$R_{TH} = R1 + R2//R5 + R3//R4$

Then considering the current source and the resistance R3 as a Norton circuit, one can convert it to Thevenin as follows $V_I = -IR3$. Then we have the voltage sources in series so

$V_{TH} = V1 + V2 -IR3$

Does this make sense? If not how do I fix it?

I agree with your answer, I guess 'I' is also a given value.

• -1: I do not believe the answer is correct. More thorough math demonstration is needed here. – Vicente Cunha Apr 5 '16 at 18:53

Redrawing originial circuit for clarity:

simulate this circuit – Schematic created using CircuitLab

Norton to Thevenin of I and R3 lead to:

simulate this circuit

Thevenin voltage is equal to open circuit voltage. Since there is no current through R1, its voltage drop is zero, leading to the following:

simulate this circuit

Circuit equations:

$$\frac{V_{th} - (V_x + V_1 - R_3I)}{R_3} + \frac{V_{th} - V_x}{R_4} = 0$$ $$\frac{(V_x + V_1 - R_3I) - V_{th}}{R_3} + \frac{V_x - V_2}{R_5} + \frac{V_x - V_{th}}{R_4} + \frac{V_x}{R_2} = 0$$

Solution is:

$$V_{th} = \frac{1}{(R_2+R_5)(R_3+R_4)}\times\\\{V_1(R_2R_4 + R_4R_5) + V_2(R_2R_3 + R_2R_4) - I(R_2R_3R_4 + R_3R_4R_5)\}$$

• R1 is open circuit? Do you know what a Norton equivalent is? – Claudio Avi Chami Apr 5 '16 at 20:17
• Yes, I do know what a Norton equivalent is. OP question is regarding to Thevenin voltage, which is equal to open circuit voltage (I believe you know that as well. Are you mistaking R3 for R1?). – Vicente Cunha Apr 5 '16 at 20:21
• No. Is two separate questions. You didn't calculate the Norton equivalent well, and R1 is not 'open'. – Claudio Avi Chami Apr 5 '16 at 20:28
• @ClaudioAviChami Done editing. The Norton to Thevenin is positively correct, and R1 has no current, nor voltage drop. – Vicente Cunha Apr 5 '16 at 20:38