Questions regarding AC circuits

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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,let`s 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 Faraday`s 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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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 $$\underline{I}$$?

 

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I hope I will have more luck with this one:

In given circuit:

http://img90.imageshack.us/img90/9085/clipboard05xw4.gif

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?

 

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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

The closest I was was equation with difference in one $$+$$ instedad of $$-$$.

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