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

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

Grade:12

2 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
10 years ago
Dear Nikhil Garg, 

At resonance the peak current through circuit is
i=E/R
peak voltage across inductance
VL=wL*(E/R)
=QE
where 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'.


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Askiitians_Expert Yagyadutt
askIITians Faculty 74 Points
10 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

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