I'm following the MIT lectures on circuits and electronics. They introduce the Thevanin method for solving linear systems in lecture 3 (https://youtu.be/RsJ1eg7XNVs?t=46m49s).

I can't seem to get same results using the Thevenin method as with the node method in this circuit when calculating the voltage e in this circuit:

circuit http://oi57.tinypic.com/2gtx7co.jpg

This is my result with the node method:

$$ 0 = (e - V) * \dfrac{1}{R_1} + e * \dfrac{1}{R_2} - I $$ $$ e = V * \dfrac{R_2}{R_1 + R_2} + I * \dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}} $$ $$ e = V * \dfrac{R_2}{R_1 + R_2} + I * \dfrac{R_1*R_2}{R_1+R_2} $$

But when I try to use the Thevenin method, I get a different answer.

First, I convert the part of the circuit with R2 and I:

circuit http://oi61.tinypic.com/2ibkk81.jpg


$$ T_{th} = I * R_2 $$ $$ R_{th} = R_2 $$

Then, after combining the voltage sources, I use voltage divider to find the voltage across Rth:

$$ e = (V - I * R_2) * \dfrac{R_2}{R_1 + R_2} $$ $$ e = V * \dfrac{R_2}{R_1 + R_2} - I * \dfrac{(R_2)^2}{R_1 + R_2} $$

I must have done something wrong here. Why do I get different values for e?

  • \$\begingroup\$ I'm not sure what's up with the answer you got with Thevenin equivalents yet, but your answer using nodal analysis is correct. \$\endgroup\$ – helloworld922 Sep 20 '15 at 1:04

I have no idea how you got to the second to last equation of yours. However, let's assume you have the circuit with a Thevenin equivalent circuit substituted in only for R2 and I (your \$R_{th}\$ and \$V_{th}\$ are correct).

Several ways to solve this new circuit:

Nodal Analysis


simulate this circuit – Schematic created using CircuitLab

Assuming current flows "clockwise" (in the direction of the drawn arrow), performing nodal analysis at node e gives:

\begin{gather} I_{R1} = I_{R2}\\ \frac{V - e}{R_1} = \frac{e - V_{th}}{R_2}\\ R_2 V + R_1 V_{th} = R_1 e + R_2 e\\ e = \frac{R_2 V + R_1 V_{th}}{R_1 + R_2} = \frac{R_2 V + R_1 R_2 I}{R_1 + R_2} \end{gather}

Which is the same answer as you got without using the Thevenin equivalent.


See Weezveez's answer. It's the same result.

Mesh Analysis

\begin{gather} V - V_{R1} - V_{R2} - V_{th} = 0\\ V - I_1 R_1 - I_1 R_2 - V_{th} = 0\\ I_1 = \frac{V - V_{th}}{R_1 + R_2}\\ e = V_{th} + V_{R2} = V_{th} + I_1 R_2\\ e = I R_2 + \frac{V - V_{th}}{R_1 + R_2} R_2\\ e = I R_2 + \frac{V - I R_2}{R_1 + R_2} R_2\\ e = \frac{R_2 V + R_1 R_2 I}{R_1 + R_2} \end{gather}

As you can see, all of these methods give the same answer.

  • \$\begingroup\$ Ah! Looks like I accidentally moved the ground node when combining the voltage sources. Your diagram makes it perfectly clear. Thank you! \$\endgroup\$ – Robz Sep 20 '15 at 4:11

What I did for the Vth is I used superposition. First I disconnected the voltage source, which creates a short in its place, making the voltage V1=VR1.


simulate this circuit – Schematic created using CircuitLab

Then, I disconnected the current source, which creates an open link in its place. Since no current can flow from the voltage source to complete a full circuit, the voltage on V2 is simply V2=V.


simulate this circuit

By superposition, Vth=V1+V2= V+ VR1 So e=(V+VR1)*R2


simulate this circuit

By voltage division: \$\frac{R2*(Vth)}{R1+R2}=\frac{R2*(V+I*R1)}{R1+R2}=\frac{R2*V}{R1+R2}+\frac{I*R2*R1}{R1+R2}\$

  • \$\begingroup\$ Cool, so superposition also yields the same as the node approach. I still want to know where I went wrong with the Thevenin method. \$\endgroup\$ – Robz Sep 20 '15 at 4:05

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