PSC, current and Node Voltage Analysis

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SUMMARY

The discussion centers on the application of node voltage analysis in linear circuits, specifically regarding the relationship between current ib and voltage v2. The user initially believed that ib should equal -v2 due to the passive sign convention but found that this led to complex fractional results. After consulting peers, they were advised that ib equals v2, which produces simpler integer results. The resistor connected to v2 is confirmed to be 1Ω, reinforcing the conclusion that ib = -v2 is incorrect in this context.

PREREQUISITES
  • Understanding of node voltage analysis in electrical circuits
  • Familiarity with the passive sign convention
  • Basic knowledge of Ohm's Law
  • Experience with linear circuit components and their configurations
NEXT STEPS
  • Study the principles of node voltage analysis in greater depth
  • Review the passive sign convention and its applications in circuit analysis
  • Practice solving linear circuits using different resistor configurations
  • Explore common pitfalls in circuit analysis to avoid mathematical errors
USEFUL FOR

Electrical engineering students, circuit design professionals, and anyone looking to enhance their understanding of node voltage analysis and circuit behavior.

Weird Fishes
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First off, I apologize if this belongs under homework help. It’s kind of a grey area I suppose...

Anyway, I just finished my linear circuits midterm. One of the questions on the midterm had a circuit we needed to solve using node analysis. That’s not the question, though. I’ve crudely redrawn the circuit:
Circuit.png

I forgot to draw on the common node, but it’s the bottom one.

Now I thought, due to the passive sign convention, that ib would be equal to -v2, since ib is in the direction of a voltage rise (v=-bR == v2=-ib). The thing is though, ib=-v2 gives really ugly fractional results and ib=v2 gives really nice integer results. Also, the one person I’ve talked to so far said that ib=v2.

I may have gotten one or two of the resistors wrong redrawing it, but I know for a fact that the v2 to ground resistor was 1Ω.

So anyway, what is correct? (and if you could point out where I went wrong, that’d be great too =P)Usually the only things I get wrong are mathematical errors, but when it’s something fundamental like this I like to take a step back and make sure I have my concepts right. Thanks in advance.
 
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Weird Fishes said:
First off, I apologize if this belongs under homework help. It’s kind of a grey area I suppose...

Anyway, I just finished my linear circuits midterm. One of the questions on the midterm had a circuit we needed to solve using node analysis. That’s not the question, though. I’ve crudely redrawn the circuit:
View attachment 21226
I forgot to draw on the common node, but it’s the bottom one.

Now I thought, due to the passive sign convention, that ib would be equal to -v2, since ib is in the direction of a voltage rise (v=-bR == v2=-ib). The thing is though, ib=-v2 gives really ugly fractional results and ib=v2 gives really nice integer results. Also, the one person I’ve talked to so far said that ib=v2.

I may have gotten one or two of the resistors wrong redrawing it, but I know for a fact that the v2 to ground resistor was 1Ω.

So anyway, what is correct? (and if you could point out where I went wrong, that’d be great too =P)Usually the only things I get wrong are mathematical errors, but when it’s something fundamental like this I like to take a step back and make sure I have my concepts right. Thanks in advance.

As drawn, -iB * 1 Ohm = V2.
 
Yeah, that’s what I thought. (obviously)

I guess we’ll see what happens when I get my midterm back...
 

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