current at resonance

Hello all;

could any one help me, i have a circuit as shown in attachment
i want to know the current through the Cp at resonance?
i know that the current capacitor is :I= -ωCp Vsinωt
so is it ωCp V , or is it zero?

At resonance the capacitive reactance equals the inductive reactance so the net reactive impedance is zero. The only impedance limiting the current is thus the series resistance.

Hi.

I doesn't like DOC files very well because it's a waste of my time to open ms word (or Open Office Writer) just to watch a simple picture. And it's also likely that some users have choosed to not have any compatible word processing application on their machine.

Rather than doc, you should use a image file directly so that users that is interested can look at the picture directly in their web browser. The best image format today is PNG.

A PNG formatted picture has some major advantages over doc:

• Smaller size wich occupies less bandwidth
• Any forum users can watch the image directly in the browser
• Some users will not want to read a doc file of different reasons. Therefore there is better chance to get help if the drawing/shematics is a PNG picture.

I'm a nice guy and have extracted your picture from the doc file and used FSIV to saved it as a png file. Try to compare the file size AND the time it takes to view the picture.

[edit]
I also should try answering your question as well (since I'm pretending to be nice here)
The steps is:
* Find the resonance frequenzy.
* Calculate the reactance for Cp using that freq.
* Use ohm's law to calculate the current, just the way you would have doe it with a resistor. The difference now is that this current will have a 90 degree phase before the voltage. (sorry I couldn't manage to put the words better together)

XL-power,

The AC sinusoidal current through Cp is V*ω*Cp . Voltage source VC locks the voltage across Cp, so its current must follow the applied voltage. The current of the other branch at resonance or not does not affect the current through Cp because they are independent parallel branches.

Ratch

At resonance the capacitance and inductance cancel each other out.

So just use ohms law on Rm.

antknee said:

At resonance the capacitance and inductance cancel each other out.

So just use ohms law on Rm.

Click to expand...

XL-power asked for current through Cp, not Cm as far I can see. That make Rm out of the equation, unless he want to calculate the total current.

crutschow said:

At resonance the capacitive reactance equals the inductive reactance so the net reactive impedance is zero. The only impedance limiting the current is thus the series resistance.

Click to expand...

thanks "crutschow" ,
yes i know this but i am asking if there is a capacitor in parallel with the RLC circuit

Grossel said:

Hi.

I doesn't like DOC files very well because it's a waste of my time to open ms word (or Open Office Writer) just to watch a simple picture. And it's also likely that some users have choosed to not have any compatible word processing application on their machine.

Rather than doc, you should use a image file directly so that users that is interested can look at the picture directly in their web browser. The best image format today is PNG.

A PNG formatted picture has some major advantages over doc:

• Smaller size wich occupies less bandwidth
• Any forum users can watch the image directly in the browser
• Some users will not want to read a doc file of different reasons. Therefore there is better chance to get help if the drawing/shematics is a PNG picture.

I'm a nice guy and have extracted your picture from the doc file and used FSIV to saved it as a png file. Try to compare the file size AND the time it takes to view the picture.

[edit]
I also should try answering your question as well (since I'm pretending to be nice here)
The steps is:
* Find the resonance frequenzy.
* Calculate the reactance for Cp using that freq.
* Use ohm's law to calculate the current, just the way you would have doe it with a resistor. The difference now is that this current will have a 90 degree phase before the voltage. (sorry I couldn't manage to put the words better together)

Click to expand...

really you are a nice and helpful man,, i use a program to draw the circuit and my thinking that not every one will have this program,, so the first choiice is to put it in doc file..
i forget the PNG...

thanks for your reply and your help
i do it :
the resonance frequency is: ω=1/√(Lm Cm )
I=V/Z
I=V/jωCp
I=jVCp/√(Lm Cm)
am i right?

Ratchit said:

XL-power,

The AC sinusoidal current through Cp is V*ω*Cp . Voltage source VC locks the voltage across Cp, so its current must follow the applied voltage. The current of the other branch at resonance or not does not affect the current through Cp because they are independent parallel branches.

Ratch

Click to expand...

thanks for your reply,,
so through you words I=V*ω*Cp
and if i want to get the current at the resonance so i have to put ω=1/√(Lm Cm )
and this give me same result as "Grossel" said
i get I at resonance to be I=jVCp/√(Lm Cm)
so if i want the power the current will be I=VCp/√(Lm Cm)

antknee said:

At resonance the capacitance and inductance cancel each other out.

So just use ohms law on Rm.

Click to expand...

yes Antknee as "Grossel" reply you i didn't want across Cm i want it across the Cp
thank you

Grossel said:

XL-power asked for current through Cp, not Cm as far I can see. That make Rm out of the equation, unless he want to calculate the total current.

