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?

 

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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?