A  B
   |‾‾|  |‾‾|
i1 ↑  R1 R2 ↓ i2

equivalent to:

       R1 A B R2
     |‾  ‾° °‾  ‾|
i1R1 ±           ∓ i2R2

as seen from across the two resistors (A-D and B-D)?

PS: I wish we had a quick way to insert pretty circuit diagrams here.

  • 3
    \$\begingroup\$ circuitlab.com is useful to make quick diagrams and circuit simulations. \$\endgroup\$ – Reid May 4 '12 at 22:27

Note: this answer was given before the question drew and specified reference points.

There is no way to answer your question. In order to know whether they are equivalent or not, you have to specify as seen from where? As seen from which two nodes?

Look at the figure. These two circuits are equivalent as seen (for instance) from A-B, or from A-D, but not (for instance) from C-A, or from C-B. That nuance is important.


  • \$\begingroup\$ Great catch! I meant as seen from A-D and from B-D. \$\endgroup\$ – Andres Riofrio May 5 '12 at 2:09

Looks right. Think about the voltage across R1 in the first circuit (i1R1, + on top, assuming bottom node is GND). Then consider the voltage from the bottom node to the right-side of R1 in the second circuit (i1R1, assuming bottom node is GND).

  • \$\begingroup\$ The reasoning is not right. You just checked that open-circuit voltages are equal. You should also check that short-circuit currents are equal. If resistor R1 in the bottom circuit (not the expression "R1" in "I1·R1", but the explicit series resistor) was different, you wouldn't detect that with your reasoning, and the two circuits would not be equivalent (as seen from A-B --see my answer--). \$\endgroup\$ – Telaclavo May 5 '12 at 0:26
  • \$\begingroup\$ The node C is completely different in the two circuits. A source transformation is not a valid technique to use when the node between the source and resistor is the one in question. Thus, I figured from C-A and C-B were irrelevant references to consider. \$\endgroup\$ – Shamtam May 5 '12 at 14:42
  • \$\begingroup\$ You didn't get my point. Forget about node names. You did not check, in your answer, that short-circuit currents are equal. \$\endgroup\$ – Telaclavo May 5 '12 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.