A generator with an adjustable frequency of oscillation is wired in series to an inductor of L = 2.50 mHand a capacitor of$\mathbf{C}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}$. At what frequency does the generator produce the largest possible current amplitude in the circuit?

$$\begin{gathered} \mathrm{L}=\text{2.50}{\text{mH=2.5×10}}^{-3}\text{H} \\ \mathrm{C}=\text{3.00μF=3}{\times 10}^{-6}\mathrm{F} \end{gathered}$$

By using the formula for the angular frequency of oscillation for the ideal circuit and the relation between angular frequency and oscillating frequency, find thefrequency at which the generator produces the largest possible current amplitude in the circuit.

Formulae are as follows:

$$\begin{gathered} \mathrm{\omega }=\frac{\text{1}}{\sqrt{\mathrm{LC}}} \\ \mathrm{\omega }=\text{2}\mathrm{\pi f} \end{gathered}$$

Where,

ω= angular frequency,

f = frequency,

L = inductance,

C = capacitance.

Angular frequency of oscillation is given by,

$$\begin{gathered} \mathrm{\omega }=\frac{\text{1}}{\sqrt{\mathrm{LC}}} \\ \mathrm{\omega }=\text{2}\mathrm{\pi f} \\ \text{2}\mathrm{\pi f}=\frac{\text{1}}{\sqrt{\mathrm{LC}}} \\ \mathrm{f}=\frac{1}{\text{2}\mathrm{\pi }\sqrt{\mathrm{LC}}} \end{gathered}$$

Rearranging the terms,

$\mathrm{f}=\frac{1}{\text{2}\mathrm{\pi }\sqrt{\mathrm{LC}}}$

Substituting the given values,

$$\begin{gathered} \mathrm{f}=\frac{1}{2\mathrm{\pi }\sqrt{2.5\times {10}^{-3}\times 3.00\times {10}^{-16}}} \\ \mathrm{f}=1.84\times {10}^3\mathrm{Hz} \\ =1.84\mathrm{KHz} \end{gathered}$$

Hence,the frequency at which the generator produces the largest possible current amplitude in the circuit is 1.84 kHz.

Using the formula for angular frequency and the relation between the angular frequency and oscillating frequency, found the required answer.