Circuit problem with two sources and three resistors (Nodal Analysis)

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Homework Help Overview

The discussion revolves around a circuit problem involving two sources and three resistors, specifically focusing on nodal analysis and mesh analysis techniques. Participants are exploring the relationships between current and voltage in the context of controlled sources.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of using nodal analysis versus mesh analysis, with some expressing confusion over the interpretation of controlled voltage sources. Questions arise regarding the correctness of specific current and voltage values derived from the circuit analysis.

Discussion Status

The conversation is ongoing, with various interpretations being explored. Some participants are questioning the accuracy of the provided answer of 0.123A, while others are sharing their own calculations and experiences with the problem. There is an acknowledgment of differing results among participants, and guidance is offered regarding the clarity of circuit diagrams.

Contextual Notes

Participants note that the problem may not align with standard coursework, and there is a mention of potential errors in the original calculations. The discussion includes references to the conventions used in circuit diagrams and the implications of those conventions on understanding the problem.

wcjy
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Homework Statement
Determine the current I (in Amperes) in the circuit below using nodal analysis.
Relevant Equations
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Ans: 0.123A
Tried mesh analysis and got it but didnt for nodal analysis
 
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40I-V0? Subtracting a voltage from a current?
 
haruspex said:
40I-V0? Subtracting a voltage from a current?
40I is from the current controlled voltage source. Isn't 40I in volts?
 
haruspex said:
40I-V0? Subtracting a voltage from a current?
The rotated square (diamond) symbol with '+ -' is a controlled voltage source.
https://1.bp.blogspot.com/-VWAqITUrvhA/UY-k_NBxwfI/AAAAAAAAAUM/BPVlWJIFgXQ/s1600/sources.png

You can tell it's a voltage source because the '+ -' indicates polarity (as opposed to an arrow for a current source indicating current's direction).

'40I' indicates the output-voltage is numerically (in volts) 40 times the value of the current labelled 'I' (in amps) on the diagram.
 
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wcjy said:
View attachment 278629

Ans: 0.123A
Tried mesh analysis and got it but didnt for nodal analysis
Can't see a problem. I also get I = 0.112(335)A and (for information) = 5.28(634) V. If you plug these values into check, you find that they give correctly balanced currents. Are you certain the answer should be 0.123A?

You could post your mesh analysis for checking. But purple on a black background is pretty eye-unfriendly and probably puts a lot of people off. (Typed-up using Latex is the preferred method.)
 
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Steve4Physics said:
Can't see a problem. I also get I = 0.112(335)A and (for information) = 5.28(634) V. If you plug these values into check, you find that they give correctly balanced currents. Are you certain the answer should be 0.123A?

You could post your mesh analysis for checking. But purple on a black background is pretty eye-unfriendly and probably puts a lot of people off. (Typed-up using Latex is the preferred method.)
the system said the correct answer was 0.123. But anyways thanks, many people got 0.112 also. probably the answer is wrong. Gonna check with my professor.

My mesh analysis was wrong. I initially got 0.112 as well but i somehow tweak it to 0.123 thinking it was correct. Looking back, the change that i did, did not make any sense.

Anyways thanks so much for helping!
 
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Steve4Physics said:
The rotated square (diamond) symbol with '+ -' is a controlled voltage source.
https://1.bp.blogspot.com/-VWAqITUrvhA/UY-k_NBxwfI/AAAAAAAAAUM/BPVlWJIFgXQ/s1600/sources.png

You can tell it's a voltage source because the '+ -' indicates polarity (as opposed to an arrow for a current source indicating current's direction).

'40I' indicates the output-voltage is numerically (in volts) 40 times the value of the current labelled 'I' (in amps) on the diagram.
Ok, I'd never come across that concept.
But the correct way to specify its value would be with units, namely, as 40IΩ.
 
haruspex said:
40I-V0? Subtracting a voltage from a current?
Steve4Physics said:
You can tell it's a voltage source because the '+ -' indicates polarity (as opposed to an arrow for a current source indicating current's direction).
Yes. Also, it's in series with a resistor, which doesn't make a lot of sense for a current source.
 
It's not nodal analysis, I prefer the source transformation (Thevenin) approach, but this is the answer. You get the erroneous solution by reversing the polarity of the dependent source [ (1+r/R2) ⇒ (1-r/R2) ].
20210225_130836.jpg
 
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haruspex said:
Ok, I'd never come across that concept.
But the correct way to specify its value would be with units, namely, as 40IΩ.
True, but by convention the units are usually omitted from the diagram and are assumed to be appropriate for the "conversion" implied by the sensed value and resulting output of the controlled source.

If units were specified on this diagram they would be more likely to take the form of Volts/Amp in order to indicate that so many volts are produced for a given current in amps, rather than being "reduced" from V/A to Ohms.

A similar convention occurs for the ##\beta ## value for transistors, where a simple apparently unitless value is given such as 200. In reality it represents the ratio of collector current to base current, the current amplification factor for the transistor, and so really should have units of Amps per Amp.
 
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  • #11
This is not something covered on the courses I teach but I was intrigued so decided to try to find out how to do such problems and have a go.
Perhaps not surprisingly, I can't get 0.123A as an answer. However, I'm wary of suggesting that's because the answer is wrong until I know for sure.

Please, @wcjy - it would be great if you could confirm one way or another when you ask your prof. Thanks :)
 
  • #12
rsk said:
This is not something covered on the courses I teach but I was intrigued so decided to try to find out how to do such problems and have a go.
Perhaps not surprisingly, I can't get 0.123A as an answer. However, I'm wary of suggesting that's because the answer is wrong until I know for sure.

Please, @wcjy - it would be great if you could confirm one way or another when you ask your prof. Thanks :)
yea the answer is wrong!
 
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