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AC circuit with a switch -- analysis

  1. Aug 7, 2016 #1
    1. The problem statement, all variables and given/known data
    [itex][/itex]
    Given the circuit of sinusoidal current (attachment1) with given data:
    [itex]\underline{Z_3}=200(3-j4)\Omega,\underline{Z_4}=100(3+j20)\Omega,\underline{Z_5}=100(3+j4)\Omega,\underline{Z}=100(2+j5)\Omega,\underline{I_{g2}}=-10(2-j)mA.[/itex]
    After the switch is closed, the increment of voltage 1-2 is given: [itex]\Delta \underline{U_{12}}=(4+j3)V[/itex].
    Find the complex apparent power of [itex]\underline{I_{g2}}[/itex] after the switch is closed.

    image1.png

    2. The attempt at a solution

    Attempt:
    By using current compensation theorem (note: I don't have to use this theorem, anything can be used to solve the problem) on the branch with the switch and impedance [itex]\underline{Z}[/itex], we get the following circuit (attachment2 - switch and impedance [itex]\underline{Z}[/itex] are replaced by [itex]\underline{I_c}[/itex]):

    image2.PNG

    In the case when switch is open, compensation current [itex]\underline{I_c}[/itex] is equal to zero, and in the case when the switch is closed, it has some unknown value.

    By using superposition theorem (note: I don't have to use this theorem, anything can be used to solve the problem), we can analyze the circuit from attachment2 by looking at [itex]\underline{I_c}[/itex] and other generators are removed. Now, we get the following circuit (attachment3):
    image3.PNG

    From this circuit, we know potentials of nodes 1 and 2 since [itex]\Delta \underline{U_{12}}=\underline{V_1}-\underline{V_2}[/itex], so we can use potential of nodes method to find the complex value of [itex]\underline{I_{c}}[/itex] and the voltage [itex]\underline{U_{23}}[/itex]. By setting the potential [itex]\underline{V_2}[/itex] to zero, and after solving the system of two linear complex equations with [itex]\underline{V_3}[/itex] and [itex]\underline{I_{c}}[/itex] as unknowns, we get:

    [tex]\underline{V_2}=0,\underline{V_1}=(4+j3)V,\underline{V_3}=(12.48+j53.4)V,\underline{I_c}=(-6.44-j41.57)mA,\underline{U_{23}}=(-12.48-j53.4)V[/tex]

    Complex apparent power of [itex]\underline{I_{g2}}[/itex] (attachment1) after the switch is closed can be found by the following equation:

    [tex]\underline{S_{I_{g2}}}^{(c)}=\underline{U_{35}}^{(c)}\cdot \underline{I_{g2}}^{*}[/tex]

    where [itex]\underline{U_{35}}^{(c)}[/itex] is the voltage across [itex]\underline{I_{g2}}[/itex] when the switch is closed, and [itex]\underline{I_{g2}}^{*}[/itex] is the complex conjugate of [itex]\underline{I_{g2}}[/itex].

    We can find the voltage [itex]\underline{U_{35}}^{(c)}[/itex] from the following equation:
    [tex]\underline{U_{35}}^{(c)}=\underline{U_{35}}^{(o)}+\Delta \underline{U_{35}}[/tex]

    where [itex]\underline{U_{35}}^{(o)}[/itex] is the voltage across [itex]\underline{I_{g2}}[/itex] when the switch is opened, and [itex]\Delta \underline{U_{35}}[/itex] is the voltage across [itex]\underline{I_{c}}[/itex] from the attachment3 and is equal to [itex]\Delta \underline{U_{35}}=(-12.48-j53.4)V[/itex] (look at attachment3).

    In order to find the voltage [itex]\underline{U_{35}}^{(o)}[/itex], we look at the circuit from attachment1, where only the generator [itex]\underline{I_c}[/itex] is removed.

    Question: Since the following parameters are not given: [itex]\underline{I_{g1}},\underline{Z_1},\underline{E_2},\underline{E_6},\underline{Z_2}[/itex], how to find the voltage [itex]\underline{U_{35}}^{(o)}[/itex]?
     
    Last edited: Aug 7, 2016
  2. jcsd
  3. Aug 11, 2016 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    I looked at this, but seemed to find too many unknowns.

    I said that with fixed current through Z1 it follows that ∆V12 will be equal to ∆V13, so I concentrated on finding ∆V13. Also, the DC sources are zero impedance at AC, so Z2 is shorted out in any AC analysis and the switch effectively places Z in parallel with Z3.

    No node is shown as ground, so perhaps consider node 3 as ground, and the problem then comes down to finding V5 with the switch closed.

    I couldn't manage to solve it, though on glancing back over my working I do see an error...
     
  4. Aug 11, 2016 #3
    @NascentOxygen ,

    Here is the solution from by book:

    The complex apparent power of [itex]\underline{I_{g2}}[/itex] after the switch is closed is [itex]\underline{{S_{I_{g2}}}^{(c)}}=(240+j20)mVA[/itex].
    I don't know if this solution is correct.
     
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