Calculating Thevenin's Equivalent for a Circuit

  • Thread starter TheRedDevil18
  • Start date
  • Tags
    Equivalent
In summary: I then worked out the voltages using ohms law to get 1.575 and 1.5 volts. Summing them I get Vth = 3.075VSo the voltage in the 1000 ohm resistor = 0.0025*1000 = 2.49V and the voltage in the 750 ohm resistor = 0.0041*750 = 3.075V. Adding the two I get Vth = 5.57V
  • #1
TheRedDevil18
408
1

Homework Statement


Solve the thevenins equivalent of the circuit
Thevinin.jpg


Homework Equations

The Attempt at a Solution



I think this is wrong but I got 750 ohms for my Rthevinin and 2.3V for my Vthevinin. I checked using KVL with the load resistor set to 500 ohms and get a current of 0.0037A through the 500 ohm resistor but when I use the thevinins circuit I get a current of 0.0018A

This is how I got Rth,
Short the Voltage source, so the 750 is in parallel with the 1500 which is equivalent to 500 ohm resistor in series with the 1000 ohm to give a 1500 ohm in parallel with the second 1500 ohm to give my Rth of 750 ohms

This is how I got Vth,
Replacing the voltage source I see that the 1500(one connected in the separate branch) and 1000 ohm resistors have no current flowing through them. That leaves just the 1500 and 750 in series with the voltage source. So the voltage across the terminals would equal to the voltage in the 750 ohm resistor which works out to be 2.3V = Vth

Can someone please tell me what I'm doing wrong ?, thanks
 
Physics news on Phys.org
  • #2
Your Rth method and result look fine, but your Vth method and result are not: There will be a current through the upper 1500 Ω and the 1000 Ω resistors.
 
  • #3
gneill said:
Your Rth method and result look fine, but your Vth method and result are not: There will be a current through the upper 1500 Ω and the 1000 Ω resistors.

If the current is flowing through all the resistors then is the voltage at the terminals measuring the voltage across the 750 ohm resistor only ?, so do I have to find the current flowing through the 750 ohm resistor and from their the voltage ?
 
  • #4
TheRedDevil18 said:
If the current is flowing through all the resistors then is the voltage at the terminals measuring the voltage across the 750 ohm resistor only ?, so do I have to find the current flowing through the 750 ohm resistor and from their the voltage ?
The voltage at the Out terminal will be the sum of the voltages across the 750 and 1000 Ohm resistors.
 
  • #5
gneill said:
The voltage at the Out terminal will be the sum of the voltages across the 750 and 1000 Ohm resistors.

I still don't think I'm getting the correct answer

I worked out the equivalent resistance by taking the 750 in parallel with the 1000 ohm resistor and added that to the 1500 ohm in series to get 1928.57 ohms which is in parallel with the 1500 ohm (branched one) to get an equivalent resistance of 843.75 ohms. I then worked out the total current to be 0.0083A(7/843.75).

I then used the current divider rules to work out the current in the 1000 ohm and 750 ohm resistors which I got to be 0.0021A(750 ohm) and 0.0015A(1000 ohm). I then worked out the voltages using ohms law to get 1.575 and 1.5 volts. Summing them I get Vth = 3.075V

Don't think I'm doing this correctly, need some help, thanks
 
  • #6
I don't see how the 750 and 1000 Ohms resistors can be taken to be parallel.

Let's rearrange the circuit diagram a bit to (hopefully) make things easier to see:

Fig1.gif
 
  • #7
gneill said:
I don't see how the 750 and 1000 Ohms resistors can be taken to be parallel.

