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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Sign up to join this communityWhat 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?
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.