Click to expand...

the total current will be across the RLC pluse across the Cp and for the former current i has done with it but am not sure of that of the Cp
,,,
will take the replies which i foind it from all of you

XL-power,

so through you words I=V*ω*Cp
and if i want to get the current at the resonance so i have to put ω=1/√(Lm Cm )
and this give me same result as "Grossel" said
i get I at resonance to be I=jVCp/√(Lm Cm)
so if i want the power the current will be I=VCp/√(Lm Cm)

Click to expand...

Why are you calculating the current at resonance? As I said in my reply, it has no effect on the current through Cp. I gave you the answer for the current through Cp. Why are you asking about power now? Only the resistor dissipates power.

Ratch

Grossel said:

XL-power asked for current through Cp, not Cm as far I can see. That make Rm out of the equation, unless he want to calculate the total current.

Click to expand...

Oh I see. It looked like a piezo so I couldn't see the point of knowing the current through Cp and guessed what was wanted.

I'd say the answer will be I= V*ω*Cp

Hi there,


Yes, if he wants the current through Cp and not Cm then the Rm,Lm,Cm circuit has nothing to do with this problem, unless of course we are to assume some small but significant generator resistance as some problems do. Then (and only then) we have to consider both parallel networks to get the right answer.

Not having to consider the other network, we get get I=V*2*pi*F*C as others have pointed out.

Ratchit said:

XL-power,

Why are you calculating the current at resonance? As I said in my reply, it has no effect on the current through Cp. I gave you the answer for the current through Cp. Why are you asking about power now? Only the resistor dissipates power.

Ratch

Click to expand...

sorry about writing the power, yes am asking for the current.
and i said resonance because i need to get the total current at the resonance.
so the current through RLC branch is just I=V/Rm
and through the Cp is I=V*ω*Cp
and the resonance does not affect the Cp branch so don't have to put ω equal to 1/√(Lm Cm) as i did before

good
thanks a lot

antknee said:

Oh I see. It looked like a piezo so I couldn't see the point of knowing the current through Cp and guessed what was wanted.

I'd say the answer will be I= V*ω*Cp

Click to expand...

nevermind,,, thanks

MrAl said:

Hi there,


Yes, if he wants the current through Cp and not Cm then the Rm,Lm,Cm circuit has nothing to do with this problem, unless of course we are to assume some small but significant generator resistance as some problems do. Then (and only then) we have to consider both parallel networks to get the right answer.

Not having to consider the other network, we get get I=V*2*pi*F*C as others have pointed out.

Click to expand...

I=V*2*pi*F*C
thank you

This a good example of an ill-posed problem.

The OP says "i want to know the current through the Cp at resonance?", but it isn't clear what "resonance" means. The Lm-Rm-Cm series circuit has a resonance frequency, but the complete circuit connected to the VC source has a different resonance frequency.

Also, as Skilling explained in his classic textbook, there are three definitions of resonance frequency given by the American Standards Association. One is the frequency of zero phase angle between current and voltage, another is the frequency of maximum capacitor voltage, and a third is the frequency of natural oscillation of current in the circuit. Usually, the first is the one used in problems like this one.

But, we still don't know if we want 1.), the resonance frequency where the current in the series Lm-Rm-Cm branch is in phase with the voltage across that branch (which is just VC), or do we want 2.), the frequency where the current into the complete circuit (which is the same as the total current from the VC source) is in phase with the voltage from the VC source.

My guess is that the second one is wanted.

In that case, we would need to calculate that resonance frequency, and then calculate the current in Cp at that frequency.

I would think that it wouldn't suffice to simply refer to a resonance frequency as ω and then say that the current in Cp is V*ω*Cp, or V*2*pi*F*Cp. I'd think that a more complicated expression would be required, involving all the components in the circuit.

Only the OP could answer that, my guess is that the first is wanted. The second could only be an academic/mathematical derivation with in my opinion little practical value and too little information to begin. In other words OP has to do that himself.

Hi again,


Yes it could be a trick question aimed at misleading the student into thinking that he has to use the entire circuit to calculate the 'resonant' frequency and that somehow it might be needed while it's a much simpler question.
Because the circuit as is contains two independent circuits, the resonance of the first has nothing to do with the current in the second.
I would have thought that if they wanted us to do the entire circuit then they would have specified at least some small generator resistance, but then again it often depends on the course and where they are in that course.

For example, if they were trying to exemplify how perfect sources do not have any series resistance then the circuit is to be analyzed as is, but on the other hand if they were trying to point out how most real life sources have at least some series resistance no matter how tiny, then we have to analyze the entire circuit to get the true solution. It depends a lot on what they were talking about in the course just prior to asking this question.
Interestingly, as a third possibility they might be handed a schematic from a real life device and asked whether or not the series resistance of the generator matters in that circuit...then it would be up to us to do a few calculations and decide.