Question

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 ?

Answer 1

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.