Thevenin's and Norton's Theorem

In summary, the circuit has to be solved when there is a 2000Ω resistor connected to A and B. Thevenin's theorem is used to find Vth and Rth. Thevenin's theorem is used to find the current through the resistor when the voltage source is open. The Norton equivalent is found by using superposition.
  • #1
Hi. It's me again. Right now I have some troubles with Thevenin's and Norton's theorem.
Here is the circuit.
problem_4_0.png

I have to find the thevenin voltage Vth and thevenin resistance Rth and solve the circuit when there is a 2000Ω resistor connected to A and B.
Then, I try to solve it...

First, for the Rth, I just treat the voltage source as a short circuit and current source as an open circuit. So, the 4000Ω resistor is disconnected.

Rth = 2000Ω

Then, for the Vth, Vth = Voc.
By using KCL,

5/2000 + 0.1×10-3 = Voc/Rth

5/2000 + 0.1×10-3 = Voc/2000

∴ Voc = 5.2 V
i.e. Vth = 5.2 V
problem_4_2.png


Is this correct?
 
Last edited:
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  • #2
ichabodgrant said:
Rth = 2000Ω

Then, for the Vth, Vth = Voc.
By using KCL,

5/2000 + 0.1×10-3 = Voc/Rth

5/2000 + 0.1×10-3 = Voc/2000

∴ Voc = 5.2 V
i.e. Vth = 5.2 V
Looks right. :smile:

So what will the Norton equivalent be?
 
  • #3
Is this correct?

Your answer is correct, I'm not sure I understand your method.

You could also use superposition as well. The voltage source is straight forward at 5 volts (open current source)

The current source is a bit more tricky( short voltage source). .1 amp thru 4K resistor is .4 volts...but you then put it thru the 2K resistor and its .2 volts in th opposite direction.

How can you have two different voltages in two parallel branches?

Must mean there is .2 volts across the current source in the opposite direction of the 4K resistor.
5 +.2 = 5.2 Volts. Or you could do the way above, but superposition is just another way to enjoy and undestand the problem.
 
  • #4
IN = 2.6 x 10-3 A and RN = Rth

I had thought about using superposition...
But when I attempted to do this question, I hadn't clearly understood these theorems.
My lecturer just taught too fast... I couldn't really get myself familiar with these theorems just from what he taught in class. So I spent a few hours on youtube, watching some lecture videos... and also asking for your kindly help :P

Thank you very much
 

1. What are Thevenin's and Norton's Theorem?

Thevenin's and Norton's Theorem are two equivalent circuit analysis techniques used to simplify complex electrical circuits into simpler and more manageable forms. They are commonly used in electrical engineering and physics to solve problems involving voltage, current, and power in a circuit.

2. How are Thevenin's and Norton's Theorem related?

Thevenin's and Norton's Theorem are mathematically equivalent, meaning they both provide the same results when analyzing a circuit. The main difference between the two is that Thevenin's Theorem simplifies a circuit into a voltage source and a series resistance, while Norton's Theorem simplifies a circuit into a current source and a parallel resistance.

3. When should Thevenin's or Norton's Theorem be used?

Thevenin's and Norton's Theorem should be used when analyzing complex circuits with multiple resistors, voltage sources, and current sources. They can be used to find the equivalent circuit for a specific part of the circuit, making it easier to calculate voltage, current, and power for that part.

4. How do you determine the Thevenin equivalent circuit?

To determine the Thevenin equivalent circuit, you must first remove the load resistor from the original circuit. Then, calculate the open-circuit voltage at the load terminals. This voltage will be the Thevenin voltage. Next, find the equivalent resistance by short-circuiting all voltage sources and open-circuiting all current sources in the original circuit. This resistance will be the Thevenin resistance.

5. Can Thevenin's and Norton's Theorem be used for non-linear circuits?

No, Thevenin's and Norton's Theorem can only be used for linear circuits. Non-linear circuits, which have elements that do not follow Ohm's law, require more complex analysis techniques to find equivalent circuits.

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