Finding Vy in a Circuit with Dependent Sources

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Discussion Overview

The discussion revolves around determining the voltage Vy in a circuit that includes dependent sources. Participants share their attempts at solving the problem, detailing their calculations and reasoning. The context is primarily homework-related, focusing on circuit analysis techniques.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates Vx as 6V using Ohm's law and then computes the current supplier as 0.6 A.
  • The same participant adds the resistance of parallel resistors to find it as 1.667 ohms and determines a voltage drop of 2V across it.
  • The participant estimates the current for the 2 ohm resistor as 1 A and combines it with the previous current to apply Ohm's law, resulting in a final value of 3.2.
  • Another participant expresses difficulty in opening the provided PDF file and requests a re-upload.
  • A third participant shares an image of the circuit in PNG format.
  • A fourth participant reposts the image in JPEG format, indicating a concern about the previous upload.
  • One participant suggests that once the current is known, the potential drop across the 2 Ω resistor should be straightforward using Ohm's law, implying that intermediate steps may be unnecessary.

Areas of Agreement / Disagreement

There is no consensus on the approach to solving the problem, as participants express different levels of detail in their calculations and reasoning. Some participants suggest that fewer steps may be needed, while others provide detailed calculations.

Contextual Notes

Participants are working with dependent sources, which may introduce complexities in their calculations. The discussion includes various representations of the circuit, but the original figure is not accessible to all participants.

Who May Find This Useful

Students or individuals interested in circuit analysis, particularly those learning about dependent sources and Ohm's law, may find this discussion relevant.

maximade
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Homework Statement


Determine Vy in the circuit of the below figure:
(In attachment)


Homework Equations


ohms law
resistance adding equations for parallel and series


The Attempt at a Solution


So I found Vx=(1.2A)(5ohms) = 6v
Afterwards I computed the current supplier: (6)(0.1)=0.6 A
Then I added the resistance of the parallel resistors and got 1.667 ohms
Then found the voltage drop across it using the 1.2 A and get 2 v.
Seeing how the parallel resistors share the same voltage, I computed the supposed current for each (without the dependent current supply) and get 1 A for the 2 ohm resistor.
Then I added the 1 A to the 0.6 A from before, and then applied ohms law (1.6 A)(2 ohms) to get 3.2.

This is my first time doing a problem regarding dependent sources, please let me know if I am missing any steps or using too many steps.

Thanks.
 

Attachments

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maximade said:

Homework Statement


Determine Vy in the circuit of the below figure:
(In attachment)


Homework Equations


ohms law
resistance adding equations for parallel and series


The Attempt at a Solution


So I found Vx=(1.2A)(5ohms) = 6v
Afterwards I computed the current supplier: (6)(0.1)=0.6 A
Then I added the resistance of the parallel resistors and got 1.667 ohms
Then found the voltage drop across it using the 1.2 A and get 2 v.
Seeing how the parallel resistors share the same voltage, I computed the supposed current for each (without the dependent current supply) and get 1 A for the 2 ohm resistor.
Then I added the 1 A to the 0.6 A from before, and then applied ohms law (1.6 A)(2 ohms) to get 3.2.

This is my first time doing a problem regarding dependent sources, please let me know if I am missing any steps or using too many steps.

Thanks.

For some reason I'm not able to open the PDF file -- it says it is damaged or has other problems. Can you try uploading it again?
 
I made it on paint, here's a PNG file.
 
Sorry for the repost, but i don't think ti got though on my last reply. Here's the JPEG*
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    11.6 KB · Views: 481
One you know the current I then the potential drop across the 2 Ω resistor should be trivial by Ohm's law. You don't need any intermediate steps with parallel resistance calculations or anything else.
 

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