Math Help
Gday everyone,
I need a bit of assistance with a math problem. If I have a 4.9uF Capacitor used in parallel across an Inductive circuit to improve the power factor on a 230v 50Hz circuit. If I then Inject a 1050Hz signal into this circuit how much current will the Capacitor see. I have so far: The Current through the Capacitor for the 50Hz signal is: Xc= 1/ 2*pi*f*C = 1/ 6.28*50*.0000049 = 1/.00154 =649.3 Ohms Ic= Vc/Xc = 230/649.3 = .354A Now for the 1050Hz signal I get Ic = 7.43A Do I now just add these two together so I get 7.43A + .354A = 7.784A ? I tried to check this with LTspice but I only get 7.43A at the Capacitor & not the sum of the two currents together. So now I am a bit confused as always? Cheers 
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Hi abraxalito,
Thanks for the reply, you made me think more about this with your comments & I seem to have a misunderstanding of how this would actually be connected to get the results needed. I am only learning & I noticed this depends on how things would be connected. All I have to work with is the following: If a 4.9uF Capacitor is used to correct the power factor of a Fluorescent lamp connected to a 230v 50Hz supply, what current will it take when the supply authority injects a 1050Hz signal onto the supply system to switch on controlled load equipment? This is all I have to work with, & like I said after reading your comments I am now unsure about a lot of things:confused: Cheers 
Glad to have got you thinking about this. In order to make progress we'd need to know something about the relative amplitude of the 1050Hz signal. I am guessing it cannot be anywhere near the 230VRMS of the 50Hz mains. Also I guess that this 1050Hz signal is only going to last for a short while  so are you worried that by having a capacitor on the supply it might get attenuated and hence not turn on the equipment like its supposed to?

Yes, I understand what your saying about the Amplitude of the 1050Hz signal but there is no value given for this. There is only the information in the above question given?
I have taken on a course to get a better understanding of electronics etc but some of the questions seem to lack information & I was going to work on a 230V amplitude for this signal. Cheers 
I did a bit of googling around and found a worst case amplitude of 60V but in practice its going to be under 10% of the mains signal. So I suggest trying with a 10VRMS signal for the 1050Hz component. The higher frequency is 21X greater so into a capacitor, this 10V signal will give practically the same current as 230V at 50Hz.

Yes I see what you mean.
I have however just noticed in the next question I have to answer it says: What value of choke Inductance is needed to limit the current to the normal 50Hz value of in the above question. So I guess they may be refering to a 230V rms value for the 1050Hz signal for this question. I can work this out once I know what to do with the original question. I can calculate the currents of the two seperate signals but I am unsure if they just sum together or not when mixed? Cheers 
Yes, I agree with your diagnosis of the poorly specified (and seemingly unrelated to the realworld) question.
Yes, you can calculate the currents separately and simply sum them  both phases will be 90 degrees as its just a capacitor load. In general though, AC currents can't be summed in magnitude only because they have associated phase. So the normal case is vector summation. When you add a choke into the equation to reduce the current to the 50Hz level, the phase of that resulting current will not be 90 degrees even though the magnitude can be made the same. 
Hi abraxalito,
Thanks for that, I see what your saying with Vector summation for AC. You mentioned that this is a capacitive load, I am not sure I understand as a Fluro light has a Ballast Coil which makes it Inductive & capacitive with the addition of the power factor correction capacitor doesn't it? I may be misunderstanding also? Cheers 
Oh, yeah, you're right. I was limiting my comments to where you said 'how much current will the capacitor see?'  not taking account of the rest of the circuit.

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