PDA

View Full Version : Algebra help!, Kirchoffs rule problem.


jKotha
Feb27-07, 06:51 PM
Hey guys, I'm having a little trouble with some basic algebra. I'm working on solving for currents using kirchoffs rule.

4E/R - 9I3/2 = I3

Trying to solve for I3... I should get:
8E/2R -9I3 = 2(I3).. and then 8E/2R = 11(I3).. and then I3= 4E/11R

However I am looking at my teachers solution and her answer was:

4E/R - 9I3/2 = I3
4E/R = 2I3/2 + 9I3/2 = 11I3/2 . How did she get the 2I3/2? It doesn't make sense to me. Once again, I am trying to solve for I3.

Please help, I really appreciate it. Thanks.

robb_
Feb27-07, 06:54 PM
Look at your first equation again. You did two things in that equation. There is your error.

jKotha
Feb27-07, 06:59 PM
I don't understand..

How did I do 2 things?
If i'm left with 8E/2R = 11(I3), then finding I3 would be dividing both sides by 11, which would not give me the correct answer.

Also, that doesn't explain where she got the 2I3/2 on the right side. Thats what I'm concerned about.

Still stuck, thanks for the help, but I don't quite understand it.

robb_
Feb27-07, 07:00 PM
How did you go from - 9I3/2 to -9I3?

jKotha
Feb27-07, 07:02 PM
I multiplied both sides of the equation by 2.

robb_
Feb27-07, 07:04 PM
Did you multiply each term by 2?

jKotha
Feb27-07, 07:11 PM
yes..

4E/R - 9I3/2 = I3

x 2

8E/2R - 9I3 = 2I3

jKotha
Feb27-07, 07:14 PM
Nevermind..
I"m stupid.. I figured it out.

Thanks man.

robb_
Feb27-07, 07:14 PM
4*2=8. right?

robb_
Feb27-07, 07:14 PM
no problem friend