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I've been attempting this problem regarding the usage of Thévenin's Theorem to find the value of Io. I am able to get the correct Thevenin Resistance, however attempting to find the Thevenin Voltage has been confusing me. When breaking the circuit at the 4kOhm resistor I tried to do a kVl loop around the bottom left square containing the 5V source and the open voltage from opening the resistor and got a value of 5 which is incorrect. I'm confused as to why it is incorrect and am wondering if it is because there is a voltage attributed to the 2mA independent source which I am not considering. I have attached the problem, any guidance would be greatly appreciated.

Thank you!

Problem Circuit.

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  • \$\begingroup\$ I would use superposition : \$V_{th} = 5+2m*(7k) + 9m*(6k)\$ \$\endgroup\$ – across Mar 20 at 16:01
  • \$\begingroup\$ Would the steps for that be leaving only one of the independent sources active at a time and seeing the voltage contribution to Vth from each, then adding them all? Just trying to make sense of the expression you've provided as the visualization of it is not clear to me. Ex/ How does 9m * 6k come about? \$\endgroup\$ – Norbert Tanacs Mar 20 at 16:17
  • \$\begingroup\$ Yes. For 9m*6k: 1) Remove 4k resistor. 2) Short 5V source and open 2m source. 3) Consider only the 9m source. T \$\endgroup\$ – across Mar 20 at 16:19
  • \$\begingroup\$ Then there is only one loop in the circuit: 9mA goes through 6k resistor and drops a voltage of 9m*6k. \$\endgroup\$ – across Mar 20 at 16:20
  • \$\begingroup\$ Notice that the thevinin port, 9mA source and the 6k resistor are in parallel. So 9m*6k is the voltage seen at the thevinin port. \$\endgroup\$ – across Mar 20 at 16:22
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1) 2mA goes through 1k, combines with 9mA, then 2+9 = 11mA goes through 6k.
2) Thus the voltage drop across 1k is 2m*1k = 2V
3) Voltage drop across 6k is 11m*6k = 66V
4) Just add up the voltages: 5V + 66V + 2V = 73V is the open circuit voltage across 4k resistor.

enter image description here

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  • \$\begingroup\$ That's a tremendously creative kVl loop to me, I never even thought about that path. Beautiful stuff. \$\endgroup\$ – Norbert Tanacs Mar 20 at 23:31
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successive simplifying images

Picture being worth a thousand words, voila!

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  • \$\begingroup\$ I don't think your rearrangement of the circuit is correct, moving from the first image to the second. \$\endgroup\$ – Elliot Alderson Mar 20 at 19:30
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    \$\begingroup\$ If you're talking about my reconnection of the 2 ma source, you're right. I did change it but there's no difference in circuit function. It works the same connected to ground as it does connected to a voltage source connected to ground. It's easier to do the following steps with that change. No matter how much current is drawn from the + side of the source it will put out 5 V, and no matter how high or low the voltage is at the - side of the current source it will source 2 ma. \$\endgroup\$ – user128351 Mar 20 at 21:21
  • \$\begingroup\$ You reached a much different answer than becca becca. Can you leave a comment for their answer explaining why it is incorrect? \$\endgroup\$ – Elliot Alderson Mar 20 at 22:23
  • \$\begingroup\$ @stretch shifting the 2mA source to ground is really clever and allowed using the source transformations! +1 for teaching me this cool trick XD \$\endgroup\$ – across Mar 21 at 0:39
  • \$\begingroup\$ @ElliotAlderson both our answers are same. Stretch found the current throug 4k resistor directly. Whereas in my answer I found the thevinin equivalent about the 4k resistor. In my answer, 73V/7k = 10.43mA is the shortcircuit current through 4k resistor, which matches with stretch's last picture:) \$\endgroup\$ – across Mar 21 at 0:45

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