Question

explain electrical resonance in an LCR series circuit . Deduce the expression for the resonant frequency of the circuit

Answer 1

LCR circuit:

At any instant the voltage equation of LCR circuit is given by:

$L\frac{di}{dt} + Ri + \frac{q}{c} = E_0 sin wt\\\\L\frac{dq^2}{dt^2} + R \frac{dq}{dt} + \frac{q}{c} = E_0 sin wt$

where i = dq/dt = instantaneous current, q = instantaneous charge

The solution of the above equation is given by:

$i = \frac{E_0 sin(\omega t - \phi)}{\sqrt{R^2 +(\omega L + \frac{1}{\omega C})^2 } }\\\\ \implies i = \frac{E_0}{Z} sin (\omega t - \phi)\\\\ \implies i = i_0 sin (\omega t - \phi)$

where i₀ = E₀/Z = peak current

and $Z = \sqrt{{R^2} + (\omega L - \frac{1}{\omega C})^2}$ is called the impedance of LCR circuit.

Condition for resonance:

When, ωL = 1/ωC then

Z = R = pure resistive = minimum

Hence, max current flows through the circuit. The emf and current are also in same phase. Thus, the circuit is called resonance circuit. The above condition is called resonance condition.

Resonance frequency (f₀):

Under resonance condition,

ω₀L = 1/ω₀C

Or ω₀² = 1/LC

Or ω₀ = 1/√LC

Or 2πf₀ = 1/√LC

Or f₀ = 1/2π√LC

which is the expression for resonance frequency of LCR circuit.