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I'm new to Thevenin's theorem and I'm trying to figure out this simple one. enter image description here

I did what's on the next figure. I'm wondering If I can join the cables as I did and I could say that the voltage across \$R_{1}\$ is also \$V_{th}\$ since they are in parallel. The answer says \$R_{eq}=R_{1}\$. But I'm getting \$R_{eq}=R_{1}+R_2\$.

enter image description here

Thank you for your time!

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    \$\begingroup\$ are you trying to compute A? \$\endgroup\$ – analogsystemsrf Apr 4 '19 at 16:27
  • \$\begingroup\$ Yes A with respect to the ground node. \$\endgroup\$ – FelipeMedLev Apr 4 '19 at 17:00
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Req = R2. See connection below.

R1 is shorted (V1 = 0) and R3 is open (I3 = 0)

enter image description here

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  • \$\begingroup\$ Ohhh. Now I get. Thanks \$\endgroup\$ – FelipeMedLev Apr 4 '19 at 17:27
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R1 ia shorted with the conducting wire in place of V1 so the parallel equivalent resistance is 0 ohm (For two resistances in parallel, equivalent resistance is $R=\ frac{R1×R2}{R1+R2}$, If any resistance is zero, equivalent resisatnce is also zero). R3 has been open circuited correctly therefore the thevenin resistance is R2 and not R1

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  • \$\begingroup\$ I don't understand the part "\$R_{1}\$ is shorted with the wire in place of \$V_1\$". Doesn't it become a closed circuit? \$\endgroup\$ – FelipeMedLev Apr 4 '19 at 16:57
  • \$\begingroup\$ For two resistances in parallel, equivalent resistance is \frac{R1×R2}{R1+R2}, If any resistance is zero, equivalent resisatnce is also zero \$\endgroup\$ – Shlok Vaibhav Apr 4 '19 at 17:32

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