Question

Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?

Answer 1

Explanation:

Step 1:

Let an LCR circuit be connected with the emf via an AC supply                     $E=E 0 \sin \omega t$

Let the Circuit Inductance be L

Let the circuit have net impedance

$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$

Where,

R = circuit  resistance  

XL = inductor  reactance  

XC = capacitor reactance

Step 2:

The magnitude of the inducer voltage is given by the

$V=L \frac{d i}{d t}$

The current in the circuit can be summarised as

$I=I_{0} \sin (\omega t+\phi)$

Where, ϕ is the difference in phase between the current and the supply voltage

Therefore, the inductor voltage can be written as

$V=L I_{0} \cos (\omega t+\phi)$

Step 3:

Thus the voltage peak value is given via the inducer

$\begin{aligned}&V=L I_{0}\\&V=\frac{E_{0}}{Z} \times L\end{aligned}$

Hence the peak voltage throughout the inducer is provided by

$V=\frac{E_{0}}{2} \times L$

Step 4:

At resonance Z = R,  

$\begin{aligned}&V=\frac{E_{0}}{Z} \times L\\&\text { If } \frac{L}{R}>1\\&V>E_{0}\end{aligned}$

Therefore, if $\frac{L}{R}>1$ magnitude at resonance, the voltage value across the inductor will be greater than the supply voltage peak value.