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

419 Points
12 years ago
Dear Nikhil Garg, At resonance the peak current through circuit isi=E/Rpeak voltage across inductanceVL=wL*(E/R)=QEwhere Q is the quality factor (wL/R). Thus at resonance voltage drop across L is Q times the applies voltage. Hence the chief chacterstics of series resonance circuit is 'Voltage magnification'.Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

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12 years ago

Hello

Nikhil

In a series RLC circuit at resonance, the current is limited only by the resistance of the circuit

$I = \frac{V}{R}$

If R is small, consisting only of the inductor winding resistance say, then this current will be large. It will drop a voltage across the inductor of

$V_\mathrm L = \frac{V}{R} \omega_0 L$

An equal magnitude voltage will also be seen across the capacitor but in antiphase to the inductor. If R can be made sufficiently small, these voltages can be several times the input voltage. The voltage ratio is, in fact, the Q of the circuit,

$\frac{V_\mathrm L}{V} = Q$

So there can be possibility the Voltage accross induction become larger than the input voltge...this is called Q-amplification at resonance ...

With regards

Yagya