Finding Thevenin Equiv for Circuit w/ R1, R2, R3, R4, R5

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SUMMARY

This discussion focuses on finding the Thévenin equivalent for a circuit involving resistors R1, R2, R3, R4, and R5, with values of R1: 5.6k, R2: 12k, R3: 2.2k, R4: 15k, R5: 1k, and a DC voltage source of 12V. The user initially calculated the equivalent resistance incorrectly but later confirmed the correct equivalent resistance as 4.781k. The discussion emphasizes the importance of calculating both the equivalent voltage (Vth) and equivalent resistance (Rth) to form the complete Thévenin equivalent circuit, with Vth calculated as 5.574V.

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DmytriE
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Hi All,

I am struggling to find the thevenin equivalent for the following circuit. I will leave out the values for the resistances because they don't really matter in this case. I'm looking for guidance on how to find the overall thevenin.

I know the R-Thevenin is the same as the R-equivalent for the circuit. So I approached the problem as I would to find the equivalent for any other circuit.

R1 is parallel with R2. So This new resistance would be (R1 * R2)/ (R1 + R2). Let's call this new resistance R12

This next part is where I get confused. I believe that R3, R12, and R5 are all in series with one another so you can add their resistances. and this new resistance would be in parallel with R4.

I then simulate the circuit using PSpice and I do not get the same answer as the simulation. Any help would be greatly appreciated!

https://docs.google.com/file/d/0B70IRvZhJgh0UDAxaWowN0tIYkU/edit?usp=sharing
 
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I cannot access the shared circuit diagram I guess?
 
Rudinhoob said:
I cannot access the shared circuit diagram I guess?

I have now fixed the problem. Everyone should be able to access it. Thanks for the heads up.
 
Could you please show how your simulation and analytical solution differ?

Edit: Just a thought, but did you calculate the equivalent voltage source as well that's needed to form the complete Thévenin equivalent circuit?
 
Last edited:
Here are the values for each resistor:
R1: 5.6k, R2: 12k, R3: 2.2k, R4: 15k, R5: 1k
VDC: 12V

R1 || R2 = 3.818k
R3, R12, R5 are all in series: 7.018k
R3.12.5 || R4: 4.781k

https://docs.google.com/file/d/0B70IRvZhJgh0eTdxZ2ZVeS1aamc/edit?usp=sharing

RTh = Vin / I1
I1 = 935uA
RTh = 12V / 935uA = 12.830k

There is a discrepancy of approximately 2.5 - 3 times...
 
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Very strange. You have done an admirable job of calculating the correct equivalent resistance (4.781K) and then for some reason you have ignored it and gone on to calculate an incorrect equivalent resistance. Why?
 
If you want to use your simulation to check your calculation of the equivalent resistance then:

- The equivalent voltage, Vth, is equal to V_O when the output is an open circuit.

Thus Vth = 15e3*371.6e-6 V = 5.574 V.

- Rth = Vth/I_O when the output is a short circuit.

Thus, in your simulation, you need to short out R4 and measure the current through R3/R5. Vth divided by this current is your equivalent resistance.
 
Thank you very much! I was having trouble understand how to find Vth. With your help and some help of my classmate I feel confident in finding VTh and RTh.
 

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