/sci/ - Science & Math » Thread #11710628

54KiB, 1220x914, Superposition example 1 of 7.png

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Anonymous Mon 25 May 18:21:36 2020 No.11710628 View Reply Original Report

Quoted By:

Hey guys I'm taking circuit analysis online and prof. isn't responding, can anyone help me understand what he did on these slides?
I understand the concept of superposition and can solve the circuit by simplifying the resistors I just don't know what math he did on slides 2 and 6 to get those matrices and I feel like I'll need to know that to solve problems where I can't just simplify resistors.

Anonymous

View Same Google iqdb SauceNAO Superposition example 2 of 7.png, 53KiB, 1219x910

Anonymous Mon 25 May 2020 18:24:44 No.11710637 Report

Quoted By: >>11710779

It's clear that the values correspond to the reciprocals of the resistors but why did he put them where they are? And why is there a -1/4 in each corner?
This is also fucking me up because he hasn't mentioned anything about how to use kirchhoff's rules on a circuit with a current source so I'm not sure how to treat it.

Anonymous

View Same Google iqdb SauceNAO Superposition example 5 of 7.png, 55KiB, 1217x910

Anonymous Mon 25 May 2020 18:25:47 No.11710646 Report

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Skipping 3 and 4 because they're pretty self explanatory, it just shows the solution for the 12V source by combining resistors

Anonymous

View Same Google iqdb SauceNAO Superposition example 6 of 7.png, 49KiB, 1215x909

Anonymous Mon 25 May 2020 18:28:05 No.11710658 Report

Quoted By: >>11710779

I feel like I'm really close on this one actually, the 4+8+4 clearly corresponds to the top loop and the 4+3 corresponds to loop i3, but I just can't understand their placements in the matrix or why he can designate each loop as having an equivalent current (either i2 or i3) considering there are forks in both that should split the current and there's a 4ohm resistor shared by both loops.

Anonymous

View Same Google iqdb SauceNAO Superposition example 7 of 7.png, 59KiB, 1215x911

Anonymous Mon 25 May 2020 18:41:21 No.11710697 Report

Quoted By:

Oh and here's the last one but it just adds the results from previous slides like you'd expect

Anonymous

Anonymous Mon 25 May 2020 18:44:00 No.11710710 Report

Quoted By: >>11710724

so Calc, some basic e&m, and linear alg, what else do I need to do this kinda stuff op?

Anonymous

Anonymous Mon 25 May 2020 18:46:28 No.11710724 Report

Quoted By:

>>11710710
Syllabus says we'll be using Laplace transforms to analyze circuits later on. There's also a lab portion and the parts were kinda expensive for some shit that only puts out 2mA max. Other than that you're pretty much spot on.

Anonymous

Anonymous Mon 25 May 2020 19:00:40 No.11710779 Report

Quoted By: >>11710849

>>11710637
It's literally just Kirchhoff's first law. From the middle point the current that flows to the left resistor is
$\frac{1}{4 \Omega} v_1$ .
The current that flows to the bottom resistor is
$\frac{1}{3 \Omega} v_1$
The current that comes from the right resistor is
$\frac{1}{4 \Omega} (v_2 - v_1)$
According to Kirchhoff's first law the sum of these is zero.

The second equation in the matrix is Kirchhoff's first law applied to the power source.

>>11710658
This is Kirchhoff's second law applied to both loops.
This is fucking highschool shit.

Anonymous

Anonymous Mon 25 May 2020 19:19:23 No.11710849 Report

Quoted By: >>11710872

>>11710779
This is starting to make sense, just a couple minor questions. In the matrix, is the left column referring to elements in the equation with coefficient v_1 and the right column to coefficients v_2?
For slide 6, how do the current loops work? I understand applying Kirchhoff's second law but why is there an extra -4 in each row? For the left resistor does that mean you do + 4i_2 - 4i_3?
Seeing it written out the way you have it's starting to make sense and I just don't understand the notation being used.
Also you must have been pretty fortunate to go to a high school that taught this. I was in mostly AP classes and didn't get to learn this until Physics 2 in college.

Anonymous

Anonymous Mon 25 May 2020 19:27:43 No.11710872 Report

Quoted By: >>11710896

>>11710849
The current flowing through the left 4Ohm resistor is i_3 - i_2 if viewed from left to right or i_2 - i_3 when viewed from right to left.

Anonymous

Anonymous Mon 25 May 2020 19:34:02 No.11710896 Report

Quoted By:

>>11710872
Is that because of the loops drawn, with the currents being in the direction of the chosen arrows on the loop? Or does it have to do with the currents branching at the nodes?

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