Engineering AC circuit with a switch -- analysis

[gruba]

Homework Statement

Given the circuit of sinusoidal current (attachment1) with given data:
\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.
After the switch is closed, the increment of voltage 1-2 is given: \Delta \underline{U_{12}}=(4+j3)V.
Find the complex apparent power of \underline{I_{g2}} after the switch is closed.

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 \underline{Z}, we get the following circuit (attachment2 - switch and impedance \underline{Z} are replaced by \underline{I_c}):

In the case when switch is open, compensation current \underline{I_c} 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 \underline{I_c} and other generators are removed. Now, we get the following circuit (attachment3):

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

\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

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

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

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

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

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

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

Question: Since the following parameters are not given: \underline{I_{g1}},\underline{Z_1},\underline{E_2},\underline{E_6},\underline{Z_2}, how to find the voltage \underline{U_{35}}^{(o)}?

 

[NascentOxygen]

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...

 

[gruba]

@NascentOxygen ,

Here is the solution from by book:

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