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Help on finding Thevenin Resistance and Current across resistor for RL Circuit?

dknguyen

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#21
So, ultimately, depending on the current loop orientation I define, that dictates how the polarity changes in response (with respect to the voltage source). Which, in other words, determines whether the voltage term is either positive (voltage drop), or negative (voltage rise). I'm so used to having the voltage term isolated from the rest of the other terms when doing mesh-current analysis that I've forgotten how it's interpreted when included with the other terms (or simply having the loop equation be 0 = voltage term + ....). And so with that, the way current runs through resistors and sources are different that care has to be taken when arranging them correctly. Sounds like I need to take some time tonight to review passive sign orientation with sources and mesh-current analysis, since both play a role on how current runs through inductors and capacitors as well.
Yeah. You can define whatever damn current loop directions you want. As long as you stay consistent with the currents and voltage drops polarities according to your convention it when tracing your current loop in your chosen direction to write out the loop equations. You can pre-separate the sources to one side if you really want to, but you need to be aware of how it works because what happens if you get a circuit where a single current loop has two sources in it that may or may not oppose each other? Then you're screwed. Or even worse, if they're dependent sources.

RLC mesh gets pretty ugly though if you stick to time domain. Solving differential equations due to all the inductor di/dT terms and dV/dT capacitor terms. I normally just go with frequency domain jwL and 1/jwC and then Laplace to turn it back into time domain or whatever. both are pretty involved for exam questions though if they want you to go all the way through the problem. But you still need to know it in case they just ask for a precursory analysis where an inductor or cap was already charging and a switch changes state which causes the inductor and cap to become a source. Either way, mesh analysis rides on that passive sign convention so you should know it.

Yeah...that's the term...passive sign convention. I had totally forgotten it was a thing.
 
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Thread starter #22
You can draw any mix of CCW or CW loops you want. I just draw them in the same direction because it makes things simpler. The point to take is that when following the loop equations use polarities that match the directions that you defined.

Loops X is indeed zero because the only current unique to loop X is flowing through R1, but there's a short across R1 so no current flows through R1, therefore Ix = 0. All the currents in the bottom and right side of that loop are accounted for in Iz and Iy.

Loop Z: Iz should be 3(Iz - Iy). You drew Iz CCW, so if you follow the loop in that direction to write the equations, Iz is positive (obviously), but in R3 Iy is flowing in the opposite direction that you are tracing the loop in, so Iy must be negative.

Loop Y is 0 = -8V + 4iy + 3(iy-iz).

The reasons for this are explained immediately above and in my last post about sources. These are the most confusing parts of basic mesh analsysis. If you can understand why these polarities are the way they are, you are good. You can define the current loops in any mix of direction you want (I just draw them all in the same direction because to me it removes variability).

What matters is to stick to the conventions you defined when writing your loop equations. If the current loop direction you defined is correct for the voltage drop of whatever component you come across, it gets a + polarity. If it's wrong (i.e. it's a source but the current direction treats it as a load), then it gets a negative sign. You can do this because, unless you are solving for a mystery source, you know ahead of time how it should behave. If you are solving for a mystery source then you don't know this which is very similar to encountering a load in your trace. You don't really know what's happening in the RLC loads. In either case, you do the same thing: just assume the voltage drop is in the direction you are tracing and if the result ends up being positive you know your assumption was right. If you were wrong, the result is negative.

If an adjacent current loop merges with the current loop you are tracing in the same direction, it gets a + polarity. If not, it gets a - polarity.

Mesh analysis is all about defining/assuming a convention and blindly keeping track of it as you write your loop equations. If it's positive, your assumption was right. If it's negative, your assumption was wrong.
Okay, well I do like the approach you take with defining current loops, as it's a method that I had not considered in the past. It certainly helps in establishing commonality and consistency too for the directions that have their respective polarities. The concept behind polarities is something I am still learning in the process, but it is as you say: understanding and mastering it will help to also master the conventions that are associated with them. RLC loads are something that I'll actually be approaching soon, upon further review of both RL and RC circuits, but making the necessary presumptions for the current loop direction does to the said voltage drop direction is certainly key too. But wait, if what you mentioned the idea of nearby loops merging with the current loops I defined is true, then does that mean that for i3(t=infinity), \[ i_3(t=\infty)\;=\;i_z\;-\;i_y \]? Or does your previous equation for i3 still hold for this steady state operation? Because in one of your previous posts, you described the current through R3 as I3 = Iz + Iy, but that was only due to me drawing Loop y in the CW position in my past drawing. Since I re-drew the current loop Y direction as CCW, should iy then be subtracted from iz to get i3?

