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Fundamental question...

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throbscottle

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I learnt this kind of thing in college, but it was so many years ago I've forgotten, and need someone to check my reasoning please! This is relevant to op-amp bias current and noise, but I'm using knife-and-fork values to check I've got the theory right.

Suppose I have two resistors in series, 9Ω and 1Ω, call the connection node A. Connect 10v across the ends of the two resistors, the 1Ω end being 0v, so 1 amp flows, voltage at A is 1v. Connect a constant current to node A. It can either source or sink, but say source because it's easier to think about. Say it's 100mA.

Am I correct in thinking that the current from the constant current source (or sink) sees the 2 resistors as though they are in parallel? So the 9Ω resistor sees 1A - 90mA = 91mA, and the 1Ω resistor sees 1A + 10mA = 1.01A ???? So the voltage at node A is 1.01V?

I've always kind of fudged around questions like this, taking the experimental approach, these simple R/V/I theoretical networks were confusing enough in class and long since forgotten, but suddenly I need to re-learn it. All help appreciated!
 
hmmm... The voltage source should have an impedance of 0Ohms, so yes the 2 resistors will look like they are in parallel. therefore surely you will get a proportionately higher current flowing through the 1Ohms resistor than the 9Ohms resistor. Surely 90mA flows through the 1 Ohm resistor and 10mA flows through the 9Ohm resistor... apart from that, I agree with your logic
 
simonbramble beat me to the punch. ;) The 1A constant current splits according to the inverse value of the resistances (It's apparent that the lower the resistance the easier the current will flow). So the 1Ω resistor sees 90ma of the current and the 10Ω resistor sees 10mA of the current. This makes the 1Ω current total 1.09A and the 9Ω current total 0.99A.
 
Oh yes, so they are - I was so snagged up in the stuff I wasn't sure about I didn't make sure I'd got the part I should be sure about right! I did do it in my head whilst driving though...

Anyway, thanks as always, nice to know my brain still works, sort of, market research hasn't killed it off altogether yet.... glngngghhh
 
May I add a short comment?

The "problem" has started with throbscottle´s formulation "Suppose I have two resistors in series...".
I think two elements can be considered to act in series or in parallel only if - at the same time - they are referenced to a particular source (voltage or current) that is connected between two of three available nodes.
 
Umm, I did say they have 10v connected across them! I know I really should have included a drawing, but I am lazy and I thought it was simple enough to describe!
 
Hello,


Yes a theoretical voltage source has 0 ohms impedance so to the current source it looks like the two resistors are in parallel, but you have to be careful with the numbers and make sure the calculation comes out right. For example, with only a 100ma current source (voltage source shorted) the total resistance is 0.9 ohms, and the voltage across the two resistors is 0.090 volts Thus the current through the 1 ohm resistor is 0.090/1=90ma, and the current through the 9 ohm resistor is 0.090/9=10ma (in this case we consider that negative). So when we connect the 10 volt source we get 1-0.010=990ma through the 9 ohm resistor and 1.09 amps through the 1 ohm resistor.

But the more general way to attack this kind of circuit is simply through the use of "Superposition". Superposition is used to simplify a circuit that has more than one source. This circuit has one voltage source and one current source, so that means superposition might help.

To use superposition to solve a circuit with two or more sources, we use the following basic procedure:
1. Short all voltage sources (this is called 'killing' a voltage source).
2. Open circuit all current sources (this is called 'killing' a current source).
3. Now that we have all sources 'killed', un-kill each source one at a time, and calculate the desired response (voltage or current), then kill that source again.
4. Add up all of the responses calculated in #3 above, that's the solution.

So applying this procedure to the circuit at hand, we would first short out the 10v source, then open circuit the current source.
We could then start by un-killing the 10v source (removing the short), then calculate the current through the 1 ohm resistor call that i1.
Next, we would again short out the 10v source, and then reconnect the current source, then calculate the current again through the 1 ohm resistor call that i2.
The solution is the total response which is simply the sum of the two currents just calculated:
iTotal=i1+i2

So in short all we had to do was apply each source one at a time and calculate the desired response using each source, then add all of those responses to get the final solution which is the total response. We just have to remember that to kill a voltage source we short it out, but to kill a current source we open circuit it. This is the theoretical procedure and for a real life practical circuit if we wanted to try this out in real life we'd have to modify it slightly.


APPLYING THIS PROCEDURE TO A REAL LIFE CIRCUIT...

One catch to doing this in real life (we usually dont do this anyway) is that when we short out a voltage source we get an infinite current, and when we open circuit a current source we get an infinite voltage, either of which will cause big problems, so we have to modify the theoretical procedure slightly if we wanted to try this out on a real life circuit with real life sources.
In the real life circuit instead of shorting out the voltage source directly, we first disconnect it from the circuit and then intead of shorting out the source itself we short out the two terminals that the source had originally connected to.
And instead of open circuiting the current source, we either disconnect it's power supply first or short it out before disconnecting it from the circuit which is required for open circuiting it.


APPLYING THIS PROCEDURE IN SPICE...

Spice has it's own issues. Often it wont let you short out a voltage source and will complain if you do, so we do the same as in real life, we disconnect one lead and short only the two original circuit terminals.
To kill a current source, often BOTH leads of the current source will have to be disconnected from the circuit and then you can short out the current source to keep the voltage developed across the current source low.
 
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Umm, I did say they have 10v connected across them! I know I really should have included a drawing, but I am lazy and I thought it was simple enough to describe!

Yes - there was no misunderstanding. Your description was sufficient.
Nevertheless, this thread and your problem description gave rise to point to this "small problem": There are many examples in circuit analysis where two elements look as they were connected in series - however they act in parallel for some specific calculations (example: RC time constant of an emitter circuitry where only a part of the emitter resistor is shunted by a capacitor).
Exercise: Draw two resistors with one common node (without any source or ground) and ask students "series or parallel"? I guess 90% will answer: series.
 
Wow - I just love this forum! So good to revisit the basics - with hand-holding :) Thanks everyone.
 
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