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Resistors in Parallel

chopnhack

Apr 28, 2014
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images_circuits_H1P1_resistor_combinations.gif


This is an algebra issue - I have written what I thought would be the correct formula for Network C - the computer doesn't like the input method, claims that it failed to parse the equation....

It liked my other two!!

Mind you, this has to be in terms of R

Parallel resistance - there are two parallel circuits.

[(R1*R2)/(R1+R2)]*2

but in terms of R:
[(R^2)(2R)]*2

This is more of a "think like a computer" issue than electrical basics knowledge.
 

Gryd3

Jun 25, 2014
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For giggles can you post the answer for A and B as well?
Could the square brackets through it off?
 

KrisBlueNZ

Sadly passed away in 2015
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Yeah, I imagine it doesn't like the square brackets. Just use parentheses for all nesting levels.
 

chopnhack

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I tried brackets, parenthesis, no parenthesis - a few variables, tried long form without algebraic reduction of terms...

Giggles:

a: R+R+R
b: R/3

I used brackets here for clarity only. ;-)
 

chopnhack

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You tried it without "^2"? Just addition, subtraction, multiplication and division symbols?
Yes, I tried this:

(R*R)/(R+R)+(R*R)/(R+R)

I tried this:

((R*R)/(R+R))*2

No avail, I am pretty confident that a parallel resistor equation is (R1*R2)/(R1+R2)
I think I will pass it up to the TA's for some assistance.
 

Gryd3

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Resistors in Parallel:

1/Rt = 1/Ra + 1/Rb + 1/Rc + ... + 1/Rx
Rt = 1 / ( 1/Ra + 1/Rb + 1/Rc + ... + 1/Rx )

(R1*R2)/(R1+R2) is the same as 1/( 1/R1 + 1/R2 ) a little algebra will allow you to change the form of the equation to something easier to work with or use.
I'm curious why is causing the answer to fail.
 

Laplace

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What if the computer recognized the parallel function ║; then you could write the answer as R+(R║2R)
 

chopnhack

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What if the computer recognized the parallel function ║; then you could write the answer as R+(R║2R)
Just tried that and still no go.. Very strange. I am waiting on some of the TA's to recognize my post. Maybe tomorrow we will have an answer! Thanks for the thought :)
 

(*steve*)

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Yes, I think it does if its the one I think it is, but it should like the others too.

I'm not sure that its too fond of extranious spaces and make sure its not expecting r instead of R.
 

Harald Kapp

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This is more of a "think like a computer" issue than electrical basics knowledge.
I think not. Your solution is not correct. consider this schematic:
bla-png.14983

R3 and R4 are in series, so the total resistance is R34=R3+R4.
R34 in turn is parallel to R2, so the total resistanceR234 is R2||R34= R2*R34/(R2+R34)
R234 in turn is in series with R1 so the total resistance of the network is R1234 = R1+ R234

Since R1=R2=R3=R4:
R34 = 2R
R234 = R*2R/(R+2R)
R1234 = R + R*2R/(R+2R) = R* (1+2/(1+2)) = R*5/3

If it were a tool question, we would have to know which computer tool you use to solve the equations.
 

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