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.

**Attempt:**

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**):

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*):

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?

## Best Answer

Let's start with a few considerations on given schematicsFirst 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

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

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.

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. \$

and given what found above in (7)

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

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

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

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

Numerically we get