I am looking at the circuit in: Apply Thevenin Theorem to a circuit with 2 sources?

The circuit and its equivalent have been given as follows in the thread:


Circuit 1

Thevenin Equivalent:


Instead of v_test, I presume I can place a resistor, resistance 10, and the current across the resistors should be the same for both the original circuit and its equivalent.

For the first circuit, the current across the added resistor is: 0.96 A

For the equivalent circuit, the current across the added resistor is: \begin{equation} i = 6.4/(3.3333+10) = 0.48 \end{equation}

what is the issue here?

  • \$\begingroup\$ Well, it seems that somebody made a mistake (are you 100% sure it was not you?) Any reason not to comment in or fix that thread instead? \$\endgroup\$ – Edgar Brown Jan 23 at 21:06
  • \$\begingroup\$ I think there is some question about whether the 10 ohm resistor in the original circuit was to be part of the Thevenin equivalent or not. It looks like the original answer assumed that it was not, but you have assumed that it is. \$\endgroup\$ – Elliot Alderson Jan 23 at 21:09
  • \$\begingroup\$ Well, the Rth is indeed equal to 3.33R but Vth is not 6.4V. The correct Vth value is 12.8V \$\endgroup\$ – G36 Jan 23 at 21:09
  • \$\begingroup\$ @G36 I just found out the same thing too. I thought the answer given in the thread was correct \$\endgroup\$ – Snifkes Jan 23 at 21:16
  • \$\begingroup\$ @Snifkes As you can see this is not always the case. And the internet is full of "errors" but you can find them in books and in datasheets also. \$\endgroup\$ – G36 Jan 23 at 21:23

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