Yeah, pretty easy stuff, but I’m still not understanding of what to do when a depended voltage source’s value is given as a multiple of a current. Without trying to describe what the circuit looks like (you’ll probably figure out pretty easily based on eq’s), here are my loop equations
Ix = I2
(2k)(I1) + 4k( I1 - I2) -2000Ix = 0
-12V +2kI2 + 4k (I2 - I1) = 0
so from there 12V = 6kI2 + 4kI1. I figured trying to plug in Ix into all I1’s and solving from there, but you still have that 2000Ix in terms of a current instead of a voltage, so I don’t see any point in trying to go further until I have a clue what to do with it.
Bah, depended = dependent. It’s a small circuit, so I guess I’ll just try to describe:
3 branches, far left has the dependent Vs = 2000Ix
middle has 4k resistor
right has 2k resistor
top left 2k resistor
top right is 12V voltage source (- 12 + ), Ix is just to the right of it
nothing on any bottom branches
I assume that by 2000Ix you mean a voltage source with the + end up that is 2000*Ix. A 2 loop circuit with the left hand loop having the current-dependent source and a 2k and a 4k in series in the loop. The 4 k is common to the two loops. The right hand loop has a 4k (common to the other loop) the 12 V source and a 2k in series in the loop.
What you do is sum up all the voltages around loops 1 and 2.
Loop 1
2000Ix - 2000I1 - 4000*(I1 - Ix) = 0
for loop 2
4000*(Ix - I1) + 12 + 2000*Ix = 0
The equations assume both currents are circulating clockwise around their respective loops. I1 is current in left hand loop, Ix is current in right hand loop.
That’s what basically what I got loop wise, just replacing I2 with Ix. This is the step I was going to take, but stopped because I have the problem with solving said equations. Simplifying I2 (or Ix) I get
6kIx - 4kI1 = 12V
for the first equation I still have
-2000Ix + 2kI1 +4kI1 -4kI2 = 0
I figure I can find I1 by taking the equation I got for I2, but then it becomes a huge clusterfuck with Ix = 12V/6k + 4kI1/6k … bah, just as I was typing this, I realised that what I thought was a cluster fuck ends up being a current + another current. Something just didn’t look right about it, but that 4k/6k just becomes a ratio for I1. Thanks, a bunch, though.
You have 3 unknowns, I1, I2 and Ix in the second equation. Ix is equal to I2 isn’t it? Your second post said that Ix is the output of the 12 V source and that’s I2.
Damnit, I went through with what I thought was right, but I’m getting I1 = 0 when working it out. I guess I am still confused. I’m getting the same loop equations as you have, so I’m OK with that, I guess I just don’t know how to solve them. Also, the 2000Ix is the dependent voltage source, so aside from not know how to solve it, once you do solve for Ix do you just multiply it by 2000 and set that to the value of the dependent voltage source?
OK, I’m seeing the problem I was having may have been a typo on the paper. The thing you’re doing that I’m still not sure is kosher is taking the 6000 before the I1, which is resistance in ohms, and the -2000 before the Ix of the DVS and comparing them. That’s all it has in the example that I’ve been doing. However, on a nearly identical problem on a practice test, the 2000Ix is given as 2000 (V/A), which would solve my major concern of the relationship between the DVS and Ix.
All of the terms in all of the equations are volts.
Your current controlled voltage source units are volts. So the product of 2000*Ix is volts. Since Ix is current, the 2000 must have the units of resistance in order for the product to be volts. And so it can be added to any resistance since it is also a resistance.