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Homework Help: Questions regarding AC circuits

  1. May 6, 2008 #1
    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:

    \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}

    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?

    Also,is it because complex numbers are just,let`s say it like this,"artificial" extension of real numbers set?

    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:


    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.

    There is also a "-" when active and reactive power for system like one illustrated in attachment is calculated.

    Attached Files:

    Last edited: May 6, 2008
  2. jcsd
  3. May 6, 2008 #2
    The second question is regarding complex power [tex]\underline{S}[/tex]:

    why expression [tex]\underline{S}=\underline{U}\;\underline{I}[/tex] does not give the correct result,instead of that it is used [tex]\underline{S}=\underline{U}\;\underline{I}^*[/tex] where [tex]\underline{I}^*[/tex] is complex-conjugate of [tex]\underline{I}[/tex]?
    Last edited: May 6, 2008
  4. May 8, 2008 #3
    I hope I will have more luck with this one:

    In given circuit:

    http://img90.imageshack.us/img90/9085/clipboard05xw4.gif [Broken]

    where is:

    [tex]R_1=10\;\Omega[/tex], [tex]R_3=2,5\;\Omega[/tex], [tex]R_2=X_C=X_L=5\;\Omega[/tex] and [tex]\underline{E}=50e^{j\frac{\pi}{2}}\;V[/tex],

    find the value of current source [tex]\underline{I}_S[/tex].

    Voltage drop on [tex]R_3[/tex] is [tex]\underline{U}=100\;V[/tex].

    First I calculate current trough [tex]R_3[/tex]: [tex]\underline{I}_{R_3}=\frac{\underline{U}}{R_3}=40\;A[/tex].

    Further,by using Superposition theorem and removing branch containing [tex]\underline{I}_s[/tex] I find that [tex]\underline{E}[/tex] produces current of [tex]j20[/tex].When I substract that value from [tex]\underline{I}_{R_3}[/tex] I get [tex]\underline{I}_S=(40-j20)\;V[/tex].

    However,correct result is [tex]\underline{I}_S=(-10+j20)\;V[/tex].

    What I do wrong?
    Last edited by a moderator: May 3, 2017
  5. Aug 27, 2008 #4
    Problem 4:

    Could someone explain to me how to write equations using 2nd Kirchoff`s Law along the closed loop in circuit where we have mutual inductance between two inductors?

    I can`t comprehend this correctly at all.

    Here is example circuit:

    http://img125.imageshack.us/img125/4779/clipboard033mw7.jpg [Broken]

    The closest I was was equation with difference in one [tex]+[/tex] instedad of [tex]-[/tex].

    I need explanation here,not just equations,I already have them.
    Last edited by a moderator: May 3, 2017
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