Yeah. You can define whatever damn current loop directions you want. As long as you stay consistent with the currents and voltage drops polarities according to your convention it when tracing your current loop in your chosen direction to write out the loop equations. You can pre-separate the sources to one side if you really want to, but you need to be aware of how it works because what happens if you get a circuit where a single current loop has two sources in it that may or may not oppose each other? Then you're screwed. Or even worse, if they're dependent sources.

RLC mesh gets pretty ugly though if you stick to time domain. Solving differential equations due to all the inductor di/dT terms and dV/dT capacitor terms. I normally just go with frequency domain jwL and 1/jwC and then Laplace to turn it back into time domain or whatever. both are pretty involved for exam questions though if they want you to go all the way through the problem. But you still need to know it in case they just ask for a precursory analysis where an inductor or cap was already charging and a switch changes state which causes the inductor and cap to become a source. Either way, mesh analysis rides on that passive sign convention so you should know it.

Yeah...that's the term...passive sign convention. I had totally forgotten it was a thing.
Interesting. Dependent sources are certainly a nightmare that, while troublesome, are just as necessary into understanding the way current loop directions work, alongside with overall circuit analysis. I actually haven't been introduced to the frequency domain yet with respect to RLC circuits, at least not yet. I do wish to learn more about that method (considering that Laplace is also involved), but if both ways are viable ways into understanding how RLC circuits (for transient responses) work, it would definitely be best to learn both of the approaches (diff. equations vs. frequency domains) you mentioned.
 
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dknguyen

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#23
I should add that in post #17, you were making more work for yourself by converting the current source to it's thevnin voltage equivalent. The current source outright gives you the loop current for that loop there.

It might not always work out that way though. I can imagine cases where you need to find the voltage drop across the current source but where you can't change it to a Thevnin voltage equivalent either. Not sure what you would do...you might have to assign a dummy mystery voltage drop across it and hope it will solve itself out. It should since the voltage produced by the current source just equals everything else in the loop.

Like if a current source bridges the merge between two loops...then you would have another equation where Isource = I_loop1 + I_loop2, or minus depending on your defining of things. But you would still need the voltage drop across it in some cases just to trace through the loop.

Interesting. Dependent sources are certainly a nightmare that, while troublesome, are just as necessary into understanding the way current loop directions work, alongside with overall circuit analysis. I actually haven't been introduced to the frequency domain yet with respect to RLC circuits, at least not yet. I do wish to learn more about that method (considering that Laplace is also involved), but if both ways are viable ways into understanding how RLC circuits (for transient responses) work, it would definitely be best to learn both of the approaches (diff. equations vs. frequency domains) you mentioned.

Frequency domain and Laplace makes some things a lot easier. You just treat inductors like a resistor with resistance (impedance) jwL and for capacitors it's 1/jwC, where w = 2*pi*f. Then Laplace or Laplace tables to turn it into the time domain if required. It's like magic.


But wait, if what you mentioned the idea of nearby loops merging with the current loops I defined is true, then does that mean that for i3(t=infinity), \[ i_3(t=\infty)\;=\;i_z\;-\;i_y \]? Or does your previous equation for i3 still hold for this steady state operation? Because in one of your previous posts, you described the current through R3 as I3 = Iz + Iy, but that was only due to me drawing Loop y in the CW position in my past drawing. Since I re-drew the current loop Y direction as CCW, should iy then be subtracted from iz to get i3?
You mean in post #13?

Your drawing takes precedence. If you defined Iz and Iy as going in opposite directions through R3, then the current through R3 will be I3 = Iz - Iy or Iy - Iz, depending on which loop equation you were looking at. It all hinges on your drawing.

In that case, I said I3 = Iz + Iy because that's how you defined current directions in that drawing. If you defined them differently, it would change because your calculated Iz and Iy would change based on how you drew them.
 
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Thread starter #24
I should add that in post #17, you were making more work for yourself by converting the current source to it's thevnin voltage equivalent. The current source outright gives you the loop current for that loop there.

It might not always work out that way though. I can imagine cases where you need to find the voltage drop across the current source but where you can't change it to a Thevnin voltage equivalent either. Not sure what you would do...you might have to assign a dummy mystery voltage drop across it and hope it will solve itself out. It should since the voltage produced by the current source just equals everything else in the loop.

