Thevenin Generator: Maximum Active Power & Calculation

In summary: Two ideal current sources connected in series with different values. Actually, this is really any circuit node that only has current sources connected to it. They would then all have to sum to zero at that node. This is KCL.
  • #1
Edy56
38
5
Homework Statement
In the electrical circuit shown determine:
Impedance Z so that it develops on it maximum active power and calculate that power.
Relevant Equations
none
komp.png

So I have calculated Zab which is 3-3j (it's correct).
Now I have to calculate (Uab)0. This is where I just get completely lost.
In my opinion:
(Uab)0=E3+I3*R3+Ic1*Xc1+E4

Note: I recognize these currents are not shown in the picture, but based on their indexes and basic logic I hope you understand where I imagine them being.

Now obviously I am missing I3 and Ic1 so I decided to try to use node potential method to determine their values.
These are my equations:

U10*(1/(jXl1+R1)+1/(-jXc)+1/R3)-U20*(1/jXl1) = E1/(jXl1+R1) +E3/R3

U20(1/(R1+jXl1)+1/(-jXc2))-U30(1/-jXc2)-U10(1/(jXl1+R1+)= J+ E1/(jXl1+R1) + E2/(-jXc2)

My biggest concern is regarding the branch with R3and E3. Because of the break at node 'a', I would assume there is no current flowing there, but I have a generator there so I am confused.
 
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  • #2
Edy56 said:
My biggest concern is regarding the branch with R3and E3. Because of the break at node 'a', I would assume there is no current flowing there, but I have a generator there so I am confused.
Current flow always has to return in the circuit to where it started, where you measure it leaving from. This is from conservation of charge, you can't just generate electrons, you can only move them. E3 is a voltage generator, it doesn't necessarily have to have current flowing through it.
 
  • #3
DaveE said:
Current flow always has to return in the circuit to where it started, where you measure it leaving from. This is from conservation of charge, you can't just generate electrons, you can only move them. E3 is a voltage generator, it doesn't necessarily have to have current flowing through it.
I have seen that in similar cases when calculating (Uab)0 they also add the generator. But that doesn't make really sense to me. There is no current flowing through it, and I get that te generator doesn't necessarily need it, but where does the voltage come from? Voltage is current * resistance.
 
  • #4
Edy56 said:
I have seen that in similar cases when calculating (Uab)0 they also add the generator. But that doesn't make really sense to me. There is no current flowing through it, and I get that te generator doesn't necessarily need it, but where does the voltage come from? Voltage is current * resistance.
When they calculate the open circuit voltage they remove the impedance in question, Z, in your case. All voltage and current sources are still included with their values. Then you calculate the voltage across the terminals where Z used to be.

E3 makes the voltage defined, whether there is any current or not. It is an ideal voltage source and makes whatever current it needs to to make it's defined voltage. Same for current sources, but they make whatever voltage is required.

There are exceptions that you should never see in a well posed problem. That is a generator that can't make the voltage (or current) it needs to, usually because the current (or voltage) require would be infinite. Or that make contradictions, like requiring a voltage to be two different values. Some examples of this are:
1) An ideal voltage source with a short circuit, a 0Ω load. This would require infinite current.
2) An ideal current source with an open circuit, an ∞Ω load. This would require infinite voltage.
3) Two ideal voltage sources connected in parallel with different values. Actually, this is really any circuit loop that only has voltage sources in it. They would then all have to sum to zero around the loop. This is KVL.
4) Two ideal current sources connected in series with different values. Actually, this is really any circuit node that only has current sources connected to it. They would then all have to sum to zero at that node. This is KCL.
 
  • #5
DaveE said:
When they calculate the open circuit voltage they remove the impedance in question, Z, in your case. All voltage and current sources are still included with their values. Then you calculate the voltage across the terminals where Z used to be.

E3 makes the voltage defined, whether there is any current or not. It is an ideal voltage source and makes whatever current it needs to to make it's defined voltage. Same for current sources, but they make whatever voltage is required.

There are exceptions that you should never see in a well posed problem. That is a generator that can't make the voltage (or current) it needs to, usually because the current (or voltage) require would be infinite. Or that make contradictions, like requiring a voltage to be two different values. Some examples of this are:
1) An ideal voltage source with a short circuit, a 0Ω load. This would require infinite current.
2) An ideal current source with an open circuit, an ∞Ω load. This would require infinite voltage.
3) Two ideal voltage sources connected in parallel with different values. Actually, this is really any circuit loop that only has voltage sources in it. They would then all have to sum to zero around the loop. This is KVL.
4) Two ideal current sources connected in series with different values. Actually, this is really any circuit node that only has current sources connected to it. They would then all have to sum to zero at that node. This is KCL.
U10*(1/(jXl1+R1)+1/(-jXc))-U20*(1/jXl1) = -E1/(jXl1+R1)

