Series parallel combination resistance.

[Simonslimline]

Just spotted this on my facebook feed :juggle2: and thought it may be a good exercise for people to have a go at.

What is the total resistance (Rt) of the circuit?




:book:

 

[rolyberkin]

Just spotted this on my facebook feed :juggle2: and thought it may be a good exercise for people to have a go at.

What is the total resistance (Rt) of the circuit?

View attachment 31519

:book:

10.5, although not completely sure, added the 10Ω,2Ω & 1Ω as in series to make 13Ω, dealt with resulting 13Ω in parallel to the 4Ω to get 3.05Ω, and then added the 5Ω, 3.05Ω and 2Ω together in series to get the RT as 10.5Ω.

Probably not that straight forward though

 

[Simonslimline]

Yes, nice and easy then really.

Resistor 4 is in parallel with resistors 1,2 & 10.

Resistor 1, 2 & 10 are in series with each other to make a total of 13 Ohms.

So if we use product over sum we get 4 x 13 = 52. 4 + 13 = 17. 52 / 17 = 3.058823529 Ohms.

So now we have 3 resistances we can add in series for the total Rt. 5 + 3.058823529 + 2 = 10.058823529 Ohms.

I have not rounded down though hence a slightly different answer.

 

[rolyberkin]

Need to get my head back into the science module as have my 302 on the 26th, a good bit of resistor revision

 

[Marvo]

........then added the 5Ω, 3.05Ω and 2Ω together in series to get the RT as 10.5Ω.

Probably not that straight forward though

Your method was spot on but unfortunately your addition let you down.

 

[telectrix]

the answer is 6. i am not a number, i am a real person.

 

[Simonslimline]

Your method was spot on but unfortunately your addition let you down.

Well spotted. I missed that. :rolleyes2:

 

[Marvo]

No problem, the difficult part was correct, we can always work on the adding up later

Seeing as you all seem to be in the mood, here's another one for you.

 

[hightower]

No problem, the difficult part was correct, we can always work on the adding up later

Seeing as you all seem to be in the mood, here's another one for you.

View attachment 31529

I get 60.13 ohms but only on phone with no paper to jot on so might have gone wrong somewhere.

 

[rolyberkin]

I get 60.13 ohms but only on phone with no paper to jot on so might have gone wrong somewhere.

60.3Ω for me also, even think I have added it up correctly this time!

 

[rolyberkin]

60.3Ω for me also, even think I have added it up correctly this time!

Or 60.13Ω even (what a wan@+$!)

 

[hightower]

Or 60.13Ω even (what a wan@+$!)

Ha was about to comment saying that our answers were different. ... too eager with the keyboard [emoji38]

 

[rolyberkin]

I blame it on my sausage sized fingers!

 

[Marvo]

I haven't had time to work it out and write out the method so I'll leave it to one of the trainees to lay it out as a proper written answer

**edit** @ Rolyberkin, I think your calculator might need calibrating

 

[hightower]

I did furthest three as series, total 66 ohms run in parallel with 22 ohms is 16.5 ohms. Then 22 plus 22 plus 16.5 for the next simplified series, total 60.5 in parallel with 22 ohms, resulting in 16.13. Final step was 22 plus 22 plus the 16.13 from the simplified circuit equaling 60.13.