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I have a question on the Kirchoff law problem, section 7.3 problem 18. On the loop portion of the defining equations, relative to a clockwise direction, I find the equation of the right loop to be -12 I2 + 8 I3 = -24. Is this correct to assume the I2 term is negative due to the counterclockwise flow of I2, as with the voltage term?
 
I have a question on the Kirchoff law problem, section 7.3 problem 18. On the loop portion of the defining equations, relative to a clockwise direction, I find the equation of the right loop to be -12 I2 + 8 I3 = -24. Is this correct to assume the I2 term is negative due to the counterclockwise flow of I2, as with the voltage term?
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That equation is the same thing I got, I just took a different direction for my KVL around the loop (e.g. <math>12 I_2 - 8 I_3 = 24</math>). However, I don't fully understand what your question is. --[[User:Rrusson|Rrusson]] 18:48, 25 August 2013 (UTC)
  
 
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Revision as of 14:48, 25 August 2013


Homework 1 collaboration area

Feel free to toss around ideas here. Feel free to form teams to toss around ideas. Feel free to create your own workspace for your own team. --Steve Bell

Here is my favorite formula:

$ f(a)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-a}\ dz. $

This is a test formula:

$ A \vec x= \vec b $ - Eun Young

I have a question on the Kirchoff law problem, section 7.3 problem 18. On the loop portion of the defining equations, relative to a clockwise direction, I find the equation of the right loop to be -12 I2 + 8 I3 = -24. Is this correct to assume the I2 term is negative due to the counterclockwise flow of I2, as with the voltage term?

That equation is the same thing I got, I just took a different direction for my KVL around the loop (e.g. $ 12 I_2 - 8 I_3 = 24 $). However, I don't fully understand what your question is. --Rrusson 18:48, 25 August 2013 (UTC)


Back to MA527, Fall 2013

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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