# Circuits dependent sources

• Engineering

## 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.

## The Attempt at a Solution

#### Attachments

• Untitled.pdf
13.8 KB · Views: 485

berkeman
Mentor

## 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.

## The Attempt at a Solution

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
11.6 KB · Views: 341
gneill
Mentor
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.