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Questions regarding AC circuits

  • Engineering
  • Thread starter R A V E N
  • Start date
  • #1
Suppose that we have a simple system like one illustrated in attachment.

If we need to find complex admittance of that system,we can write:

\underline{Y}=G+jB=\frac{1}{\underline{Z}}=\frac{1 }{R+jX_L}\cdot\frac{R-jX_L}{R-jX_L}=\frac{R-jX_L}{R^2+X_L^2}=\frac{R}{R^2+X_L^2}+j\frac{-X_L}{R^2+X_L^2}

from where we can see that it is [tex]B=\frac{-X_L}{R^2+X_L^2}[/tex],althought it is [tex]B=\frac{X_L}{R^2+X_L^2}[/tex].

Why is this "-" just neglected,what is physical explanation of that?

Or it is just hardcore mathematical laws against imperfect physical reality?

Also,is it because complex numbers are just,let`s say it like this,"artificial" extension of real numbers set?

Probably the explanation is that while one physical parameter is rising(susceptance [tex]B[/tex]),the other is lowering(inductive reactance [tex]X_L[/tex]) and vice-versa,like it is in Faraday`s law of induction:


the magnetic field which is produced by induced current(which is in turn produced by induced electromotive force [tex]e[/tex]) is in oposition with the change of outer flux [tex]\phi[/tex](sorry if my technical english sounds a bit clumsy).

But what if there is capacitor instead of inductor?
In that case there is no confusion like this.

There is also a "-" when active and reactive power for system like one illustrated in attachment is calculated.


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Answers and Replies

  • #2
The second question is regarding complex power [tex]\underline{S}[/tex]:

why expression [tex]\underline{S}=\underline{U}\;\underline{I}[/tex] does not give the correct result,instead of that it is used [tex]\underline{S}=\underline{U}\;\underline{I}^*[/tex] where [tex]\underline{I}^*[/tex] is complex-conjugate of [tex]\underline{I}[/tex]?
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  • #3
I hope I will have more luck with this one:

In given circuit:

http://img90.imageshack.us/img90/9085/clipboard05xw4.gif [Broken]

where is:

[tex]R_1=10\;\Omega[/tex], [tex]R_3=2,5\;\Omega[/tex], [tex]R_2=X_C=X_L=5\;\Omega[/tex] and [tex]\underline{E}=50e^{j\frac{\pi}{2}}\;V[/tex],

find the value of current source [tex]\underline{I}_S[/tex].

Voltage drop on [tex]R_3[/tex] is [tex]\underline{U}=100\;V[/tex].

First I calculate current trough [tex]R_3[/tex]: [tex]\underline{I}_{R_3}=\frac{\underline{U}}{R_3}=40\;A[/tex].

Further,by using Superposition theorem and removing branch containing [tex]\underline{I}_s[/tex] I find that [tex]\underline{E}[/tex] produces current of [tex]j20[/tex].When I substract that value from [tex]\underline{I}_{R_3}[/tex] I get [tex]\underline{I}_S=(40-j20)\;V[/tex].

However,correct result is [tex]\underline{I}_S=(-10+j20)\;V[/tex].

What I do wrong?
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  • #4
Problem 4:

Could someone explain to me how to write equations using 2nd Kirchoff`s Law along the closed loop in circuit where we have mutual inductance between two inductors?

I can`t comprehend this correctly at all.

Here is example circuit:

http://img125.imageshack.us/img125/4779/clipboard033mw7.jpg [Broken]

The closest I was was equation with difference in one [tex]+[/tex] instedad of [tex]-[/tex].

I need explanation here,not just equations,I already have them.
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