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Electric Circuits - Find the Equivalent Resistance (Dependent Source within circuit)

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Homework Statement



[PLAIN]http://img856.imageshack.us/img856/8508/circuit.png [Broken]

Homework Equations



Loop Analysis and Node Analysis

The Attempt at a Solution



The Question states I have to find the equivalent resistance within the circuit. But, since there is a dependent source withing the circuit, I can't suppress and instead need to add a test voltage within the circuit ( I did and it's on the left), which will produce a test current, i. What I'm having trouble with here is what approach should I use to solve this?
 
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Answers and Replies

  • #2
gneill
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I think that the choice of method is a toss-up in terms of effort required. Mesh KVL will yield 3 or 4 equations to solve (you can remove the 30 Ohm resistor at the input and then put it in parallel with the result at the end, leaving you with three loops to deal with rather than four), or solve for the node voltages in terms of V and then find the required branch currents.

For the nodal method one node is constrained by the test voltage V, and another by the controlled source connected to that node. So just one node equation is required up front (Vx) along with the ancillary constraints on the other nodes, but then you've got to go back and fill in the branch currents using the node voltages in order to find the current that the test voltage is pushing into the first node.

I guess the thing to do is pick a path and start slogging.
 
  • #3
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I think that the choice of method is a toss-up in terms of effort required. Mesh KVL will yield 3 or 4 equations to solve (you can remove the 30 Ohm resistor at the input and then put it in parallel with the result at the end, leaving you with three loops to deal with rather than four), or solve for the node voltages in terms of V and then find the required branch currents.

For the nodal method one node is constrained by the test voltage V, and another by the controlled source connected to that node. So just one node equation is required up front (Vx) along with the ancillary constraints on the other nodes, but then you've got to go back and fill in the branch currents using the node voltages in order to find the current that the test voltage is pushing into the first node.

I guess the thing to do is pick a path and start slogging.
So if I chose, Loop analysis, I would have 4 meshes, and just apply KCL to each loop. I'll have 4 equations with 6 unknowns then wouldn't I? I would have Vx, V, I, I for top loop, I for bottom right loop, and I for loop to the left of the bottom right loop. With the bottom left loop containing V and I. Unless I'm understanding you wrong here, which loops would you use here to write KCL?
 
  • #4
gneill
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As I said, you can remove the initial 30 Ohm resistor leaving 3 loops to analyze, but remember to put it back (in parallel) later.

You will have one main equation for each loop. Use the mesh currents flowing through the "Vx resistor" to replace Vx in your top loop.
 
  • #5
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As I said, you can remove the initial 30 Ohm resistor leaving 3 loops to analyze, but remember to put it back (in parallel) later.

You will have one main equation for each loop. Use the mesh currents flowing through the "Vx resistor" to replace Vx in your top loop.
Yeah I know you said that but it's a little confusing to me so I'd rather leave it in there without causing any unnecessary trouble for me later on if I forget something.
 
  • #6
gneill
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Yeah I know you said that but it's a little confusing to me so I'd rather leave it in there without causing any unnecessary trouble for me later on if I forget something.
That's fine. It just gives you one more equation to handle.
 
  • #7
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Would these 4 equations be correct?

KCL Top Mesh:

30(i1 - i2) + 15(i1 - i3) + 2Vx = 0

KCL Bottom Left Mesh:

30i - V = 0

KCL Bottom Middle Mesh:

15(i3 - i1) + 15(i3 - i2) + 30(i3 - i) = 0

KCL Bottom Right Mesh:

15i2 + 15(i2 - i3) + 30(i2 - i1) = 0

Where the currents are: i1 = Top Mesh going clockwise; i2 = Bottom Right Mesh going clockwise; i3 = Middle Bottom Mesh going clockwise; and i = Bottom Left Mesh going clockwise. You said I have to replace Vx in the Top Mesh with the current that flows the resistor Vx is assigned to in the picture. I understand that much, but when writing the equations, they differed from each of the 2 Bottom loops that pass through that resistor. Should I replace Vx with 15(i2 - i3) or 15(i3 - i2)?
 