U20(1/(R1+jXl1)+1/(-jXc2))-U30(1/-jXc2)-U10(1/(jXl1+R1)= J+ E1/(jXl1+R1) + E2/(-jXc2)

are these correct then? I removed E3 and R3
 
  • #6
Edy56 said:
U10*(1/(jXl1+R1)+1/(-jXc))-U20*(1/jXl1) = -E1/(jXl1+R1)

U20(1/(R1+jXl1)+1/(-jXc2))-U30(1/-jXc2)-U10(1/(jXl1+R1)= J+ E1/(jXl1+R1) + E2/(-jXc2)

are these correct then? I removed E3 and R3
No. I don't see any method to your process, like you're guessing. That won't work.
 
  • #7
DaveE said:
No. I don't see any method to your process, like you're guessing. That won't work.
I am not guessing. I am using the node potential method.
 
  • #8
Edy56 said:
I am not guessing. I am using the node potential method.
I might be making a mistake somewhere, but I do not see it nor do I understand it. So if you could tell me where I am wrong so I can try to see why I would be thankful because I had been at this problem for 5 hours now and I really do not see where I am making mistakes.
 
  • #9
How were you able to calculate a numerical value for Zab when XC1, XC2, and XL1 have no numerical values?

What nodes are the voltages U10 and U20? If those are the voltages at nodes 1 and 2 why would you call those voltages U10 and U20 rather than U1 and U2?

Rather than just presenting final equations, show how you derived them starting with the fundamental steps of a nodal analysis. That way, we could see where you made a mistake.
 
  • #10
The Electrician said:
How were you able to calculate a numerical value for Zab when XC1, XC2, and XL1 have no numerical values?

What nodes are the voltages U10 and U20? If those are the voltages at nodes 1 and 2 why would you call those voltages U10 and U20 rather than U1 and U2?

Rather than just presenting final equations, show how you derived them starting with the fundamental steps of a nodal analysis. That way, we could see where you made a mistake.
okay so I do have numerical values, I did not show them because I am just interested in just knowing how to solve for (Uab)0.

As for the nodes, I don't know what to tell you. That's just how we are told to solve problems. Here is the "formula":

nodes.png
 
  • #11
Edy56 said:
U10*(1/(jXl1+R1)+1/(-jXc))-U20*(1/jXl1) = -E1/(jXl1+R1)

U20(1/(R1+jXl1)+1/(-jXc2))-U30(1/-jXc2)-U10(1/(jXl1+R1)= J+ E1/(jXl1+R1) + E2/(-jXc2)

are these correct then? I removed E3 and R3
U10*(1/(jXl1+R1)+1/(-jXc1))-U20*(1/(R1+jXl1)) = -E1/(jXl1+R1)

I've shown in red corrections to two mistakes I see.
 
  • #12
solved it now, thank you
 
  • #13
The Electrician said:
U10*(1/(jXl1+R1)+1/(-jXc1))-U20*(1/(R1+jXl1)) = -E1/(jXl1+R1)

I've shown in red corrections to two mistakes I see.
Okay i have another problem. I have to determine complex power of the Xc1.
I have calculated that U10=5 (this Is correct)
I know that Xc1=5
So logically I=U10/-jXc1=j
Sc=1/2*-jXc1*|I|^2
Sc=1/2*-j5
Sc=1/2*-j5 - this Is not correct, whyy I am supposed to get 25*(-j)/2
 

1. What is a Thevenin generator?

A Thevenin generator is a simplified model of a complex electrical circuit that can be used to analyze and predict the behavior of the circuit. It is made up of a voltage source and a series resistance, and is used to determine the maximum active power that can be delivered to a load.

2. How is the maximum active power calculated using a Thevenin generator?

The maximum active power is calculated by finding the load resistance that will result in the maximum power transfer from the Thevenin generator. This is done by finding the load resistance that is equal to the Thevenin resistance, and then using the formula P = (V_th)^2 / (4*R_th), where V_th is the Thevenin voltage and R_th is the Thevenin resistance.

3. What is the significance of calculating the maximum active power using a Thevenin generator?

Calculating the maximum active power using a Thevenin generator allows for the efficient design and optimization of electrical circuits. It helps determine the optimal load resistance for maximum power transfer, and can also be used to predict the behavior of the circuit under different load conditions.

4. Can a Thevenin generator be used for both DC and AC circuits?

Yes, a Thevenin generator can be used for both DC and AC circuits. However, the calculations for maximum active power may differ slightly for AC circuits due to the presence of reactance in the circuit.

5. Are there any limitations to using a Thevenin generator for circuit analysis?

While a Thevenin generator is a useful tool for circuit analysis, it is a simplified model and may not accurately represent the behavior of complex circuits. Additionally, it assumes linear behavior and does not take into account non-linear elements such as diodes or transistors.

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