2
\$\begingroup\$

I'm trying to self-study this problem and I found a bizarre problem with an unusual circuit.

enter image description here

I have never faced this in all the problems that I have solved. What do they mean by finding the equivalent resistance between b and e. I don't even know if this is possible. If you look carefully, you'll see that there is a terminal c between terminal b and e. I have no idea how this could be solved.

\$\endgroup\$
  • 2
    \$\begingroup\$ No one cares about the other terminals. Ignore them. \$\endgroup\$ – Ignacio Vazquez-Abrams Sep 25 '14 at 6:48
  • 2
    \$\begingroup\$ To underscore what the various answers have implied: this drawing is deliberately drawn so that it's hard to see what's going on. Often the best first step is to redraw it so that the current flows are clearer, usually by putting one of the terminals that you're interested in at the top and the other terminal that you're interested in at the bottom. Don't forget, the drawing doesn't dictate the physical location of the resistors, just the connections between them. As long as you don't change what's connected to what you can move things around so that it's easier to analyze. \$\endgroup\$ – Pete Becker Sep 25 '14 at 12:45
7
\$\begingroup\$

Look at the path from b to e: if you go up along the \$5\Omega\$, \$3\Omega\$ and \$4\Omega\$ resistors, up to there you have a total resistance of \$(5+4+3)\Omega=12\Omega\$. This resistance is in parallel to the resistance in the lower path up to that point, which is \$4\Omega\$. Evaluating the parallel connection of these two resistances gives

$$R_1=\frac{12\Omega\cdot 4\Omega}{(12+4)\Omega}=3\Omega$$

To this resistance you have to add the \$12\Omega\$ resistor leading to point e. So you get a total resistance between b and e

$$R=R_1+12\Omega=15\Omega$$

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ It doesn't make much sense to me. What happened to the 10Ω resistor? It also seems that the path from b to e is (5 + 4 + 3 + 12) = 24Ω. \$\endgroup\$ – George Chalhoub Sep 25 '14 at 6:58
  • 1
    \$\begingroup\$ @georgechalhoub: The 10 Ohms resistor is not in any path from b to e. And you have to add the 12 Ohms resistor only after having computed the parallel resistance formed by the two paths described in my answer, because it is common to both paths. Look at the lower connection point of the 12 Ohms resistor. Both paths from b to e meet there. Up to that point, the upper path has a resistance of 5+3+4 Ohms and the lower path has 4 Ohms. These paths are in parallel, so you need to compute this total resistance, and then you add the 12 Ohms resistor. \$\endgroup\$ – Matt L. Sep 25 '14 at 7:01
  • \$\begingroup\$ Oh, I see it all now. The a and c terminals tricked me. I thought terminal a could be not removed. One more question, did we remove completely 10Ω resistor from our circuit? If yes, why could we do it? \$\endgroup\$ – George Chalhoub Sep 25 '14 at 7:11
  • \$\begingroup\$ @georgechalhoub: Yes, you might as well remove the 10 Ohms resistor. Imagine a voltage between points b and e. Would there be any current through the 10 Ohms resistor? \$\endgroup\$ – Matt L. Sep 25 '14 at 7:13
9
\$\begingroup\$

The first helpful step is to re-draw the schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Here I've ignored all other terminals because they don't matter. Once you can prove to yourself that the two circuits are equivalent, then prove to yourself that R6 has no effect and can be dropped. Re-drawing the circuit again,

schematic

simulate this circuit

The answer should become obvious now.

| improve this answer | |
\$\endgroup\$
4
\$\begingroup\$

Looking just at B, C, and E ...

enter image description here

| improve this answer | |
\$\endgroup\$

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.