Elementary Linear Circuits - Dependent Sources

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  • Thread starter Devtycoon
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Homework Statement



the problem is a circuit analysis. I can try to make the circuit but the image i created shows all the resistors and voltage sources.

DE_1_10.png

imgurl

Find i,v,is,vs

Homework Equations



[itex]\sum[/itex]loop vi=0
[itex]\sum[/itex]node ij=0

The Attempt at a Solution



I can not make a voltage loop as of yet. That would be my first attempt. Each node that I make an equation for results in more variables than what I have equations for. I am stuck on a very fundamental level here I think.
 
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Answers and Replies

  • #2
NascentOxygen
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Hi Devtycoon. Welcome to the Physics Forums.

You don't have the right URL to display your pic. Let me fix that .....

DE_1_10.png
 
  • #3
NascentOxygen
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What are some of the equations you've been writing down?
 
  • #4
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KCL for each node. I know there is a amperage going into/out of the 3 ohm resistor that I can not account for. I uploaded some work I have done with better images this time.
 
  • #5
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I also can't find a good place to put my ground. Placing my ground in different places will change my equations.
 
  • #6
NascentOxygen
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We don't need a whole panoply of images! One would be ample, with relevant notations added.

First step: how do you decide the direction to mark the current in the 4 Ω resistor?
 
  • #7
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My convention I want to stick to is assume that current flows from positive to negative terminal
 
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  • #8
NascentOxygen
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My convention I want to stick to is assume that current flows from positive to negative terminal
Well, you won't get far if you don't assume that! :smile:

So, in what direction will the current be through the 4Ω resistor?
 
  • #9
NascentOxygen
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I also can't find a good place to put my ground. Placing my ground in different places will change my equations.
You don't need a ground, but you can consider the lowest horizontal wire as ground if you wish.
 
  • #10
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So, in what direction will the current be through the 4Ω resistor?

I will have the current going from the top of the 4Ω resistor (positive terminal) down through the bottom (negative terminal)
 
  • #11
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You don't need a ground, but you can consider the lowest horizontal wire as ground if you wish.

So the positive terminal of the battery is connecting to the ground? Sorry, I am new to circuits.
 
  • #12
NascentOxygen
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So the positive terminal of the battery is connecting to the ground? Sorry, I am new to circuits.
Sure! It doesn't care where you call ground.

You don't need ground, so maybe leave it out if it's going to confuse you.

Now, some of these equations .....?
 
  • #13
NascentOxygen
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I will have the current going from the top of the 4Ω resistor (positive terminal) down through the bottom (negative terminal)
Correct. I asked because one of the pics you posted earlier seemed to show current going the other way.
 
  • #14
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Ya I messed that up. I will begin at the node with 2i going in and i going out. It seems that I overlooked this situation. If 2 i is going in and i is going out then the 3 ohm resistor must have 1i going into that to have a KCL law satisfied.
 
  • #15
NascentOxygen
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Ya I messed that up. I will begin at the node with 2i going in and i going out. It seems that I overlooked this situation. If 2 i is going in and i is going out then the 3 ohm resistor must have 1i going into that to have a KCL law satisfied.
Right.
 
  • #16
NascentOxygen
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Ya I messed that up. I will begin at the node with 2i going in and i going out. It seems that I overlooked this situation. If 2 i is going in and i is going out then the 3 ohm resistor must have 1i going into that to have a KCL law satisfied.
Right.
 

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