But isn't it the way we are taught networks. I have always used this procedure of multiplying current by -1. Yes, you do sound logical when you talk about other approaches yielding the same result.But if you conclude that your answer doesn't make sense, then all you can conclude is that your answer is wrong. You can't just change your answer to patch what doesn't make sense because your answer is your answer and if it is wrong then it is wrong. You have to go back and figure out what is wrong. There is absolutely no basis for believing that multiplying a wrong answer by -1 is going to somehow magically make it right.
Can we practically make a negative resistance without the help of tunnel diode (I have very little knowledge about both negative resistance and tunnel diode) as is the case here?What would a negative resistance mean? All it would mean is that if you apply a positive test voltage to the circuit using a test source that current is driven back into the source. Is that possible? Sure, provided the circuit you are connecting the source to has a way to deliver power to the test source and that the more voltage you apply the more current is driven into the test source, which usually means that you have some kind of dependent source in the circuit.
Why bother?Can we practically make a negative resistance without the help of tunnel diode (I have very little knowledge about both negative resistance and tunnel diode) as is the case here?
I don't follow. Are you saying that you are taught that if your answer is doesn't make sense to multiply it by -1 in order to correct it? Does THAT make sense?But isn't it the way we are taught networks. I have always used this procedure of multiplying current by -1.
Yes, although because a tunnel diode is a passive component we don't get a large-signal negative resistance but only a small-signal negative resistance around a certain large-signal operating point. Basically, as we start increasing the signal voltage our signal current increases. But at some point it peaks and after that, as we continue to increase the signal voltage, the signal current actually decreases for a while. Then we get a minimum and, as we continue increasing the signal voltage the signal current once again increases. Within the region where the signal current decreases with increasing signal voltage we can model the device as having a negative resistance.Can we practically make a negative resistance without the help of tunnel diode (I have very little knowledge about both negative resistance and tunnel diode) as is the case here?
Different thing entirely.Why bother?
One way to interpret a negative sign is to look at the current. In the beginning you assumed that the current is going in one direction, you gave that direction a positive sign, you have a positive current. Now you solved the equations and you have negative sign. You can give that negative sign to the current, you now have negative current. What does that mean? It means that your earlier assumption was wrong. You assumed that current is going to direction A, when in fact the current is going in direction B. So just switch current direction and the negative sign will go away.
In your specific example you don't say in what direction the current I is. I will assume that the current I is leaving terminal a. That makes current I in clockwise direction. Your solution is: V=(R/11)(-I). So just switch the direction of current I. The new direction of current I is counter clockwise, now I is positive and solution becomes: V=(R/11)(I)
V/I=R/11
I don't follow. Are you saying that you are taught that if your answer is doesn't make sense to multiply it by -1 in order to correct it? Does THAT make sense?
That's my doubt right there.Why bother?
One way to interpret a negative sign is to look at the current. In the beginning you assumed that the current is going in one direction, you gave that direction a positive sign, you have a positive current. Now you solved the equations and you have negative sign. You can give that negative sign to the current, you now have negative current. What does that mean? It means that your earlier assumption was wrong. You assumed that current is going to direction A, when in fact the current is going in direction B. So just switch current direction and the negative sign will go away.
In your specific example you don't say in what direction the current I is. I will assume that the current I is leaving terminal a. That makes current I in clockwise direction. Your solution is: V=(R/11)(-I). So just switch the direction of current I. The new direction of current I is counter clockwise, now I is positive and solution becomes: V=(R/11)(I)
V/I=R/11
Let us back track a little.No, WBahn that's not what I mean. I thought the same what shteii01 has stated above.
That's my doubt right there.
Yes this is very much simple network analysis no diodes involved. I was just curious about the practical significance of this negative resistance thing. Yes I have a little knowledge about diodes but this particular problem is not remotely related to diode.Let us back track a little.
I think you mentioned diode... I assumed that we were looking at material from second year circuits analysis class, no diodes. What material are we looking at? Are you studying diodes and are we looking at the material that is teaching you about modeling diode in some specific case?
Only if he is satisfied with having the wrong answer.Since this is simple stuff. I think you can use my explanation then.
Alright. I think I see what you mean.Only if he is satisfied with having the wrong answer.
There is more than a minor difference between having an equivalent resistance of (R/11) and having an equivalent resistance of (-R/11).
Don't believe me?
Set R=22Ω and then connect your Thevenin equivalent circuit consisting of Rth=22Ω/11=2Ω and connect it to a 10V source connected in series with a 1Ω resistor. How much power is dissipated in the 1Ω resistor. Now do the same but use a Thevenin equivalent circuit consisting of Rth=-22Ω/11=-2Ω.
Makes a different, doesn't it?
But which is right?
Well, plug the same source and resistor into the original circuit and do the analysis. Which is right?
It's definitely a subtle point, but an important one.Alright. I think I see what you mean.
by Jake Hertz
by Aaron Carman
by Aaron Carman