A variety of questions regarding AC circuits

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Homework Statement

In this circuit

(In attachment.)

find the value which is showed on ammeter A,if voltmeter is showing 50\;V.
Values are R=R_2=X_L=X_C=10\;\Omega and R_1=5\;\Omega

The Attempt at a Solution

First I calculate current thought branch with R_2 and X_C:

\underline{I}_2=\frac{\underline{U}}{-jX_C}=j5\;A

then knowing that potential difference at the ends of branches(one with R_2 and X_C and the other with R_1 and X_L) is the same,I proceed:

\underline{U}_1=\underline{U}_2

\frac{\underline{I}_1}{R_1+jX_L}=\frac{\underline{I}_2}{R_2-jX_C}

\underline{I}_2=\frac{R_1+jX_L}{R_2-jX_C}\underline{I}_2=-3.75-j1.25\;A

Overall current \underline{I} in circuit is the sum of two currents from two branches,so

\underline{I}=\underline{I_1}+\underline{I_2}=-3.75+j3.75

module of this value is value showed on ammeter

|\underline{I}|=5.303

However,correct solution is 3\sqrt{5}\approx6.708

Where is the mistake?

 

[R A V E N]

The second question is a theoretical one.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?

 

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Can moderator please delete this doubled topic?