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Circuit problem - need a sanity check

  • Engineering
  • Thread starter DrummingAtom
  • Start date
  • #1
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Homework Statement


Find Vx. With R1, R2 and i3 given.

*In my attachment, I renamed the given values.*

Homework Equations



KCL, KVL, Ohm's law

The Attempt at a Solution



For KVL, I solved for V3 in L2 then subbed it in L1 to give:

3Vx - Vx + V1 = 0

From here, I tried to get my KCL equations to have i1 in terms of i3 so I could replace V1 with R1 and i1. But the KCL won't let me do that.. in fact, I can't get any of those KCL equations in different terms they just cancel each other out in every case.

I'm lost if this is an error in the problem or I'm doing something wrong. Thanks for any help.
 

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Answers and Replies

  • #2
gneill
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I think you're making your life more difficult than need be; the two resistors are in series and thus form a single branch and have a single current. So you really only need one node (where R2 and the two current supplies join) in order to write your KCL equation.

The two current supplies feed into that node, and one path leads out (through R2). Call the latter current i2. What's Vx in terms of this current i2?
 
  • #3
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I think you're making your life more difficult than need be; the two resistors are in series and thus form a single branch and have a single current. So you really only need one node (where R2 and the two current supplies join) in order to write your KCL equation.

The two current supplies feed into that node, and one path leads out (through R2). Call the latter current i2. What's Vx in terms of this current i2?
Vx = i2R2 or = 5i2 with the given value.
 
  • #4
gneill
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Vx = i2R2 or = 5i2 with the given value.
Okay, that makes one equation. What's the KCL equation for the node?
 
  • #5
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Node = i3+i4-i2 = 0

Then plugging that into Vx = 5(i3 + i4).
 
Last edited:
  • #6
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The main thing I'm still confused about was my original KCL equations. They showed no solutions, does that mean it's in series?
 
  • #7
gneill
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Node = i3+i4-i2 = 0

Then plugging that into Vxi2 = 5(i3 + i4).
Careful, you've made an assumption about the direction of current i2 when you wrote the expression for Vx. Better make sure that you remain consistent!
 
  • #8
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Hmm, I'm still confused about when I can make assumptions on the direction.

So, it would be:

Vx = 5(-i3 - i4)
 
  • #9
gneill
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The main thing I'm still confused about was my original KCL equations. They showed no solutions, does that mean it's in series?
A few things spring to mind when I look at your original KCL equations. First, note that the two resistors are in series and thus form a single branch. This would eliminate your node "A" from consideration. Second, you don't really need to write the equation "C", since if you select it as the choice for the reference node then its equation will be redundant, duplicating information in the others. Third, you've introduced two currents (i1 and i2) in a single branch; that's bound to lead to either redundancy or unsolvable equations.

To write KCL equations to solve a circuit, start by selecting a reference node (or "ground" node as it's sometimes called). Then identify the other independent nodes (In this circuit there's only one). Write the KCL equation for each independent node. If there's more than one independent node you'll want to assign variables for node voltages and use them along with KVL on the branches between nodes to replace the unknown currents with voltage/resistance expressions. Then solve for the node voltages.
 
  • #10
gneill
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Hmm, I'm still confused about when I can make assumptions on the direction.

So, it would be:

Vx = 5(-i3 - i4)
When you wrote the equation Vx = i2*R2 you tacitly made the assumption that i2 is flowing from left to right through R2 and thus INTO your node "B". Thus all the currents are taken as flowing into node B, and the KCL for that node is:

i2 + i3 + i4 = 0

Now you can replace i2 and i4 with expression involving Vx (which is, after all, the variable you're hoping to solve for!). You've got your Vx = i2*R2 (so solve for i2) and your i4 = 3*Vx.
 
  • #11
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Ok, I was going that route but wouldn't it be 3Vx = i4R4? That was the only reason I chose not to do that because I thought I would be bringing in another unknown variable..
 
  • #12
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Thanks for your help by the way, I really appreciate it.
 
  • #13
gneill
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Ok, I was going that route but wouldn't it be 3Vx = i4R4? That was the only reason I chose not to do that because I thought I would be bringing in another unknown variable..
I only see resistors R1 and R2 in that circuit. i4 = 3Vx.
 

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