# AC circuit with a switch -- analysis

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1. Aug 7, 2016

### gruba

1. The problem statement, all variables and given/known data

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 $\underline{U_{35}}^{(o)}$?

Last edited: Aug 7, 2016
2. Aug 11, 2016

### 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...

3. Aug 11, 2016

### 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.