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 ?

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Queentdawg9261 Queentdawg9261

07-12-2017
Physics

contestada

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 ?

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Kalahira Kalahira · Answer 1 · 2017-12-17T23:21:02+01:00

3.32ÂµF and 1.64ÂµF

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 ÂµF

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 ÂµF

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 ÂµF. Rounding to 3 significant figures gives us 3.32ÂµF and 1.64ÂµF.