Question

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

Answer 1

Let a LCR circuit is connected across an AC supply with the emf E = E₀ sin ωt.
Let the inductance in the circuit be L
Let the net impedence of the circuit be $Z=\sqrt{R^2+(X_{\mathrm{L}}-X_{\mathrm{C}}{)}^2}$
Where,
R = resistance in the circuit
X_L = reactance due to inductor
X_C = reactance due to capacitor
The magnitude of the voltage across the inductor is given by
$V=L\frac{\mathrm{d}i}{\mathrm{d}t}$
The current in the circuit can be written as $I=I_0\sin (\omega t+\varphi )$
Where, ϕ is the phase difference between the current and the supply voltage
Thus, the voltage across the inductor can be written as
$V=L{I}_0\cos (\omega t+\varphi )$
Thus peak value of the voltage across the inductor is given by
$$\begin{gathered} V=L{I}_0 \\ \Rightarrow V=\frac{E_0}{Z}\times L \end{gathered}$$
Therefore, the peak voltage across the inductor is given by $V=\frac{E_0}{Z}\times L$
At resonance Z = R,
$V=\frac{E_0}{R}\times L$,
If $\frac{L}{R}>1$
V > E₀
Therefore if magnitude of $\frac{L}{R}>1$ at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.