Series parallel combination resistance.

Marvo · Feb 16, 2016

Nicely done. I know this stuff gets boring, once the penny has dropped you can do them with your eyes closed.

Here's another that might require a little more thought. Assume for simplicity that all resistors in the cube are 10ohms.

What is the actual resistance between the two points highlighted by the red arrows

hightower · Feb 16, 2016

Marvo said:
Nicely done. I know this stuff gets boring, once the penny has dropped you can do them with your eyes closed.

Here's another that might require a little more thought. Assume for simplicity that all resistors in the cube are 10ohms.

View attachment 31535

What is the actual resistance between the two points highlighted by the red arrows

8.27 ohms is my guess. Tried to simplify circuit, so to start it can go one of 3 paths, so drew that as 3 in parallel. Then can go one of 6 paths, so 6 in parallel (miss counted hence scribble), then can finish one of 3 paths, so 3 in parallel. This gives a 3er, a 6er, and a 3er all connected in series. Picture showing my messy working out attached.

Not sure if I've approached this the right way though?

Ps ignore the bottom right, was messing with motors before, and bottom left is a game of swanny we're about to start.

telectrix · Feb 16, 2016

i'm still trying to find 12 10 ohm resistors to cobble together in a cube.

Marvo · Feb 16, 2016

Funny you should say that Tel. When i was an apprentice and one of our college lecturers gave us this problem it had me a bit stumped and that's exactly what I did. I built the cube, measured the resistance and worked backwards and figured out the method of doing it.

I'd actually forgotten about it and this thread reminded me so I drew it out and posted it but I see now the answer's all over the internet and I could have saved myself some time and pinched a drawing from elsewhere.

You've got the correct method Hightower, you need to draw it out as three parallel networks that are in series as per your sketch. After that it should be plain sailing to the answer. I'll let the other trainees have a bash and if I've got time tomorrow I'll solve it and confirm the answer.

rolyberkin · Feb 16, 2016

Okay so I needed a print out and the kids markers to solve this one, the Mrs thinks I have gone completely mad!

The three blue lines are in parallel, they lead to six red again in parallel and then these lead in turn to three yellow, so a 3 x + 6 x + 3 x parallel circuit. I definitely won't forget how to work out resistors now!

3.3Ω + 1.6Ω + 3.3Ω = 8.2Ω

Marvo · Feb 17, 2016

Theft of kids crayons gets a 'like' from me

You also have it right but it would be better to work to a minimum of two decimal places so your 1.6 ohms would be better as 1.67 (1.6666666 reoccurring rounded up). Not sure what the standard is nowadays, we always worked to 3 decimal places when I was at school.

rolyberkin · Feb 17, 2016

Marvo said:
Theft of kids crayons gets a 'like' from me

You also have it right but it would be better to work to a minimum of two decimal places so your 1.6 ohms would be better as 1.67 (1.6666666 reoccurring rounded up). Not sure what the standard is nowadays, we always worked to 3 decimal places when I was at school.

I think it is my calculator perhaps set up wrong or on the wrong setting, will have a look, the thing is so bloody confusing I keep having to watch youtube videos on how to work it!