Like if a current source bridges the merge between two loops...then you would have another equation where Isource = I_loop1 + I_loop2, or minus depending on your defining of things. But you would still need the voltage drop across it in some cases just to trace through the loop.
Ahh, you're right about that. I don't know why, but I feel like when it comes to voltage sources, they're instinctively easier to understand and work with than current sources, but seeing that the current source was available there should've made for a nice opportunity. As for the current source bridging two loops, I do agree that that would introduce another equation to be included in the mesh analysis. Whether that makes things easier or harder to understand is likely subjective for anyone working with that.

Frequency domain and Laplace makes some things a lot easier. You just treat inductors like a resistor with resistance (impedance) jwL and for capacitors it's 1/jwC, where w = 2*pi*f. Then Laplace or Laplace tables to turn it into the time domain if required. It's like magic.




You mean in post #13?

Your drawing takes precedence. If you defined Iz and Iy as going in opposite directions through R3, then the current through R3 will be I3 = Iz - Iy or Iy - Iz, depending on which loop equation you were looking at. It all hinges on your drawing.

In that case, I said I3 = Iz + Iy because that's how you defined current directions in that drawing. If you defined them differently, it would change because your calculated Iz and Iy would change based on how you drew them.
Hmm, I would most certainly like to learn more about that method for treating inductors like resistors with impedance, except that sadly I haven't been introduced to that concept yet. Does that method usually simplify things when it comes to understanding RLC circuits?

As for your post, yes, that's exactly what I meant. So in that case, if I defined the current loop Iz as going in a clockwise direction, and the loop y as going in a counterclockwise direction, would that also lead to the I3 = Iz + Iy as well? I just want to make sure because on the original post (post #1) I brought up, the original path of the I3 current across the R3 resistor was going downwards. So I'm not sure if that affects anything we've done up to this point?
 

dknguyen

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#25
As for your post, yes, that's exactly what I meant. So in that case, if I defined the current loop Iz as going in a clockwise direction, and the loop y as going in a counterclockwise direction, would that also lead to the I3 = Iz + Iy as well? I just want to make sure because on the original post (post #1) I brought up, the original path of the I3 current across the R3 resistor was going downwards. So I'm not sure if that affects anything we've done up to this point?
Yes. If they were opposite direction it would be either Iz - Iy or Iy - Iz. Positive result means the direction matches that of the left operand. Negative means the right operand. Basically same as always: positive = assumption right, negative = assumption wrong.

Hmm, I would most certainly like to learn more about that method for treating inductors like resistors with impedance, except that sadly I haven't been introduced to that concept yet. Does that method usually simplify things when it comes to understanding RLC circuits?
It beats solving differential equations which is what you do if you stay in the time domain. You work off Laplace transform tables to minimize the amount of calculus you need to do and only if you actually need to bring the answer back into the time domain.
 
Thread starter #26
Yes. If they were opposite direction it would be either Iz - Iy or Iy - Iz. Positive result means the direction matches that of the left operand. Negative means the right operand. Basically same as always: positive = assumption right, negative = assumption wrong.
Wait, so I'm confused about this matter. I do understand the part you mention about the opposing directions that current loops can take, which is what I decided to do for the t<0 case (the one with the voltage source intact only). I set the iz current loop to go CW, and iy in the CCW, so that both are in opposite directions and that I can follow their own respective conventions and polarities. Upon redoing the calculations again, it does turn out that both iz and iy are both positive, along with ix being positive as well. However, what I'm a bit lost on is what you mention about the "positive result". Were you stating with respect to i3 being positive? Or iz and iy?
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dknguyen

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#27
It's obscured here because the current loops are defined to be in the same direction and you calculated they were both positive. But it matters when currents are opposite or if you calculated some of them to be negative.

So let's say one loop was CW and the other was CCW:

In visual form: Draw an arrow beside R3 representing I3. Draw it completely backwards if you want. All loop currents that have the same direction have a positive operator, all opposite have a negative operator in front. Then substitute in the calculated loop currents WITH their calculated polarity. You basically assumed a direction for I3 when you do this. If I3 is positive you assumed right. If not you assumed wrong. No different than the rest.

You form the I3 equation with signs and operators for the variables that assume direction, then plug in loop currents WITH the calculated polarity on top of that.

Long form:

If you go I3 = Iy - Iz, you are implicitly assuming that I3 is in the same direction as Iy as drawn, not as calculated.