Let's rearrange the circuit diagram a bit to (hopefully) make things easier to see:

View attachment 81462

If I try it this way, then working out the equivalent resistance

1500 + 1000 = 2500 ohms, which is in parallel to the 1500 ohm resistor. Working the parallel combination I get 937.5 ohms + the 750 ohm in series I get Req = 1687.5 ohms. So the total current would be I = 7/1687.5 = 0.0041A. Using the current divider rule I get I2 and hence the current in the 1000 ohm resistor to be 0.0041*1500/2500 = 0.0025A and the current in the 750 ohm resistor would equal the total current which is 0.0041A. So the voltage in the 1000 ohm resistor = 0.0025*1000 = 2.49V and the voltage in the 750 ohm resistor = 0.0041*750 = 3.075V. Adding the two I get Vth = 5.57V

Not sure if that's correct but it still doesn't agree with the current I got using KVL. I know Vth should be 4.63V but I'm not sure how
 
  • #8
TheRedDevil18 said:
I still don't think I'm getting the correct answer

I worked out the equivalent resistance by taking the 750 in parallel with the 1000 ohm resistor and added that to the 1500 ohm in series to get 1928.57 ohms which is in parallel with the 1500 ohm (branched one) to get an equivalent resistance of 843.75 ohms. I then worked out the total current to be 0.0083A(7/843.75).

I then used the current divider rules to work out the current in the 1000 ohm and 750 ohm resistors which I got to be 0.0021A(750 ohm) and 0.0015A(1000 ohm). I then worked out the voltages using ohms law to get 1.575 and 1.5 volts. Summing them I get Vth = 3.075V

Don't think I'm doing this correctly, need some help, thanks
 
  • #9
TheRedDevil18 said:
Using the current divider rule I get I2 and hence the current in the 1000 ohm resistor to be 0.0041*1500/2500 = 0.0025A
You want to check the details of how to apply the current divider rule.
 
  • #10
The easiest way to solve this is by superposition.

Create the first Thevenin equivalent from the 1500 and 750 resistors in lower left sub-circuit.
Create the second Thevenin equiv from the 1500 Ohm resistor at the top.
Now by superposition, connect the two Thevenin equivalents to the Output terminal.
From this simple circuit it is easy to calculate the Thevenin voltage and resistance.
 
  • #11
DannySmythe said:
The easiest way to solve this is by superposition.

Create the first Thevenin equivalent from the 1500 and 750 resistors in lower left sub-circuit.
Create the second Thevenin equiv from the 1500 Ohm resistor at the top.
Now by superposition, connect the two Thevenin equivalents to the Output terminal.
From this simple circuit it is easy to calculate the Thevenin voltage and resistance.

Thanks, I will look into that method
 
  • #12
gneill said:
You want to check the details of how to apply the current divider rule.

Oh yes, sorry mistake that should be 0.0041*1500/4000. I now get what should be the correct answer of Vth = 4.61V

Thanks gneill :smile:
 
  • #13
TheRedDevil18 said:
Oh yes, sorry mistake that should be 0.0041*1500/4000. I now get what should be the correct answer of Vth = 4.61V

Thanks gneill :smile:
Looks good to me.
 

Related to Calculating Thevenin's Equivalent for a Circuit

What is Thevenin's equivalent problem?

Thevenin's equivalent problem is a method used in circuit analysis to simplify complex circuits into a single equivalent circuit. This allows for easier analysis and calculation of circuit parameters.

What are the assumptions made in Thevenin's equivalent problem?

Thevenin's equivalent problem assumes that the circuit is linear, meaning that the components have constant values and follow Ohm's Law. It also assumes that there are no dependent sources in the circuit.

How is Thevenin's equivalent circuit calculated?

The Thevenin equivalent circuit is calculated by first determining the open-circuit voltage (Voc) and the short-circuit current (Isc) of the original circuit. Then, the equivalent resistance (Req) is calculated by dividing Voc by Isc. The Thevenin voltage (Vth) is equal to Voc, and the Thevenin resistance (Rth) is equal to Req.

Why is Thevenin's equivalent problem useful?

Thevenin's equivalent problem is useful because it simplifies complex circuits into a single equivalent circuit, making it easier to analyze and design circuits. It also allows for easier troubleshooting and predicting circuit behavior.

What are some limitations of Thevenin's equivalent problem?

Thevenin's equivalent problem is limited in its applicability to linear circuits and cannot be used for circuits with dependent sources. It also assumes that the circuit operates under steady-state conditions and does not take into account the effects of time-varying signals.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
10
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
1K
  • Engineering and Comp Sci Homework Help
2
Replies
42
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
999
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
Back
Top