I need help with the following AC circuit problem:

Given the circuit (attachment 1) with known 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.

enter image description here


By using current compensation theorem on the branch with the switch and impedance Z we get the following circuit (attachment 2 - switch and impedance Z are replaced by Ic):

enter image description here

In the case when switch is open, compensation current Ic is equal to zero, and in the case when the switch is closed, it has some unknown value.

By using superposition theorem, we can analyze the circuit from attachment 2 by looking at Ic and other generators are removed. Now, we get the following circuit (attachment 3):

enter image description here

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 Ic and the voltage U23. By setting the potential V2 to zero, and after solving the system of two linear complex equations with V3 and Ic 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 Ig2 (attachment 1) 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 Ig2 when the switch is closed, and $$\underline{I_{g2}}^{*}$$ is the complex conjugate of Ig2.

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 Ig2 when the switch is opened, and $$\Delta \underline{U_{35}}$$ is the voltage across Ic from the attachment 3 and is equal to $$\Delta \underline{U_{35}}=(-12.48-j53.4)V$$ (look at attachment 3).

In order to find the voltage $$\underline{U_{35}}^{(o)},$$ we look at the circuit from attachment 1, where only the generator Ic 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 $$\underline{U_{35}}^{(o)}?$$

Is there another method to find the complex apparent power of Ig2 after the switch is closed?


1 Answer 1


Let's start with a few considerations on given schematics enter image description here

First we are asked to relate \$ \Delta U_{12}\$ to switch closure, let's try to get them closer running along the highlighted branches:

\$U_{12}=U_1-U_2=\$ but \$U_2=U_3+E_2-Z_1\,I_{g1}\$ and substituting

\$U_{12}=U_1-U_3-E_2+Z_1\,I_{g1}\$ and going to variations

\$\begin{align}\Delta U_{12} &= U_{12}^{(c)}-U_{12}^{(o)} &=U_1^{(c)}-U_3^{(c)}-E_2+Z_1\,I_{g1} & \\ & &- U_1^{(o)}+U_3^{(o)}+E_2-Z_1\,I_{g1} &= \Delta U_{13} \end{align}\$

given constant \$E_2\$, \$I_{g1}\$ and obviously \$Z_1\$ we got to

  1. \$\Delta U_{12}= \Delta U_{13}\$
  2. Now we remove \$Z_1\$ since any impedance series connected to a current generator does not affect the network itself but only drop across generator.
  3. Then dualy we remove \$Z_2\$ for it is parallel connected to a voltage generator and does not affect the network itself but only current supplied by generator.
  4. Now it is time to split current generator \$I_{g1}\$ into two identical ones series connected, common point can be connected anywhere we like since no current is going to flow thru.

enter image description here

  1. Now left \$I_{g1}\$ founds itself parallel connected to a voltage source and can hence be removed.
  2. Finally we add \$E_2\$ and \$E_6\$ to get an equivalent \$E=E_6-E_2\$ enter image description here

On this circuit, given constant Ig1 it is easy to relate \$\Delta U_{13}\$ and \$\Delta U_{35}\$ variations using a simple voltage divider

  1. $$\Delta U_{13}=-\frac{Z_5}{Z_5+Z_4}\Delta U_{35}$$

Now it still looks like we have two many unknowns (E and Ig1 compared to the single relation on \$\Delta U_{12}=\Delta U_{13}\$ we have) but this is not true. Our circuit may be thought driven by a single equivalent generator.

enter image description here

where \$Z_e=Z_3 || (Z_4+Z_5)\$ and \$J\$ is a linear combination of E and Ig1 which we don't even need to calculate.

Now it's just plain sailing

\$\left\{\begin{align} U_{35}^{(o)} &= Z_e & (J+I_{g2}) \\ U_{35}^{(c)} &= (Z_e || Z) & (J+I_{g2}) \end{align}\right. \$

  1. $$\Delta U_{35}=(Z_e||Z-Z_e)(J+I_{g2})=-\frac{Z_e^2}{Z_e+Z}(J+I_{g2})$$

and given what found above in (7)

  1. $$\Delta U_{12}=\Delta U_{13}=-\frac{Z_5}{Z_5+Z_4}\Delta U_{35}$$

and combining (8) and (9) we get total current driven to node 3

  1. $$J+I_{g2}=\frac{(Z_5+Z_4)(Z_e+Z)}{Z_5Z_e^2}\Delta U_{12}$$

and hence \$U_{35}^{(c)}\$ which is needed to compute request Ig2 power.

  1. $$U_{35}^{(c)}=(Z_e || Z)(J+I_{g2})=\frac{Z(Z_5+Z_4)}{Z_5Z_e}\Delta U_{12}$$

  2. $$S_{Ig2}^{(c)}=U_{35}^{(c)}I_{g2}^*$$

Numerically we get

enter image description here

Note: an error had crept into my line 8. enter image description here and propagated till the end. Now it is fixed.

  • \$\begingroup\$ Your method gives the following result: $$\underline{{S_{I_{g2}}}^{c}}=(-124177.65+j26908.137)mVA.$$ Correct result is $$\underline{{S_{I_{g2}}}^{c}}=(240+j20)mVA.$$ Could you check this calculation? \$\endgroup\$
    – user300048
    Aug 29, 2016 at 18:10
  • \$\begingroup\$ Check my updates :) You could have spot it yourself, that would be good practice. \$\endgroup\$
    – carloc
    Aug 29, 2016 at 19:26

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