Electronic – Result with Thevenin method doesn’t equal node method

circuit analysisthevenin

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

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

Where:

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

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?

Best Answer

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

schematic

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.

Superposition

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.

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