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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Ответ:
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.
Ответ:
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