  • #8
gneill
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Would these 4 equations be correct?

KCL Top Mesh:

30(i1 - i2) + 15(i1 - i3) + 2Vx = 0

KCL Bottom Left Mesh:

30i - V = 0
i3 also flows through that 30 Ohm resistor.
KCL Bottom Middle Mesh:

15(i3 - i1) + 15(i3 - i2) + 30(i3 - i) = 0

KCL Bottom Right Mesh:

15i2 + 15(i2 - i3) + 30(i2 - i1) = 0

Where the currents are: i1 = Top Mesh going clockwise; i2 = Bottom Right Mesh going clockwise; i3 = Middle Bottom Mesh going clockwise; and i = Bottom Left Mesh going clockwise. You said I have to replace Vx in the Top Mesh with the current that flows the resistor Vx is assigned to in the picture. I understand that much, but when writing the equations, they differed from each of the 2 Bottom loops that pass through that resistor. Should I replace Vx with 15(i2 - i3) or 15(i3 - i2)?
You've chosen clockwise currents, so a voltage drop in the indicated direction across the "Vx resistor" requires 15(i3 - i2).
 
  • #9
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i3 also flows through that 30 Ohm resistor.


You've chosen clockwise currents, so a voltage drop in the indicated direction across the "Vx resistor" requires 15(i3 - i2).
Ok, the overall purpose here is to get the test current, i, in terms of the test voltage, V, so that I'll end up with a ratio for the equivalent Resistance in the circuit right?
 
  • #10
gneill
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Ok, the overall purpose here is to get the test current, i, in terms of the test voltage, V, so that I'll end up with a ratio for the equivalent Resistance in the circuit right?
That was your stated plan... and it's a good one.
 
  • #11
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I'm almost completely lost at the moment after writing those equations down I can't seem to solve any at all. And I get a current for i1 = 0 Amps. Is that even possible?

Is node analysis the easier approach here rather than working with 4 equations in loop analysis?
 
  • #12
gneill
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I'm almost completely lost at the moment after writing those equations down I can't seem to solve any at all. And I get a current for i1 = 0 Amps. Is that even possible?

Is node analysis the easier approach here rather than working with 4 equations in loop analysis?
Yes, i1 = 0 is certainly possible.

Both methods have their strong points and weak points. You can get a single equation for the node voltage Vx pretty easily, and since the others are constrained nodes, it's just substitute and solve. But after you've got the node voltages (in terms of the test voltage, of course), you have to go back and, using them, find the required branch currents so that you can use KCL on the input node to find the input current.

Mesh analysis allows you to solve for the input current directly, but you start off with four (or three) equations to wrestle. A matrix method (Cramer's rule perhaps) would make it manageable.
 
  • #13
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Yes, i1 = 0 is certainly possible.

Both methods have their strong points and weak points. You can get a single equation for the node voltage Vx pretty easily, and since the others are constrained nodes, it's just substitute and solve. But after you've got the node voltages (in terms of the test voltage, of course), you have to go back and, using them, find the required branch currents so that you can use KCL on the input node to find the input current.

Mesh analysis allows you to solve for the input current directly, but you start off with four (or three) equations to wrestle. A matrix method (Cramer's rule perhaps) would make it manageable.
My equations are:

30i - 30i3 = V

45i1 - 60i2 + 15i3 = 0

-30i1 + 60i2 - 15i3 = 0

-15i1 - 15i2 + 60i3 - 30i = 0

I can add equations 2 and 3 right away to get 15i1 = 0, so i1 = 0. But from then on I'm stuck, no matter what I try I can't seem to solve for another current.
 
  • #14
gneill
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Your equations look fine, so it's just a matter of algebra at this point. Everyone has their own way of solving such systems. I tend to use a matrix method (Cramer's rule) when there are more than three equations in three unknowns.

You could pick one equation and solve for one variable (which is not the one you ultimately want to find) in terms of the other variables. Then substitute that expression for all the occurrences of that variable in the other expressions. Discard the expression you picked and move on to the others. Repeat the process until you're left with the final equation in the variable you want to solve for.
 

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