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Homework Help: A variety of questions regarding AC circuits

  1. May 4, 2008 #1
    1. The problem statement, all variables and given/known data
    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]


    3. 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?
     

    Attached Files:

  2. jcsd
  3. May 4, 2008 #2
    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?
     

    Attached Files:

    Last edited: May 4, 2008
  4. May 5, 2008 #3
    Can moderator please delete this doubled topic?
     
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