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What I did was I found the $I_{short circuit(AB)}$ = 10/5 = 2A by shorting AB.

How am I supposed to find $R_{AB}$ in order to solve for $V_{Thevenin}$? It can't be 5//10 right?

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    $\begingroup$ looks like homework... also this belongs in https://electronics.stackexchange.com/ $\endgroup$
    – Pete W
    Apr 30 at 13:35
  • $\begingroup$ No, it's not 5||10. Replace sources with their ideal resistances. Voltage sources are 0Ω and Current sources are ∞. $\endgroup$ May 1 at 19:09
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For an open circuit,Thevenin Equivalent Voltage is the Open circuit voltage.I use Node Voltage method to find the Thevenin Equivalent Voltage. Voc = Va - 0 [(Vs - Va)/R1] + Is = (Va/R2) Va = [Vs+(R1Is)] / [1+(R1/R2)] Va = [10+(51)] / [1+(5/10)] Va = (10+5)/[(10+5)/10] Va = (15/15)(10/1) Voc = Va = 10V Therefore, Thevenin’s Equivalent Voltage is 10V.

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