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Basic circuitry, adding up resistors

Tiis

Dec 28, 2014
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Hi, I'm practising for redoing my exam about basic circuitry DC. I've begun from the beginning with simplifying circuits. Stuck at the following bit:

http://imgur.com/a/D8zor

I've added notes to each picture to name what I did. I think the result at the end is wrong, if I add the resistors in the final schematic I get a ridiculously big answer, I am 100% certain my teacher wouldn't give me a question with an answer like that.
 

Tiis

Dec 28, 2014
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If I'm reading your values correctly, I get 8.226 ohms.
I think my school required that all answers are in fractions. I get 3736/455, but I thought it was way too ugly to be the correct answer.
 

DoubleDogDan

Jul 20, 2014
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3736/455 = 8.211, so we're both either doing it right & making the same error!
 

davenn

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The resistance total is a little higher than the 8.211 Ohms stated in previous post

Tiis, note you have made a printing error in your diagrams

in t he first diag, you have 4 Ω and 2 Ω circled but in future diagrams you have it as 4/3 instead of 4/2
its going to alter your final result a little

I wont tell you the answer yet, I will let you recalc it and see what you get

Doubledogdan Please DONT post any more answers

cheers
Dave
 
Last edited:

Tiis

Dec 28, 2014
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The resistance total is a little higher than the 8.211 Ohms stated in previous post

Tiis, note you have made a printing error in your diagrams

in t he first diag, you have 4 Ω and 2 Ω circled but in future diagrams you have it as 4/3 instead of 4/2
its going to alter your final result a little.

Doubledogdan Please DONT post any more answers

cheers
Dave

In order to replace the 4:2 with one resistance. You take 4 * 2 and divide it with 4 + 2. (4 * 2) / (4 + 2) = 8/6 = 4/3. The 4 Ω and 2 Ω together can be replaced with 4/3 Ω.
 

davenn

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In order to replace the 4:2 with one resistance. You take 4 * 2 and divide it with 4 + 2. (4 * 2) / (4 + 2) = 8/6 = 4/3. The 4 Ω and 2 Ω together can be replaced with 4/3 Ω.

but that doesn't gel with the way you worked out the others, which looked ok

ok, lets start at the start again and replace you 4/2 (4/3) and all the others with actual calculated values
and redo your work in steps for your images 2, 3 and 4
 

Gryd3

Jun 25, 2014
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I think my school required that all answers are in fractions. I get 3736/455, but I thought it was way too ugly to be the correct answer.
That will depend on your school...
When you start adding parallel resistances, the 'exact' number can get quite messy unless the question is very carefully calculated to provide a 'simple' answer.
It's also great that you calculate it down to a fraction that provides a value with a few decimal places of accuracy, but in the real world, that kind of accuracy will rarely (If not ever) exist.
For now, hang onto your fractions, but you should probably confirm if the answer should be a fraction or not... as almost all of the answers I calculated in physics or Electronics required an answer with no more than 5 significant digits.. I can't imagine such a complex fraction would be the desired answer if the same thing can be expressed accurately enough with a simple ( 6.43Ω ) **Note this value is made up.

I ran the numbers from your first picture, and so far, am confident in your skills. Do as Davenn suggested to double check your work and we can all move forward from there.

Proverb of the Day for @DoubleDogDan
"give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime"
This is homework help. Merely providing the answer without showing work, or encouraging the op to come to their own conclusion will not help them in their pursuit and may hinder the learning process. Of course, this would be different if they provide an answer already and simply want verification if it's done right.
 

Tiis

Dec 28, 2014
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That will depend on your school...
When you start adding parallel resistances, the 'exact' number can get quite messy unless the question is very carefully calculated to provide a 'simple' answer.

Yeah, I'll ask it after winter break is over. Other people, some using special software confirmed answer too, so I'm sure I got it correctly then. Which is weird, because the last time I practised this I got all different kinds of answers, made me doubt the answer I got now.
 

Tiis

Dec 28, 2014
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but that doesn't gel with the way you worked out the others, which looked ok

ok, lets start at the start again and replace you 4/2 (4/3) and all the others with actual calculated values
and redo your work in steps for your images 2, 3 and 4
Someone helped me work it out: https://i.imgur.com/gmAPeTx.png

Pretty confident now I was apparently correct. Surprise, surprise.
 

Gryd3

Jun 25, 2014
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Yeah, I'll ask it after winter break is over. Other people, some using special software confirmed answer too, so I'm sure I got it correctly then. Which is weird, because the last time I practised this I got all different kinds of answers, made me doubt the answer I got now.
Depends what you mean by different kinds of answers...
If you get a different answer at any time when you 're-do' the questions then there is absolutely something missing. You should find out what you are doing differently each time and if it's correct of not.
If you are consistent with the answer, then you have formed a reliable method of analyzing the diagram. This may or may not be right... if you get a correct answer, it's safe to assume it's right. If you get the wrong answer, something in your process is wrong and no matter how many times you redo it, you will get it wrong until you can identify what is being done wrong.

If in doubt, double-check your own work. If you still have issues, show us an example problem with your work and solution and we can help identify any missing or incorrect processes.

Best of luck!
 

Gryd3

Jun 25, 2014
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I didn't get what you did
It's hard to tell exactly how the math was done cuz I can't see it :s
But he appeared to work his way in from the right hand side, and would only simplify a single series or parallel relationship at a time. Assuming the numbers are correct... as I said, it's hard to read. I would have broken the diagram down in similar steps.
 

Tiis

Dec 28, 2014
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I didn't get what you did


Gryd3 ... can you do the calc's and PM me your answer ... maybe I boo boo'd

That one might indeed be a little small. It's not mine though I have to say. Here's another guy's take on it: http://i.imgur.com/WDrEIRB.jpg

The actual calculations aren't shown in this picture either, but the answer checks out, it's the same one I got and the other people got.
 

Gryd3

Jun 25, 2014
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OK ... I may have to recalc ;)
Davenn. Sorry for not PMing this... I cant find out how to ad images to a personal message :S

And I'm at work, so I did what I could with GIMP... hope that:
A) It helps.
B) Im correct.
C) Does not undermine teaching the op.

The method I used for adding parallel resistors was: Rt = ( R1^-1 + R2^-1 + ... + Rn^-1 ) ^ -1


working.gif
 

davenn

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C) Does not undermine teaching the op.

no probs ... the thread has run its course :)

Yes I used the same steps must have made a typo ... my eyes are really bad these days :(
redoing my calc's now
 

Gryd3

Jun 25, 2014
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no probs ... the thread has run its course :)

Yes I used the same steps must have made a typo ... my eyes are really bad these days :(
redoing my calc's now
That's a very good reason not to put the answer up on the thread right away! A typo could cause quite the headache ;)
No prob there Dave. To be honest, you made me nervous doing my work if you had a different answer than I .
 

davenn

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LOL

my first answer the other day was 10.2414 Ω

I found my typos and ended up with 8.2107 rounded to 8.211 Ω yeah hahahaha

over the last 12 months I am really struggling to read/see with a bad double visioning problem
that prescription glasses are not really correcting much.
As I type this my eyes are dropping in and out of both double vision and focus making the
keyboard a blur in front of me ... the screen is just as bad to read :(
 
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