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A variety of questions regarding AC circuits

  • Thread starter R A V E N
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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?
 

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Answers and Replies

  • #2
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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:

[tex]\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}[/tex]

from where we can see that it is [tex]B=\frac{-X_L}{R^2+X_L^2}[/tex],althought it is [tex]B=\frac{X_L}{R^2+X_L^2}[/tex].

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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  • #3
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Can moderator please delete this doubled topic?
 

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