Algebra help, Kirchoffs rule problem.

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The discussion revolves around solving for current (I3) using Kirchhoff's rule in an algebraic equation. The original equation presented is 4E/R - 9I3/2 = I3, but confusion arises regarding the manipulation of terms, particularly how to transition from -9I3/2 to 2I3/2. After some back and forth, the user realizes their mistake in multiplying the entire equation by 2, leading to the correct form of the equation. Ultimately, the user successfully resolves their confusion and acknowledges their error. The thread highlights the importance of careful algebraic manipulation in circuit analysis problems.
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Hey guys, I'm having a little trouble with some basic algebra. I'm working on solving for currents using kirchhoffs 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.
 
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Look at your first equation again. You did two things in that equation. There is your error.
 
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.
 
Last edited:
How did you go from - 9I3/2 to -9I3?
 
I multiplied both sides of the equation by 2.
 
Did you multiply each term by 2?
 
yes..

4E/R - 9I3/2 = I3

x 2

8E/2R - 9I3 = 2I3
 
Nevermind..
I"m stupid.. I figured it out.

Thanks man.
 
4*2=8. right?
 
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no problem friend
 

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