Question

The frequency of output signal of LC oscillator circuit is \(100 \mathrm{~Hz}\) with capacitance value \(0.1 \mu \mathrm{F}\). If value of capacitance is taken as \(0.2 \mu \mathrm{F}\), the frequency of output signal (A) decreases by \((1 / \sqrt{2})\) (B) increases by \((1 / \sqrt{2})\) (C) decreases by \((1 / 2)\) (D) increases by \((1 / 2)\)

Short answer

The frequency of the output signal decreases by \((1 / \sqrt{2})\) when the capacitance is doubled to $0.2 \mu \mathrm{F}$.

Step-by-step solution

Write down the given information

The given information is: Initial frequency, \(f_{1} = 100 \text{ Hz}\) Initial capacitance, \(C_{1} = 0.1 \ \mu \text{F}\) New capacitance, \(C_{2} = 0.2 \ \mu \text{F}\) We do not know the value of the inductance 'L', but since it does not change in this problem, we can express the change in frequency as a ratio between the two capacitance values.

Frequency formula for LC circuit

The formula for the resonant frequency of an LC circuit is given by: \[f = \frac{1}{2 \pi \sqrt{LC}}\] In our case, the inductance 'L' is constant, so we can write the initial and new frequencies as: \[f_{1} = \frac{1}{2 \pi \sqrt{LC_{1}}}\] \[f_{2} = \frac{1}{2 \pi \sqrt{LC_{2}}}\]

Find the ratio of the frequencies

Divide the equation for \(f_{1}\) by the equation for \(f_{2}\): \[\frac{f_{1}}{f_{2}} = \frac{\frac{1}{2 \pi \sqrt{LC_{1}}}}{\frac{1}{2 \pi \sqrt{LC_{2}}}} = \frac{\sqrt{LC_{2}}}{\sqrt{LC_{1}}}\] Since \(L\) is constant, it cancels out from both numerator and denominator: \[\frac{f_{1}}{f_{2}} = \frac{\sqrt{C_{2}}}{\sqrt{C_{1}}}\] Now, plug in the given values for \(C_{1}\) and \(C_{2}\) and solve for the ratio: \[\frac{f_{1}}{f_{2}} = \frac{\sqrt{0.2 \ \mu \text{F}}}{\sqrt{0.1 \ \mu \text{F}}} = \sqrt{2}\]

Match the result with the given options

The ratio of the frequencies is \(\sqrt{2}\), which means that when the capacitance is doubled, the frequency decreases by a factor of \(1/\sqrt{2}\). Thus, the correct answer is: (A) decreases by \((1 / \sqrt{2})\)