Elementary Linear Circuits - Dependent Sources

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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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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.
 
I also can't find a good place to put my ground. Placing my ground in different places will change my equations.
 
My convention I want to stick to is assume that current flows from positive to negative terminal
 
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NascentOxygen said:
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)
 
NascentOxygen said:
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.
 
Devtycoon said:
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 ...?
 
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.
 
Devtycoon said:
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.
 
Devtycoon said:
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.