Electrical Engineering - RLC Circuits and Resonance - Discussion

Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 2)
2.
In a series RLC circuit that is operating above the resonant frequency, the current
lags the applied voltage
leads the applied voltage
is in phase with the applied voltage
is zero
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Nabin Baniya said:   6 years ago
Above resonance frequency, the circuit is inductive.

ARVIND said:   7 years ago
The reactive voltage VL and VC, does it cancel each other at resonance? Can anyone explain?

Pruthviraj said:   8 years ago
Resonant frequency(fr) = 1/(2 * 3.14 * (LC)1/2,
Take that root.
Given that RLC circuit is above the fr then we know XL > XC in case fr is more then current must be low and that case current should be lags by applied voltage.

Sandeep sana said:   10 years ago
Resonant frequency XL>XC. So current lags the applied voltage.

Sandeep wasnik said:   10 years ago
At resonant condition.

XL = XC @ resonant frequency.

Resonant frequency more if XL>XC and less if XL<XC.

Here it is more and it means XL is high and hence L is high (XL = 360FL).

L always takes lags the current.

Hence the current lags the applied voltage.

Varshney said:   10 years ago
Above resonance frequency current lags voltage because at this situation XL>XC.

RAJ said:   1 decade ago
At resonance I in phase with V.

Aggrey kere said:   1 decade ago
At resonance, Xl =XC. Thus there no phase angle. Therefore; no lug or lead.

Manindra said:   1 decade ago
At resonant frequency, (Xl=Xc).

Above resonant frequency, (Xl>Xc).

Below resonant frequency (Xl<Xc).

Abhi said:   1 decade ago
At frequencies above the resonance frequency the series RLC circuit. Behaves as an inductive load (evident from equivalent impedance equation) hence current will lag.


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