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Homework Help: AC circuits - some questions

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

    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}_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?
     

    Attached Files:

    Last edited: May 4, 2008
  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:

  4. May 5, 2008 #3
    OK,I managed to resolve first problem,but what about second one?
     
  5. May 5, 2008 #4
    Probably the explanation is that while one physical parameter is rising(susceptance [tex]B[/tex]),the other is lowering(inductive reactance [tex]X_L[/tex]) and vice-versa,like it is in Faraday`s law of induction:

    [tex]e=-\frac{d\phi}{dt}[/tex]

    the magnetic field which is produced by induced current(which is in turn produced by induced electromotive force [tex]e[/tex]) is in oposition with the change of outer flux [tex]\phi[/tex](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.
     
    Last edited: May 5, 2008
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