In my circuits class, I am presented with the following exercise.


simulate this circuit – Schematic created using CircuitLab

The problem is to find the voltage across resistors A and B. The most likely solution (given the section) is to use simplification by analysis. I tried to solve for B first. My attempt included cutting the circuit in two parts (all of the resistors and just B), simplifying the part of the circuit that was not B, and then solving the resulting circuit. This yielded a parallel circuit.

I know that is not supposed to yield a parallel circuit which implies a direct voltage path from the source to resistor B.

The book as has a similar example, but in this one the two ohm resistor is to the right of the 3 ohm resistor, making simplification vastly easier. (I seem to have missed the lecture on this.)

Could someone be so kind as to point me in the right direction?

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    \$\begingroup\$ I got it! The first part, at least. Thanks anyways. Just to be nice I'll post what I did when I get the time.... \$\endgroup\$ – user1833028 Oct 21 '13 at 18:12
  • \$\begingroup\$ Normal simplification works. 3 || 6 = 2 Ohms, 2+2= 4 Ohms again 4 || 4 is 2 Ohms. So, voltage of A w.r.t Common Point (6V source negative terminal) is 6V*2/(2+4) = 2V, Current through 2 Ohms +3||6 circuit is 0.5Amps.So, voltage across 3|6 Ohms resistor is 2*0.5A = 1V. So, Va=2V, Vb=1V==> VAB = 2-1 =1V \$\endgroup\$ – user19579 Oct 22 '13 at 6:08

Your approach, if not your execution, is sound .

If you're familiar with Thevenin equivalent circuits, you know that you can replace the entire circuit to the left of the B resistor with a voltage source in series with a resistor. Once you've done this, voltage division will give you the voltage across B.

For A, replace all the resistors to the right of A with an equivalent resistor which is in parallel with A. Then, once again, apply voltage division to find the voltage across A.

  • \$\begingroup\$ Thevenin circuits are in the chapter after next. I've got A so far... I apparently was caught on an error in the simplification process itself. I hope. \$\endgroup\$ – user1833028 Oct 21 '13 at 18:14

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