AC circuit analysis problem

Thread Starter

Doros

Joined Dec 17, 2013
144
Dear all hello and the best for 2022,

during my self paced training I am doing in circuit analysis with the help of a book, I faced the attached problem which I can notsolve. I believe that the frequency should be given in order the problem to be solved.

Is frequency missing and the problem canot be solved, or I am missing something?

many thanks for your help
 

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RBR1317

Joined Nov 13, 2010
713
What if you just pick a frequency, and solve the problem for that? Then pick a different frequency, and again solve the problem. Then compare the answers...
 

Thread Starter

Doros

Joined Dec 17, 2013
144
Thanks RBR1317,

Actually I run a multisim simulation, and the capacitor value depends on the frequency. Only at 79 HZ you have power factor 1 meaning the circuit is purely resistive

Could the problem miss the frequency value?

thanks
 

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Jony130

Joined Feb 17, 2009
5,487
Could the problem miss the frequency value?
It looks like it they forget to add this information. Because from a circuit theory about how the LC circuit behaves at resonance.
We can solve for C. For example, if F = 50Hz we have XL = 3.14Ω --->C = 1/(2*pi * 3.14Ω * 50Hz) ≈ 1000μF

So what is the name of the book?
 

Ramussons

Joined May 3, 2013
1,404
This may be another question to a set of conditions given earlier.

v(t) and i(t) will be in phase at a resonant frequency determined by C and the 100 mH capacitor.
 

RBR1317

Joined Nov 13, 2010
713
Is the frequency missing? I supposed it is
This seems to be an AC power circuit, so you may be expected to use the AC power frequency, typically either 50 Hz or 60 Hz. Did the author make a blanket statement somewhere near the beginning of the chapter about what power frequency you were expected to use?
 

Thread Starter

Doros

Joined Dec 17, 2013
144
Yes but you end up spending hours for something that can not be solved in the first place.

The frequency to have the correct answer, C=400μF, needs to be 79Hz.

Thanks for your input
 

MrAl

Joined Jun 17, 2014
11,389
Hello there,

This problem can be solved for all frequencies.
Recall that the current is in phase with the voltage in a purely resistive load, therefore the cap and inductor effects must cancel out leaving a pure resistance.

For the case of C=400uf i get a bit over 79Hz, about 79.6 Hertz, but we can actually solve for C symbolically:
C=1/(w^2*L)

Knowing L=10mH this becomes:
C=100/w^2
where w=2*pi*f

and so we can find C for any frequency f from this or solve for f with a given C.
For solving for w we get:
w^2=100/C
and of course:
w=10/sqrt(C)
and with C=400e-6 we end up with:
w=10/0.02
so:
w=500
and so:
f=500/(2*pi)
f=79.6 rounded to three significant figures.
 
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