Math Help

[wdcw]

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

 

[abraxalito]

Do I now just add these two together so I get

7.43A + .354A = 7.784A ?

Can't quite follow what you're trying to achieve here. Those two currents are with two different stimuli applied to the circuit - one with 50Hz and the other with 1050Hz.

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?

So in LTSpice do you have the two voltage sources (50Hz and 1050Hz) in series? Better post up the simulation schematic so we can see what you're up to

 

[wdcw]

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

Cheers

 

[abraxalito]

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?

 

[wdcw]

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

 

[abraxalito]

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.

 

[wdcw]

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

 

[abraxalito]

Yes, I agree with your diagnosis of the poorly specified (and seemingly unrelated to the real-world) 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.

 

[wdcw]

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

 

[abraxalito]

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.

 

[wdcw]

No worries, can I still just sum the two currents together seeing that the circuit is Inductive & Capacitive.


Cheers

 

[abraxalito]

When you sum the inductor current with the capacitor's, remember to use vectors as their phases won't match LTSpice will do this for you though - just mouse over the voltage source and a current symbol will appear - click to graph the overall current from that source.

 

[wdcw]

abraxalito

I need to find the total current through the Capacitor when a 1050Hz signal is injected into a 230V signal.
So I am not sure a phasor diagram is going to find that, is it?

I have no value for L so I am not sure what IL or XL is.

Basically two voltages with different frequencies when mixed, do there currents add together?

Cheers

 

[abraxalito]

You have the C value, that's 4.9uF. Given that the question isn't asking about current into the inductive part (which means both phases are staying at 90), you can add together the current at 50Hz and that at 1050Hz. I haven't tried two different frequency sources in LTSpice myself, don't know if that works.

A phasor diagram when there's only currents in the same phase seems a bit overkill To answer your final question - its 'yes' when the currents are in phase and the load is a linear one (like a capacitor as you've got here).

 

[wdcw]

Yes, I see.

Thanks for your time abraxalito, it's greatly appreciated.

Thanks Again

Cheers

 

[abraxalito]

You're welcome, its blown a cobweb or two out of my rusty brain in the area of phasors...