What is the total reactance of a series r-l-c circuit at resonance?

Answer 1

At resonance, the impedance of L and C completely cancel each other out, so you only have R left. If your circuit does not have any gain, then you should be able to determine the RMS current very easily by looking at the RMS voltage input divided by R.

Answer 2

"Series" circuit means that all components share the same current (I).

Then, the total voltage drop of the series (U) is the sum of the voltage drops of each component:

U = UR + UL + UC

valid in time and frequency domains.

In Laplace domain (s is the complex frequency unit):

UR = R IR

UL = s L IL,UC = IC / (sC)

Then, by combining equations:

U = UR + UL + UC= R IR + s L IL + IC / (sC)

Since R, L and C are in series,

IR = IL = IC

Then, the voltage equality can be grouped as:

U = [ R + s L + 1 / (sC) ] I = ZEQ I

where ZEQ is the total equivalent impedance of the series in Laplace transform domain.

If you need the same expression in Fourier domain, just replace:

s=jω,

where j is the imaginary unit (square root of -1) and ω is the gyro frequency ([rad/s]).

Answer 3

The L and C impedances cancel each other out, so you have only R at resonance. Thus the phase angle will be zero.

Answer 4

The vector sum of inductive and capacitive reactance, at resonance, is zero. Meaning that the impedance of the circuit is equal to its resistance.

Answer 5

parallel R

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