How Can AC Circuits Be Analyzed Using Ammeters and Voltmeters?

[R A V E N]

Homework Statement

In circuit given 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}_1=\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)\;A

Module of this value is value showed on ammeter:

|\underline{I}|=5.303\;A

However,correct solution is:

3\sqrt{5}\;A\approx6.708\;A

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?

 

[R A V E N]

OK,I managed to resolve first problem,but what about second one?

 

[R A V E N]

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.