When two capacitors are connected in parallel across a 12.3 v rms, 1.46 khz oscillator, the oscillator supplies a total rms current of 560 ma. when the same two capacitors are connected to the oscillator in series, the oscillator supplies an rms current of 124 ma?

3.32µF and 1.64µF

Since, you haven't actually asked a question, I am going to make a guess on what the question is based upon the data provided. My educated guess is "What are the values of the two capacitors?"

The formula for the Capacitive reactance is

X = 1 / (2*pi*f*C)

where

X = reactance

f = frequency

C = capactance

Let's solve for C

X = 1 / (2*pi*f*C)

CX = 1 / (2*pi*f)

C = 1 / (2*pi*f*X)

Now with the capacitors in parallel, we have a reactance of:

I = V/X

IX = V

X = V/I

X = 12.3/0.56

X = 21.96428571

So the capacitance is:

C = 1 / (2*pi*f*X)

C = 1 / (2*pi*1460*21.96428571)

C = 4.96307x10^-6 = 4.96307 µF

And with the capacitors in series we have a reactance of:

X = V/I

X = 12.3/0.124

X = 99.19354839

So the capacitance is:

C = 1 / (2*pi*f*X)

C = 1 / (2*pi*1460*99.19354839)

C = 1.09896x10^-6 = 1.09896 µF

Now we can setup two equations with 2 unknowns.

4.96307 = x + y

1.09896 = 1 / (1/x + 1/y)

y = 4.96307 - x

1.09896 = 1 / (1/x + 1 / (4.96307 - x))

1.09896 = 1 / ((4.96307 - x) / (x (4.96307 - x)) + x / (x (4.96307 - x)))

1.09896 = 1 / (((4.96307 - x) + x) / (x (4.96307 - x)))

1.09896 = 1 / (4.96307 / (x (4.96307 - x)))

1.09896 = x (4.96307 - x) / 4.96307

5.45422 = x (4.96307 - x)

5.45422 = 4.96307x - x^2

0 = 4.96307x - x^2 - 5.45422

0 = - x^2 + 4.96307x - 5.45422

We now have a quadratic equation. Use the quadratic formula to solve, getting roots of 3.320460477 and 1.642609523. You may notice that those 2 values add up to 4.96307. This is not coincidence. Those are the values of the two capacitors in µF. Rounding to 3 significant figures gives us 3.32µF and 1.64µF.