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## Homework Statement

In this circuit

*(In attachment.)*

find the value which is showed on ammeter A,if voltmeter is showing [tex]50\;V[/tex].

Values are [tex]R=R_2=X_L=X_C=10\;\Omega[/tex] and [tex]R_1=5\;\Omega[/tex]

## The Attempt at a Solution

First I calculate current thought branch with [tex]R_2[/tex] and [tex]X_C[/tex]:

[tex]\underline{I}_2=\frac{\underline{U}}{-jX_C}=j5\;A[/tex]

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

[tex]\underline{U}_1=\underline{U}_2[/tex]

[tex]\frac{\underline{I}_1}{R_1+jX_L}=\frac{\underline{I}_2}{R_2-jX_C}[/tex]

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

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

[tex]\underline{I}=\underline{I_1}+\underline{I_2}=-3.75+j3.75[/tex]

module of this value is value showed on ammeter

[tex]|\underline{I}|=5.303[/tex]

However,correct solution is [tex]3\sqrt{5}\approx6.708[/tex]

Where is the mistake?