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## Homework Statement

In circuit given 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}_1=\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)\;A[/tex]

Module of this value is value showed on ammeter:

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

However,correct solution is:

[tex]3\sqrt{5}\;A\approx6.708\;A[/tex]

Where is the mistake?

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