In a series LC circuit at resonance the inductive and capacitive reactances equal to cancel each other out. Doesen't that mean that the voltage and current are in phase with no lead or lag?

Is there any way in which a series resonant LC circuit (operating at resonance) can have current which leads or lags the voltage?

 

This part of the Ebook should answer your question - http://www.allaboutcircuits.com/vol_2/chpt_6/3.html

 

In a series LC circuit at resonance the inductive and capacitive reactances equal to cancel each other out. Doesen't that mean that the voltage and current are in phase with no lead or lag?

Is there any way in which a series resonant LC circuit (operating at resonance) can have current which leads or lags the voltage?

You basically answered your own question. The simple answer is no for the reason you gave.

It's easiest to think in terms of the phasor representation using complex impedances. To have a phase lead or lag, the impedance needs to be complex or imaginary, but as you pointed out it is real for an LC circuit at resonance.

So, you need to either modify the circuit so that it is not a simple series LC, or tune off the resonance peak. The bottom line is the effective impedance needs to have a nonzero imaginary component in order for current and voltage to not be in phase.

For a series RLC circuit, the impedance is:

\( Z=R + X_L + X_C = R + {\rm j} \omega L+ {{1}\over{{\rm j} \omega C}}\)

where j is the square root of negative one. Remember that j=-1/j which is what allows the inductive and capacitive reactances to cancel.

 

"where j is the square root of negative one. Remember that j=-1/j which is what allows the inductive and capacitive reactances to cancel".

j= -1/j not a truth statement.

 

j= -1/j not a truth statement.

DEFINITION: j is a number which when multiplied by itself equals -1

Start with an obvious equality: j = j

Multiply one side by -1/(-1) which equals unity: j = j * (-1)/(-1)

Convert the -1 in the denominator to j*j using the definition of j: j = -j/(j*j)

Note that j/j equals unity since any number divided by itself is unity (excluding 0/0): j = -1/j

Hence, j = -1/j is a truth statement.

 

The simplest definition of j is (in my opinion).

\(j=\sqrt{-1}\)

 

The simplest definition of j is (in my opinion).

\(j=\sqrt{-1}\)

What is the alternative? Not only is it the simplest definition, but I thought it was the only definition. I'd be interested to hear of any alternatives, if they exist.

 

The simplest definition of j is (in my opinion).

\(j=\sqrt{-1}\)

I concur with the latter statement....Happy fourth of JULY..........

 

I concur with the latter statement....Happy fourth of JULY..........

So, which "former statement" do you not concur with?

How are we helping the OP with these statements that seem to imply there is an unspoken issue?

The definition given is exactly the one I gave. I gave a two different verbal definitions that are identical to this one. The second one was given as a preferred wording to prove the identity I gave, and which you said was wrong.

The following responses seem to make a subtle implication that something is wrong. Even if this is not the intent, it is not helpful to the OP.

 

What is the alternative? Not only is it the simplest definition, but I thought it was the only definition. I'd be interested to hear of any alternatives, if they exist.

Sorry, Steve.
I took this:

the definition of j: j = -j/(j*j)

out of context and read j = -j/(j*j) as being presented as the definition of j.
I reread the post and realized that the equation was the result of the mathematical manipulation you were performing to prove the "truth statement", which I do not and did not dispute.

 

Sorry, Steve.
I took this:
out of context and read j = -j/(j*j) as being presented as the definition of j.
I reread the post and realized that the equation was the result of the mathematical manipulation you were performing to prove the "truth statement", which I do not and did not dispute.

No problem Ron. Yeah, that would be a needlessly confusing definition.

I just want to make sure we aren't giving the OP the wrong message, that's all.