# The Equivalent Resistance

1. Oct 9, 2011

### roam

1. The problem statement, all variables and given/known data

What is the equivalent resistance between points a and b:

[PLAIN]http://img706.imageshack.us/img706/8615/circuitq.jpg [Broken]

3. The attempt at a solution

The correct answer is Req=25 Ω. Looking into the circuit the 20 Ω is in series with the 10 Ω which is in parallel with the series combination of 5 Ω and 10 Ω resistos a, which is in parallel with the last 10 Ω resistor:

Req = 20 + [10 || (10+5) || 10]

$R_{eq} = 20+ (\frac{1}{\frac{1}{10} + \frac{1}{15} + \frac{1}{10}}) = 23.75 \ \Omega$

My answer is slightly different from the model answer, so is something wrong with my calculation?

Last edited by a moderator: May 5, 2017
2. Oct 9, 2011

### Staff: Mentor

The 5Ω resistor is not in series with the middle 10Ω resistor; there's another resistor (the rightmost 10Ω resistor) sharing the node where they connect. For components to be in series they must be alone in a line and so pass the same, identical current.

I suggest that in this case you start at the "back end" of the circuit (the rightmost end) and work towards the front.

Last edited: Oct 9, 2011
3. Oct 9, 2011

### Spinnor

I think it should,

20 + 10 || ( [ 10 || 10] + 5 )

4. Oct 9, 2011

### WJSwanson

With a circuit like this, it's often extremely helpful to isolate your equipotential surfaces. This allows you to determine quickly whether various circuit elements are in series or in parallel.

5. Oct 9, 2011

### roam

Thank you guys, 20 + 10 || ( [ 10 || 10] + 5 ) worked. And the diagram was very helpful. :)

6. Oct 9, 2011

### WJSwanson

I'm glad the diagram helped. Frankly, I've had my fair share of circuit analysis (including a class devoted to it, as well as my linear algebra text which actually had an entire section devoted to matrix analysis of circuits), and I still do things that a lot of intro-level students often consider "hokey" or otherwise reserved for people who aren't very good at the topic -- like diagramming out all the equipotentials on a circuit in order to aid in evaluating equivalent capacitances and resistances.

The reason is that, simply put, visual representations are almost always helpful if they're rendered correctly. Just remember that the smartest thing you can do as a student isn't to try and solve a problem in a way that's harder for you because other people prefer to solve it that way. The smartest thing you can do as a student is to use whichever techniques help you to solve the problems that are put before you. There is absolutely no shame in setting up a diagram and marking it up, or drawing a picture to help you conceptualize the problem; in fact, the opposite is true. To quote my old intro modern physics instructor, Dr. Zhao:

"I don't get why people don't want to draw a picture. Does it embarrass them or something? PLEEEAAAASE! That's how I know someone knows what he's doing. Why do I want someone who spits numbers at me instead of thinking about what I'm asking him? That's just silly. When I see that picture with those pretty notes on it, I know that guy has his head on straight."

7. Oct 10, 2011

### roam

That's true. I never really got it until I saw your diagram, it clarified everything. Thanks a lot.