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Question
Can the peak voltage across the inductor be greater than the peak voltage of the source in an LCR circuit?
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Solution
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_L -X_c)^2`
Where,
R = resistance in the circuit
XL = reactance due to inductor
XC = 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 > E0
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.
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