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enter image description here

\$ 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?

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

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  • \$\begingroup\$ -1: I do not believe the answer is correct. More thorough math demonstration is needed here. \$\endgroup\$ – Vicente Cunha Apr 5 '16 at 18:53
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Redrawing originial circuit for clarity:

schematic

simulate this circuit – Schematic created using CircuitLab

Norton to Thevenin of I and R3 lead to:

schematic

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:

schematic

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:

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

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  • \$\begingroup\$ R1 is open circuit? Do you know what a Norton equivalent is? \$\endgroup\$ – Claudio Avi Chami Apr 5 '16 at 20:17
  • \$\begingroup\$ 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?). \$\endgroup\$ – Vicente Cunha Apr 5 '16 at 20:21
  • \$\begingroup\$ No. Is two separate questions. You didn't calculate the Norton equivalent well, and R1 is not 'open'. \$\endgroup\$ – Claudio Avi Chami Apr 5 '16 at 20:28
  • \$\begingroup\$ @ClaudioAviChami Done editing. The Norton to Thevenin is positively correct, and R1 has no current, nor voltage drop. \$\endgroup\$ – Vicente Cunha Apr 5 '16 at 20:38

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