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

#### Solution

Let a LCR circuit is connected across an AC supply with the emf *E* = *E*_{0} sin *ωt*.

Let the inductance in the circuit be *L*

Let the net impedence of the circuit be Z = `sqrt(R^2 + (X_L -X_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(di)/(dt)`

The current in the circuit can be written as `I = I_0 sin (wt + Ø)`

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

Thus, the voltage across the inductor can be written as

`V = LI_0`

⇒ `V = (E_0)/(Z)xxL,`

Therefore, the peak voltage across the inductor is given by `V = (E_0)/RxxL`

if `L/R xx L,`

if L/R > 1*V* > *E*_{0}

Therefore if magnitude of `L/R >1` at resonance the value of the voltage across the inductor will bw greater than the peak value of the supply voltage.