6}}}} $$
This simplifies to:
$$ \frac{F_{	ext{new}}}{F} = \frac{\sqrt{L \cdot 0.1 	imes 10^{-6}}}{\sqrt{L \cdot 0.2 	imes 10^{-6}}} $$
Since the inductance $$L$$ and the constants are the same, we get:
$$ \frac{F_{	ext{new}}}{F} = \sqrt{\frac{0.1}{0.2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} $$
So, the new frequency $$F_{	ext{new}}$$ is:
$$ F_{	ext{new}} = \frac{F}{\sqrt{2}} \mathrm{~Hz} $$
Therefore, the correct option is:
Option A: $$\frac{\mathrm{F}}{\sqrt{2}} \mathrm{~Hz}$$

Answer

$$\frac{\mathrm{F}}{\sqrt{2}} \mathrm{~Hz}$$

Answer

$$\frac{\mathrm{F}}{\sqrt{3}} \mathrm{~Hz}$$

Answer

$$\frac{\mathrm{F}}{2} \mathrm{~Hz}$$

Answer

$$2 \mathrm{~F} \mathrm{~Hz}$$

AC fundamentals

The frequency of the output signal of an LC oscillator circuit is '$\mathrm{F}' Hz with a capacitance of 0.1 \mu \mathrm{F}. If the value of the capacitor is increased to 0.2~ \mu \mathrm{F}$, then the frequency of the output signal will be

Choose the correct answer: