quick check

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
Ok so going back and analyzing this Op Amp and applying what we have talked about:

1)


http://img140.imageshack.us/img140/6998/555ahf.jpg
  • The charge from the current source flows in the red path
  • The voltage of V1 appears across the other two in parallel in blue
  • The current from V1 travels in the orange path and none of it travels down into the resistor in series with V2 because the path that is marked in orange has least resistance
  • The current from V2 travels in the purple path for the same reason stated in the last premise for the current from V1
What is the voltage in green marked V? Surely it can't be V2 because of the resistor that is present.. but if I try to find the voltage at the point, V = IR, I would need to know the current flowing through the resistor... but I don't know the current, so it seems to be sort of a contradiction; so perhaps finding V? would not be a trivial task and I would need to use the node method or something.

2)

Back to the original problem of this thread
http://img154.imageshack.us/img154/5195/57507543.jpg


We said:
  • The current from the current source clashes heads with the current created from the voltage source, so the effective current created by the voltage source is 5/3 + 5/2 - 4... YET the current through the resistors is still 5/3 through the 3 ohm resistor and 5/2 through the 2 ohm resistor.
My question is how is this difference of current made up for? i.e. If I have only (5/3 + 5/2 - 4)A coming out of the voltage source, and none of the current from the current source is contributing to the current running through the resistors (5/3 and 5/2 for the 3 and 2 ohm resistors, respectively).

Does the voltage source simply make up for this current loss due to the current source to sustain the steady current through the resistors? In other words is the actual current coming out of the voltage source
5/3 + 5/2 - 4 + 4?
 
Last edited:

The Electrician

Joined Oct 9, 2007
2,971
Ok so going back and analyzing this Op Amp and applying what we have talked about:

1)


http://img140.imageshack.us/img140/6998/555ahf.jpg
  • The charge from the current source flows in the red path
  • The voltage of V1 appears across the other two in parallel in blue
  • The current from V1 travels in the orange path and none of it travels down into the resistor in series with V2 because the path that is marked in orange has least resistance
  • The current from V2 travels in the purple path for the same reason stated in the last premise for the current from V1
Ok. One misconception you have which you need to get rid of is this:

All of the current does not flow in the path of least resistance.

To say this implies that there is more than one path, one of which presumably has the least resistance. If there are several paths, of different resistances, current will flow in all of the paths; more will flow in the paths of lower resistance and less in the high resistance paths. How much flows in each path will depend on just what the resistances are. That's what circuit analysis is about; figuring out how the current divides among several paths.

I should add, however, that if one of several (parallel) paths is a short circuit, then all of the current will flow in the short; otherwise the current will divide among the various paths. This is a special case.

Now, for the purpose of calculating the output voltage Vo, you can delete the current source I and the resistor R1, because they are in parallel with v1, and they won't change the voltage across V1. It is the voltage of V1 (I and R1 have no effect) that determines the current through R2, and that is what is involved in determining Vo.

What is the voltage in green marked V? Surely it can't be V2 because of the resistor that is present.. but if I try to find the voltage at the point, V = IR, I would need to know the current flowing through the resistor... but I don't know the current, so it seems to be sort of a contradiction; so perhaps finding V? would not be a trivial task and I would need to use the node method or something.
You're right; it isn't V2. You have to take account of both V1 and V2. V1 will contribute something to V and so will V2. You could use the superposition method or you could use the nodal method.

2)

Back to the original problem of this thread
http://img154.imageshack.us/img154/5195/57507543.jpg


We said:

* The current from the current source clashes heads with the current created from the voltage source, so the effective current created by the voltage source is 5/3 + 5/2 - 4... YET the current through the resistors is still 5/3 through the 3 ohm resistor and 5/2 through the 2 ohm resistor.

My question is how is this difference of current made up for? i.e. If I have only (5/3 + 5/2 - 4)A coming out of the voltage source, and none of the current from the current source is contributing to the current running through the resistors (5/3 and 5/2 for the 3 and 2 ohm resistors, respectively).

Does the voltage source simply make up for this current loss due to the current source to sustain the steady current through the resistors? In other words is the actual current coming out of the voltage source
5/3 + 5/2 - 4 + 4?
You may recall (or you could go back and check) that I said something suggesting that there could be more than one point of view about how the currents divide in this circuit. I said that if you assume that all of the 4 amps from the current source pass through the voltage source, you will get the right answer.

It's really a only a conceptual puzzle. When you use the superposition method, and short the 5 volt source, it's the case that the 4 amps from the current source flows through the short (which is in the location of the former 5 volt source), and none through the resistors. Then you remove the current source and calculate the current through the resistors due to the 5 volt source. The final answer sums the contributions from the two sources, and the current in the 5 volt source is 5/3 + 5/2 - 4; this is true regardless of any other assumptions, even though from the first part of the method the 4 amps flowed through the (reduced) voltage source.

The question of whether the current in the resistors all comes from the voltage source, and then the 4 amps from the current source flows "backwards" into the voltage source, or whether 4 amps from the current source flows into the resistors with the difference made up by the voltage source, is moot for the purpose of calculating currents in the resistors.

In the real world, the 4 amps from the current source would flow into the resistors with the deficit made up by the voltage source. If only the 4 amp current source is connected, the voltage across the two resistors would be 4.8 volts, and 4 amps would flow in the two of them. Then connecting the 5 volt source would bring the voltage across the two resistors up to 5 volts. In order to do this, an additional .166667 amps would have to be provided, and this is what the 5 volt source would do.

But to solve for the current in the resistors, the crucial point to understand is that the voltage across the resistors remains 5 volts no matter what the current source does. It's this point that leads one to suppose that all the current in the resistors comes from the 5 volt source, and that's what one part of the superposition method assumes.

Assuming that all of the current in the resistors first comes from the 5 volt source, and that later the 4 amps from the current source all goes into the 5 volt source, reducing its output current, gives the correct solution, even though in the overall circuit behavior those separate parts aren't observable. (The solution I'm referring to here is the question of what the currents in the resistors are.)

It's very typical in analyzing circuits that we break up the problem up into parts that are more easily solvable, but that in the real world, everything is happening at once. The linearity of the problem allows this separation into parts, and gives the correct solution.

The real world division of the currents would be somewhat modified by the actual resistance of the wires; in the academic problem we assume the wires have zero resistance.
 

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
Points well taken - Until Monday I am mainly concerned with ideal circuit analysis and not necessarily what happens in the real world, so I hear you.

So to reaffirm what you mentioned about the special case of zero resistance, in my colored drawing, when the current from V1 reaches the node connecting R2 and R3 (marked with V?), it all goes through the short to the left of R4 because between going down through R3 and the short-distance short with zero resistance straight ahead, it will all go straight ahead.

Same with V2.

I'm currently redoing every single problem concerning the basic circuit anaylsis method and MOSFET problems, then it's on to first and second-order circuits and impedance which are the last few topics to review before the final.
 
Last edited:

The Electrician

Joined Oct 9, 2007
2,971
Well, in this case the "short" at the node marked V, which is also the minus input of the opamp, is a virtual short.

What this means is that the opamp output forces a current through R4 of such magnitude as to make the voltage at V equal to zero volts, which is what the voltage there would be if there were a real short to ground at that point.

So, the orange current does not go down through R3 because of the virtual short at the minus input. It goes through R4 and not directly to ground as it would if there were a real short to ground at the minus input.

But the effect of preventing the orange current from going into R3 is the same as if there were a real short.
 
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