Capacitor Capers

1. Nov 29, 2005

fizixx

Hello all, I have a capacitor problem that I can't seem to get a handle on. Maybe someone out there has some ideas.
Ok.....you have an AC source. It provides 25V (rms) with frequency of 580 Hz.
There is a capacitor connected across its terminals. When a second cap. is connected in parallel to the first one, the source current increases by 0.23 A.
The object of the game here is to find the capacitance of the second cap. I'm just not seeing something about this problem, so I thought I'd post for some ideas.
Thank you ***
fiz~

2. Nov 29, 2005

Staff: Mentor

What's the relationship between the peak current to and from a capacitor and the peak voltage across it? (When the second capacitor is added, that increase in total current is associated with that second capacitor.)

3. Nov 29, 2005

fizixx

It is not known. This is all the information I have for this. At first I thought it a fairly straightforward problem using reactance somehow, and the rules for caps in parallel, but I seem to quickly run into dead-ends. I have inquired about additional info, or 'missing' info, but am assured the problem is complete.

I'm doubtful, but am trying to think it through, but I'm really out of ideas, and thought some fresh thinking could lend a hand.

Thanks for the inquiry Doc.

4. Nov 29, 2005

Staff: Mentor

When I asked "What's the relationship between the peak current to and from a capacitor and the peak voltage across it?" I was trying to get you to state the relevant physics of capacitors, not looking for additional info. You have all the info needed.

You can certainly use capacitive reactance to solve this problem. What's the reactance?

5. Nov 29, 2005

fizixx

Oh I see.....ok.....thanks for clarifying that a bit. The cap reactance is:

X_c = 1/[2(pi)fC]

f = freq = 580 Hz
C = (capacitance) unknown for either cap.

6. Nov 29, 2005

Staff: Mentor

All good. Now, given the current and the voltage, find X_c for the second capacitor. Then use it to find the capacitance.

7. Nov 29, 2005

fizixx

That's where I'm stuck I guess......

Do you mean V/I = X_c?

==> 25V/0.23 A = X_c ?? And then put in for what X_c is and use it to find "C"??

This doesn't make sense to me if this is your implication.

8. Nov 29, 2005

mezarashi

A capacitor has what is known as impedence just like a resistor, except that it is entirely complex. The complex part of impedence is termed as reactance. The reactance of a capacitor can be defined as:

$$X_c = \frac{1}{2\pi f C}$$

Now if we treat this reactance just like you would treat a resistor's resistance. How would you solve the problem? You want to find the reactance of the second capacitor, because with it, finding the capacitance is just plug in the equation.

9. Nov 30, 2005

fizixx

Thank you for the thought, but here's wehre I'm seeing a dead end:
From V = IR
==> V = I(X_c) (sorry everyone, I don't know hnow to use the nice looking math-type and fonts.)
I'm assuming this is what you are alluding to. Then since X_c = 1/(2(pi)fC)
==> V = I/(2(pi)fC)
-then- C = I/(2(pi)fV)
But..(if this IS the correct aproach)..part of my problem is...the current....surely I is not the 0.23 A stated in the problem....is it?
This should be straightfoprward, but to me it is not, and I'm pretty well mired down with this one.

10. Nov 30, 2005

Staff: Mentor

Yes.

Yes.

Why not?

11. Nov 30, 2005

Staff: Mentor

It is. When the first capacitor is connected across the power source, some (unknown) current is drawn. When the second capacitor is added in parallel, an additional current of 0.23 A is drawn. That's the current flowing to and from that second capacitor.

12. Nov 30, 2005

fizixx

Thanks Doc, and everyone......yes, you're right, and I don't know why I couldn't get past that, but it finally clicked that that is what's going on. Classic case of making something harder than it has to be.

I arrived at a reasonable answer of C =2.52 uF

I appreciate the help!

:)