Question
Explain electrical resonance in an $\text{LC}$ parallel circuit. Deduce the expression for the resonant frequency of the circuit.
✓
Answer
Consider a capacitor of capacitance $C,$ and an inductor of large self$-$inductance $L$ and negligible resistance, connected in parallel across a source of sinusoidally alternating emf from below figure. Let the instantaneous value of the applied emf be
$e = e_0 \sin ωt$
Let $iL$ and $ic$ be the instantaneous currents through the inductor and capacitor respectively. As the current in the inductor lags behind the emf in phase by $\pi / 2$ radian,
$i_L=\frac{e_0}{X_{ L }} \sin \left(\omega t-\frac{\pi}{2}\right)=-\frac{e_0}{X_{ L }} \cos \omega t$
where $X_L$ is the inductive reactance.
As the current in the capacitor leads the emf by a phase angle of $\pi / 2$ radian,
$i c=\frac{e_0}{X_C} \sin (\omega t+\pi / 2)=\frac{e_0}{X_C} \cos \omega t$
where $X_c$ is the capacitive reactance.
The instantaneous current drawn from the source is
$i=i_L+i c=e_0\left(\frac{1}{X_C}-\frac{1}{X_{ L }}\right) \cos \omega t$
If $X_L=X_C, i=0$. Thus, no current is drawn from the source if $X_L=X_C$. In such a case, alternating current goes on circulating in the $\text{LC}$ loop, though no current is supplied by the source. This condition is called parallel resonance and the frequency of ac at which it occurs is called the resonant frequency $\left( f _{ r }\right)$.
The condition for resonance is
$X _{ L }= X _{ C }$
$\therefore \omega_{ r } L=\frac{1}{\omega_{ r } C } \therefore \omega_{ r }=\frac{1}{\sqrt{L C}}$
$\therefore$ Resonant frequency, $f_{ r }=\frac{\omega_{ r }}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}}$
In practice, every inductor possesses some resistance and hence even at resonance, some current is drawn from the source. Also, the resonant frequency is different from that for zero resistence.
The resonance curve shows the variation of current $(i)$ and impedance with the frequency of the ac supply, from figure $(b).$ At resonance the current supplied by the source is minimum and the impedance of the circuit is maximum.
$e = e_0 \sin ωt$
Let $iL$ and $ic$ be the instantaneous currents through the inductor and capacitor respectively. As the current in the inductor lags behind the emf in phase by $\pi / 2$ radian,
$i_L=\frac{e_0}{X_{ L }} \sin \left(\omega t-\frac{\pi}{2}\right)=-\frac{e_0}{X_{ L }} \cos \omega t$
where $X_L$ is the inductive reactance.
As the current in the capacitor leads the emf by a phase angle of $\pi / 2$ radian,
$i c=\frac{e_0}{X_C} \sin (\omega t+\pi / 2)=\frac{e_0}{X_C} \cos \omega t$
where $X_c$ is the capacitive reactance.
The instantaneous current drawn from the source is
$i=i_L+i c=e_0\left(\frac{1}{X_C}-\frac{1}{X_{ L }}\right) \cos \omega t$
If $X_L=X_C, i=0$. Thus, no current is drawn from the source if $X_L=X_C$. In such a case, alternating current goes on circulating in the $\text{LC}$ loop, though no current is supplied by the source. This condition is called parallel resonance and the frequency of ac at which it occurs is called the resonant frequency $\left( f _{ r }\right)$.
The condition for resonance is
$X _{ L }= X _{ C }$
$\therefore \omega_{ r } L=\frac{1}{\omega_{ r } C } \therefore \omega_{ r }=\frac{1}{\sqrt{L C}}$
$\therefore$ Resonant frequency, $f_{ r }=\frac{\omega_{ r }}{2 \pi}=\frac{1}{2 \pi \sqrt{L C}}$
In practice, every inductor possesses some resistance and hence even at resonance, some current is drawn from the source. Also, the resonant frequency is different from that for zero resistence.
The resonance curve shows the variation of current $(i)$ and impedance with the frequency of the ac supply, from figure $(b).$ At resonance the current supplied by the source is minimum and the impedance of the circuit is maximum.
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