# Resistors in parallel and series

## Homework Statement ## The Attempt at a Solution

I added annotations to show my thought process

this last part was a good tricky question. but here's was my thought process.

since i need to find the points between A and B I need to do arithmetic on the points between A and B which is all the numbers in ohms

1 and 3 and 6 and 2 are in series. 4 ,12, 5 and 20 are in parallel.

with that said, 1+3+6+2 = 12 ohms

4*12/(4+12) = 48/16 = 3
5*20/(5+20) 100/25 = 4

12+3+4 = 19 ohms but its wrong.

any guidance will be appreciated

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SammyS
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## Homework Statement

[ IMG]http://i33.photobucket.com/albums/d86/warnexus/resistors-1.jpg[/PLAIN]

## The Attempt at a Solution

I added annotations to show my thought process

this last part was a good tricky question. but here's was my thought process.

since i need to find the points between A and B I need to do arithmetic on the points between A and B which is all the numbers in ohms

1 and 3 and 6 and 2 are in series. 4 ,12, 5 and 20 are in parallel.

with that said, 1+3+6+2 = 12 ohms

4*12/(4+12) = 48/16 = 3
5*20/(5+20) 100/25 = 4

12+3+4 = 19 ohms but its wrong.

any guidance will be appreciated
The 1Ω and 3Ω are in series, but they're not in series with any other single resistor; certainly not in series with the 6Ω resistor.

Similarly, the 2Ω resistor is not in series with any other single resistor.

What constitutes resistors being in parallel?

What constitutes resistors being in series?

The 1Ω and 3Ω are in series, but they're not in series with any other single resistor; certainly not in series with the 6Ω resistor.

Similarly, the 2Ω resistor is not in series with any other single resistor.

What constitutes resistors being in parallel?

What constitutes resistors being in series?
resistors in series if the components take on one path
resistors in parallel if the component are connected at each end but the component can take on multiple paths

2, 5, 20, 4 ,6 and 12 are in parallel. 1 and 3 are in series

thanks I understand now!

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SammyS
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resistors in series if the components take on one path
resistors in parallel if the component are connected at each end but the component can take on multiple paths
There's a bit more to it than that.
2, 5, 20, 4 ,6 and 12 are in parallel. 1 and 3 are in series
Mostly, that's incorrect. -- That's over simplified.

Since it's handy, let's look at the circuit in this problem for example.

First of all, we you are asked to find the equivalent resistance between points A & B . This could be done in a laboratory by placing a battery across A & B and measuring the voltage, V, across A & B as well as the current, I, passing through point A (or through point B -- same current either place.) Then use Ohm's Law: REq=V/I .

Current passing through point A has many possible paths it can take in order to finally arrive at point B. However, looking at the big picture, there are two main paths.
1. The path through point C, then through that mess of resistors, then through point D and finally through B.

2. The path through point E, then through that other mess of resistors, then through point F and finally through B.​
So, current entering this set of resistors can pass either through the leg with points C & D or pass through the leg with points E & F. No current passing through point C can get to B without first passing through D. Similarly, no current passing through point E can get to B without first passing through F.

So, the leg of the circuit from C to D is in parallel with the leg of the circuit from E to F.

The leg from E to F: 2Ω, 5Ω, & 20Ω resistors.
Any current from E that passes through the 2Ω resistor must then somehow pass through that combination of 5Ω, & 20Ω resistors before passing through F.

Any current passing through the 2Ω resistor then either passes through the 5Ω resistor or through the 20Ω resistor before going on to F. Therefore, the

So, the 5Ω & 20Ω resistors are in parallel.

In summary, the 2Ω resistor is in series with the parallel combination of the 5Ω & 20Ω resistors.​

The C D leg is a bit more complicated mainly due to the three resistors, the 4Ω, 6Ω, & 12Ω resistors. The three resistors are in parallel. Current arriving to left left of this trio can pass through any one of the three before joining up & continuing on to point D.

the oversimplification got me the answer. =] but i do see where you are getting at. thanks for the guidance.

SammyS
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the oversimplification got me the answer. =] but i do see where you are getting at. thanks for the guidance.
That's some coincidence.

How did you combine those values?

So, what did you get for the answer?