Finding Thevenin Equivalent Voltage for a Circuit with Multiple Voltage Sources

In summary, the conversation discusses finding the total resistance between points A and B in a circuit with two voltage sources, using the maximum power transfer theorem to determine the Thevenin equivalent voltage, and choosing a reference node for nodal analysis. One approach is to short circuit the voltage sources and calculate the resulting resistance, while another approach involves setting up nodal equations. It is important to choose a suitable reference node for nodal analysis.
  • #1
Ash_M
4
0

Homework Statement


See attached image
Screen Shot 2016-12-01 at 22.11.06.png


Homework Equations


Thevenin Equivalent

The Attempt at a Solution


a) Find the total resistance between A and B by short circuiting the voltage sources:

Rt = 100Ω in parallel with the two 150Ω

Two 150Ω in series: R = 300Ω
1/Rt = 1/100 + 1/300

Rt = 75Ω

I am having difficulty finding the thevenin equivalent voltage.

b) Don't entirely understand the question so I am unable to make an attempt

Any help would be greatly appreciated.
Thanks
 
Physics news on Phys.org
  • #2
Ash_M said:
I am having difficulty finding the thevenin equivalent voltage.
What have you tried? What's the usual procedure?
b) Don't entirely understand the question so I am unable to make an attempt
Hint: Maximum power transfer theorem.
 
  • Like
Likes Ash_M
  • #3
gneill said:
What have you tried? What's the usual procedure?

Hint: Maximum power transfer theorem.

I have no idea so I have not tried anything. Usually for a system with a single voltage source I would try to find the open circuit voltage between A and B, however the two voltage sources are confusing me.

Oh so would I take the Rload to be 50Ω, and since for maximum power the thevenin resistance = load resistance, change one of the resistors to give me a thevenin resistance of 50Ω?
 
  • #4
Ash_M said:
I have no idea so I have not tried anything. Usually for a system with a single voltage source I would try to find the open circuit voltage between A and B, however the two voltage sources are confusing me.
You can always do a "KVL walk" from A to B and sum up the potential changes along the way.
Oh so would I take the Rload to be 50Ω, and since for maximum power the thevenin resistance = load resistance, change one of the resistors to give me a thevenin resistance of 50Ω?
Yes.
 
  • Like
Likes Ash_M
  • #5
gneill said:
You can always do a "KVL walk" from A to B and sum up the potential changes along the way.

Yes.

I don't quite understand what you mean by a "KVL walk", but i tried going round the top of the circuit and noting how the potential would change I also wasn't sure if the current in the entire system could be calculated as I = 12/400 . Doing that would mean I get a voltage of 0V at A and then 0V at B which can't be correct.
 
  • #6
Ash_M said:
I don't quite understand what you mean by a "KVL walk", but i tried going round the top of the circuit and noting how the potential would change I also wasn't sure if the current in the entire system could be calculated as I = 12/400 . Doing that would mean I get a voltage of 0V at A and then 0V at B which can't be correct.
:smile: You did well, and a "KVL walk" is just what you did.

0 V is a perfectly valid result.
 
  • Like
Likes Ash_M
  • #7
gneill said:
:smile: You did well, and a "KVL walk" is just what you did.

0 V is a perfectly valid result.
Haha thanks so much!
 
  • #8
gneill said:
:smile: You did well, and a "KVL walk" is just what you did.

0 V is a perfectly valid result.

I took a different approach to this question, calculating Vb from 0 = (Vb-9)/300 + (Vb-(-3))/100, so Vb = 0. Is there a way to get a nodal equation for Va?
 
  • #9
Conor_B said:
I took a different approach to this question, calculating Vb from 0 = (Vb-9)/300 + (Vb-(-3))/100, so Vb = 0. Is there a way to get a nodal equation for Va?
You've used A as the reference node for your first equation, so it's potential is defined to be zero. That is, Va = 0 is the node equation for A :smile:
 
  • #10
gneill said:
You've used A as the reference node for your first equation, so it's potential is defined to be zero. That is, Va = 0 is the node equation for A :smile:

What do you mean by reference node?
 
  • #11
Conor_B said:
What do you mean by reference node?
A node whose potential is taken as 0V for reference. All the voltages in the circuit are measured with respect to the reference node. It is also called as the "circuit ground".
 
  • #12
Conor_B said:
What do you mean by reference node?
When you do nodal analysis you are determining the potentials of various nodes of the circuit with respect to some reference node. The first step is always to choose the node that will be the reference node for all the potentials.

There's often an obvious or convenient node to choose for this role (what constitutes "obvious" or "convenient" and how to recognize a good choice for the reference node is something that comes with experience). In this case node A is suitable since it not only forms one of the "output" terminals of the Thevenin equivalent, but is located at the junction of the only two voltage sources in the circuit.
 
  • #13
cnh1995 said:
A node whose potential is taken as 0V for reference. All the voltages in the circuit are measured with respect to the reference node. It is also called as the "circuit ground".

gneill said:
When you do nodal analysis you are determining the potentials of various nodes of the circuit with respect to some reference node. The first step is always to choose the node that will be the reference node for all the potentials.

There's often an obvious or convenient node to choose for this role (what constitutes "obvious" or "convenient" and how to recognize a good choice for the reference node is something that comes with experience). In this case node A is suitable since it not only forms one of the "output" terminals of the Thevenin equivalent, but is located at the junction of the only two voltage sources in the circuit.

Ok. Cool. Thanks Guys. Any time we did a problem in lectures, the ground was always marked in, so I never realized you could 'pick' a ground.
 

What is a Thevenin Equivalent Circuit?

A Thevenin Equivalent Circuit is a simplified representation of a complex electrical circuit that can accurately predict the behavior of the original circuit. It consists of a single voltage source and a single resistor that are equivalent to the original circuit, when viewed from a specific set of terminals.

How is a Thevenin Equivalent Circuit calculated?

To calculate the Thevenin Equivalent Circuit, the circuit is first simplified by removing all loads and shorting all voltage sources. Then, the equivalent resistance is calculated by finding the total resistance between the terminals. The equivalent voltage is calculated by measuring the voltage across the open terminals. The voltage source is then placed in series with the equivalent resistance to create the Thevenin Equivalent Circuit.

What are the limitations of a Thevenin Equivalent Circuit?

A Thevenin Equivalent Circuit is only accurate for linear circuits, meaning that the components must follow Ohm's Law. It also assumes that the circuit is in a steady-state condition and that the components are not temperature dependent. Additionally, it does not take into account the effects of inductance or capacitance.

Why is a Thevenin Equivalent Circuit useful?

A Thevenin Equivalent Circuit is useful because it simplifies complex circuits, making them easier to analyze and understand. It also allows for quick and accurate calculations of voltage and current at different points in the circuit. It is also helpful for troubleshooting and designing new circuits.

What are some real-life applications of Thevenin Equivalent Circuits?

Thevenin Equivalent Circuits are commonly used in electronic systems such as power supplies, amplifiers, and filters. They are also used in telecommunications, where it is necessary to analyze the behavior of a circuit with varying loads. Thevenin Equivalent Circuits are also used in circuit simulation software to model and predict the behavior of circuits before they are built.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
2K
  • Engineering and Comp Sci Homework Help
2
Replies
42
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
921
  • Engineering and Comp Sci Homework Help
Replies
9
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
17
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
Back
Top