Thevenin Voltage Question Help

[mmmboh]

Hi, this isn't homework, just a practice problem (for which the solution is given)I found online that I am trying to prepare for my midterm next week:

My question is why can't I simple solve for i_s using the left loop. So 5V-500i_s-400(6ma)=0, which gives i_s=5.2ma...alternatively I didn't really have to use KCL, I could have just said that the voltage around the loop must equal zero, so the voltage across the 500 ohm resistor is 5V-2.4V=2.6V, and that's what I would think the thevenin voltage is, but apparently it can't be done that way and the actual thevenin voltage is 1V. Would someone mind explaining this? I think I may be understanding this because I am misunderstanding current sources, can someone clear things up for me please?
Thank you.

 

[Zayer]

Because of the current source you can't apply KVL at that loop, unless you know the voltage across the current source
However,this is simple circuit you can usehttp://en.wikipedia.org/wiki/Source_transformation"

 

[mmmboh]

Would you mind explaining why KVL doesn't work when you have a current source?

and you can't use KVL, but KCL still works?

 

[xcvxcvvc]

Hi, this isn't homework, just a practice problem (for which the solution is given)I found online that I am trying to prepare for my midterm next week:

My question is why can't I simple solve for i_s using the left loop. So 5V-500i_s-400(6ma)=0, which gives i_s=5.2ma...alternatively I didn't really have to use KCL, I could have just said that the voltage around the loop must equal zero, so the voltage across the 500 ohm resistor is 5V-2.4V=2.6V, and that's what I would think the thevenin voltage is, but apparently it can't be done that way and the actual thevenin voltage is 1V. Would someone mind explaining this? I think I may be understanding this because I am misunderstanding current sources, can someone clear things up for me please?
Thank you.

You cannot use KVL conveniently with current sources, because you still need to have a voltage across the current source, which you are arbitrarily defining as zero when it is not.

I'd set up KCL at the top-right node:
$$\frac{V_{th}-5}{500} + .006 + \frac{V_{th}}{500}=0$$

 

[mmmboh]

Do you mean at node a? because that was what was done in the solution, but I don't understand it because the 6ma isn't flowing out of it, well no current is flowing in and out of it, couldn't you just use KCL at the node between the 500 and 400 ohm resistors?

If you pretended currents were coming in and out of node a, wouldn't it be i2, 6ma, and (i2+6ma-is), not is which is (5-Vth)/500?

 

[xcvxcvvc]

Do you mean at node a? because that was what was done in the solution, but I don't understand it because the 6ma isn't flowing out of it, well no current is flowing in and out of it, couldn't you just use KCL at the node between the 500 and 400 ohm resistors?

If you pretended currents were coming in and out of node a, wouldn't it be i2, 6ma, and (i2+6ma-is), not is which is (5-Vth)/500?

Yes, at node a. I detect that you are confused about the scope of a node. The entire conductor in the top right - not just that little circle connected to nothing - is node a. Therefore, a current is flowing from node a to the top-left 500 ohm resistor, a forced current of 6mA is flowing out of node a, and a current is flowing out of node a to the bottom-right 500 ohm resistor.

Remember, that is just the directional convention of nodal analysis - just because I say all currents are flowing "out of" the node, that does not mean all currents are zero. It means some of your solutions may be negative meaning the current is really flowing into the node.

Here is a little more on nodal analysis:
it makes sense to write that all the current entering the node must equal all the current leaving the node. If we define that current leaves the node toward all resistors and that a branch with a current source's current is in the direction of the current source, we can write this for node a:
R1 = top left 500 ohm
R2 = far right 500 ohm
$$I_{in}=I_{out}$$
$$I_{R1}+I_{src}+I_{R2} = 0$$
Using $$I = \frac{V}{R}$$ the above equation becomes:
$$\frac{V_{R1}}{R1}+I_{src}+\frac{V_{R2}}{R2} = 0$$
We know that voltage is defined as being across something. So Vab = Va - Vb. We also know that the node where the current is coming toward a resistor is positive relative to the other node. Therefore, R1's voltage is defined as node a - 5 volts. The other node is defined as the voltage at node a - 0(ground):
$$\frac{V_{th}-5}{R1}+I_{src}+\frac{V_{th}}{R2} = 0$$

 

[mmmboh]

Hey thanks that was a great answer. But I am still a little confused about the node thing, I was always taught a node is a point connecting two or more wires, but even by your definition, I don't see how Is is leaving node a, isn't it actually going to enter node a? and I think I am still confused about this because if node is is the right third of the circuit I don't see how the 6ma is leaving it, unless node a is the right half :S

 

[xcvxcvvc]

Hey thanks that was a great answer. But I am still a little confused about the node thing, I was always taught a node is a point connecting two or more wires, but even by your definition, I don't see how Is is leaving node a, isn't it actually going to enter node a? and I think I am still confused about this because if node is is the right third of the circuit I don't see how the 6ma is leaving it, unless node a is the right half :S

I have circled in a different color every node in that circuit.
http://img682.imageshack.us/img682/5310/nco1fl.jpg

You are basically saying that the flow of charge into the green conductor in the top right must equal the flow of charge out of it(current is the flow of charge per unit time). If you're confused, think of them as water pipes where the flow of water is current. Voltage sources are pumps, wires are pipes that can support A LOT of water flow(let us ignore the situation where a voltage source is shorted, because this analogy doesn't work well there.), and resistors are pipes with smaller diameters that restrain the flow of all the pipes. You can see then that whatever water is flowing into each of those circled pipes must equal the water flowing out of them. Otherwise, water would be created or destroyed.

 

[mmmboh]

Thanks that was a great explanation :)

 

[xcvxcvvc]

Thanks that was a great explanation :)

And to continue the analogy, we assume that the node we are analyzing has a higher "pressure" (voltage) than the nodes on the other side of the resistors. Therefore, the water(current) will flow from high pressure to low pressure - it will flow to the resistor. We choose that convention to use every time to create a consistent environment not prone to error. So since we assume that node has higher voltage, we say (higher voltage - lower voltage)/resistance = current flowing out of node. Note: when you solve for this current, it may be negative! All this means is that the node's voltage is actually lower and that the current is flowing into the node.