# Thevenin Generator: Maximum Active Power & Calculation

• Engineering
• Edy56
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.
Edy56
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

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.

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.

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.

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.

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

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.

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.

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.

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.

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":

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.

solved it now, thank you

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

## What is Thevenin's Theorem and how is it applied to find the maximum active power?

Thevenin's Theorem states that any linear electrical network with voltage sources and resistances can be simplified to a single voltage source (Thevenin voltage, Vth) in series with a single resistance (Thevenin resistance, Rth). To find the maximum active power delivered to a load, the load resistance (RL) should be equal to the Thevenin resistance (Rth). This condition is known as the Maximum Power Transfer Theorem.

## How do you calculate Thevenin voltage (Vth) in a circuit?

Thevenin voltage (Vth) is calculated by determining the open-circuit voltage across the terminals where the load is connected. This involves removing the load resistor and calculating the voltage across the open terminals using standard circuit analysis techniques such as voltage division, mesh analysis, or nodal analysis.

## What steps are involved in finding the Thevenin resistance (Rth)?

To find the Thevenin resistance (Rth), follow these steps:1. Turn off all independent voltage sources (replace them with short circuits) and turn off all independent current sources (replace them with open circuits).2. Calculate the equivalent resistance across the open terminals where the load was connected. This equivalent resistance is the Thevenin resistance, Rth.

## How do you determine the maximum active power that can be delivered to the load?

The maximum active power (Pmax) delivered to the load occurs when the load resistance (RL) is equal to the Thevenin resistance (Rth). The maximum power can be calculated using the formula:$P_{max} = \frac{V_{th}^2}{4R_{th}}$where Vth is the Thevenin voltage and Rth is the Thevenin resistance.

## Why is the concept of maximum power transfer important in electrical engineering?

The concept of maximum power transfer is important because it helps in designing efficient systems where the goal is to transfer the maximum amount of power from a source to a load. This is crucial in applications such as communication systems, power delivery, and electronic circuit design, where maximizing efficiency and performance is essential.

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