# Resistors in Parallel

Discussion in 'Electronics Homework Help' started by chopnhack, Sep 3, 2014.

1. ### chopnhack

1,573
354
Apr 28, 2014

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.

2. ### Gryd3

4,098
875
Jun 25, 2014
For giggles can you post the answer for A and B as well?
Could the square brackets through it off?

3. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Yeah, I imagine it doesn't like the square brackets. Just use parentheses for all nesting levels.

chopnhack likes this.
4. ### chopnhack

1,573
354
Apr 28, 2014
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. ;-)

5. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
You tried it without "^2"? Just addition, subtraction, multiplication and division symbols?

chopnhack and Gryd3 like this.
6. ### chopnhack

1,573
354
Apr 28, 2014
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.

7. ### Gryd3

4,098
875
Jun 25, 2014
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.

8. ### Laplace

1,252
184
Apr 4, 2010
What if the computer recognized the parallel function ║; then you could write the answer as R+(R║2R)

chopnhack likes this.
9. ### chopnhack

1,573
354
Apr 28, 2014
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

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,482
2,830
Jan 21, 2010
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.

11. ### KrisBlueNZSadly passed away in 2015

8,393
1,271
Nov 28, 2011
Yeah, let us know the outcome

12. ### Harald KappModeratorModerator

11,306
2,588
Nov 17, 2011
I think not. Your solution is not correct. consider this schematic:

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.

#### Attached Files:

• ###### bla.png
File size:
2.3 KB
Views:
221
KrisBlueNZ likes this.