# Homework Help: Questions regarding AC circuits

1. May 6, 2008

### R A V E N

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 $$B=\frac{-X_L}{R^2+X_L^2}$$,althought it is $$B=\frac{X_L}{R^2+X_L^2}$$.

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,lets say it like this,"artificial" extension of real numbers set?

Probably the explanation is that while one physical parameter is rising(susceptance $$B$$),the other is lowering(inductive reactance $$X_L$$) and vice-versa,like it is in Faradays law of induction:

$$e=-\frac{d\phi}{dt}$$

the magnetic field which is produced by induced current(which is in turn produced by induced electromotive force $$e$$) is in oposition with the change of outer flux $$\phi$$(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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Last edited: May 6, 2008
2. May 6, 2008

### R A V E N

The second question is regarding complex power $$\underline{S}$$:

why expression $$\underline{S}=\underline{U}\;\underline{I}$$ does not give the correct result,instead of that it is used $$\underline{S}=\underline{U}\;\underline{I}^*$$ where $$\underline{I}^*$$ is complex-conjugate of $$\underline{I}$$?

Last edited: May 6, 2008
3. May 8, 2008

### R A V E N

I hope I will have more luck with this one:

In given circuit:

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

where is:

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

find the value of current source $$\underline{I}_S$$.

Voltage drop on $$R_3$$ is $$\underline{U}=100\;V$$.

First I calculate current trough $$R_3$$: $$\underline{I}_{R_3}=\frac{\underline{U}}{R_3}=40\;A$$.

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

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

What I do wrong?

Last edited by a moderator: May 3, 2017
4. Aug 27, 2008

### R A V E N

Problem 4:

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

I cant 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 $$+$$ instedad of $$-$$.

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

Last edited by a moderator: May 3, 2017