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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.At resonance the capacitance and inductance cancel each other out.
So just use ohms law on Rm.
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.
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.
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.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)
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.
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
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.