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

6 years ago

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Answers : (2)

										
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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6 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

6 years ago

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