Question

What do you mean by electrical resonance? An LC circuit is in a condition of resonance. If C=$1.0\times {{10}^{-6}}F$ and $L=0.25H$, then find the frequency of oscillation of the circuit.

Answer 1

Hint: LC circuit is a closed loop circuit with two elements only and those are capacitor and an inductor. It does not have resistance in the circuit. It resonates at a natural frequency. Therefore it stores electrical energy.

Complete step by step solution:

Electrical resonance: In A.C. circuit with inductor and capacitor, when the frequency of A.C is gradually changed at certain frequency, the impedance becomes minimum. This condition is called resonance and such a circuit is called a “resonant circuit”.

Circuit: A series combination of inductor of inductance L, capacitor of capacitance C connected with source of alternating e.m.f. forms a series LC circuit as shown below,

Formula of inductive reactance is,

${{X}_{L}}$$=2\pi fL$

Formula of capacitive reactance is,

${{X}_{C}}=\dfrac{1}{2\pi fC}$

At resonance, ${{X}_{L}}={{X}_{C}}$ -(1)

Put the value of capacitive reactance and inductive reactance in equation (1).

Therefore, 

$2\pi fL=\dfrac{1}{2\pi fC}$

$2\pi {{f}^{2}}L=\dfrac{1}{2\pi C}$

Therefore,

${{f}^{2}}=\dfrac{1}{{{(2\pi )}^{2}}LC}$

$f=\dfrac{1}{2\pi \sqrt{LC}}$

Where, C = capacitor

L = inductor

Numerical: Given data-

C=$1.0\times {{10}^{-6}}F$and $L=0.25H$

We know that, frequency of oscillation of the LC circuit in resonance is given by,

$f=\dfrac{1}{2\pi \sqrt{LC}}$

$f=\dfrac{1}{2\pi \sqrt{1\times {{10}^{-6}}\times 0.25}}$

$f=318.3Hz$

Therefore, Frequency of oscillator of the LC circuit $f=318.3Hz$.

Note: Frequency of oscillation of the LC circuit  is dependent on capacitor and inductor only. It is independent of resistance. The frequency at which capacitive reactance and inductive reactance are same in magnitude and opposite in sign is called resonant frequency. Unit of frequency is Hz.