Then if you calculated Iy to be negative and Iz positive it means your assumption of Iy's direction was wrong, but let's ignore that and push ahead...

If |Iy| > |Iz| then plugging it in gets a negative I3 which means the direction of I3 is opposite of Iy as drawn, not as calculated (or same as Iz as drawn). Same thing happens if Iy is calculated positive and Iz is calculated positive but Iy < Iz.
 
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Thread starter #28
If you go I3 = Iy + Iz, you are implicitly assuming that I3 is in the same direction as Iy.

If you go I3 = Iz + Iy, you are implicitly assuming that I3 is in the same direction as Iz.

If the calculated I3 < 0 then that means your assumption, whatever it was, was wrong.

It's obscured here because the current loops are defined to be in the same direction and you calculated they were both positive. But it matters when currents are opposite or if you calculated some of them to be negative. Do you need drawn examples of every combination?

If one loop was CW and the other was CCW:

If you go I3 = Iy - Iz, you are implicitly assuming that I3 is in the same direction as Iy as drawn, not as calculated.
Oh dear, so with the assumptions you've stated in regards to i3 = iy +iz (being that both values are positive), I carelessly calculated the wrong value of i3 for this case. It is true, as you say, that the way current loops are defined and directed certainly affect what the resulting i3 value will be.

It's obscured here because the current loops are defined to be in the same direction and you calculated they were both positive. But it matters when currents are opposite or if you calculated some of them to be negative.

So let's say one loop was CW and the other was CCW:

In visual form: Draw an arrow representing I3. Draw it completely backwards if you want. All loop currents that have the same direction have a positive operator, all opposite have a negative operator in front. Then substitute in the calculated loop currents WITH their calculated polarity. You basically assumed a direction for I3 when you do this. If I3 is positive you assumed right. If not you assumed wrong. No different than the rest.

You form the I3 equation with signs and operators for the variables that assume direction, then plug in loop currents WITH the calculated polarity on top of that.

Long form:

If you go I3 = Iy - Iz, you are implicitly assuming that I3 is in the same direction as Iy as drawn, not as calculated.

Then if you calculated Iy to be negative and Iz positive it means your assumption of Iy's direction was wrong, but let's ignore that and push ahead...

If |Iy| > |Iz| then plugging it in gets a negative I3 which means the direction of I3 is opposite of Iy as drawn, not as calculated (or same as Iz as drawn). Same thing happens if Iy is calculated positive and Iz is calculated positive but Iy < Iz.
Ah, now that makes sense, especially when you frame the loop currents having their respective operators and the actual calculated loop currents with their own polarities (either positive or negative). But then, by that token, we must've set up the current loop equation for Loop z wrong, because while Loop z is drawn in the opposite direction of i3, the i3 would be set up as: i3 = iy + (-iz) = (0.444 A) + (-1.33 A) = negative, indicating that my assumption of iz direction was wrong, right?

That must be the curveball you've been highlighting I think, which means I need to re-draw and re-calculate iz and iy in their correct directions to get their correct polarities and ultimately the right i3 value?
 

dknguyen

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#29
Ah, now that makes sense, especially when you frame the loop currents having their respective operators and the actual calculated loop currents with their own polarities (either positive or negative). But then, by that token, we must've set up the current loop equation for Loop z wrong, because while Loop z is drawn in the opposite direction of i3, the i3 would be set up as: i3 = iy + (-iz) = (0.444 A) + (-1.33 A) = negative, indicating that my assumption of iz direction was wrong, right?
Sort of...you would have known your assumption for Iz was wrong the second it popped out of the calculation as a negative number. if you need to display it as an answer then you can correct the direction at that point if required, but if you need to continue using the value in other calcuations in the mesh analysis then...

That must be the curveball you've been highlighting I think, which means I need to re-draw and re-calculate iz and iy in their correct directions to get their correct polarities and ultimately the right i3 value?
...you don't need to recalculate or re-draw if your assumption was wrong. Just continue working with all the values as is. This is probably best anyways since it will just cause confusion in your conventions if you make correct direction assumptions here and there. If you were consistent it should carry all the way through to the end. That's kind of why you treat it mechanically and blindly like a computer. The polarity of your calculated voltages and currents will work together to cancel out any wrong assumptions you made in their direction on your drawing. Whenever you combine the two together, then you know the actual direction but you only need to do this when you display the final result so the person reading it has an easier time.

You have to make a distinction between the polarity as drawn (direction of the currents that you assumed) from the polarity of the calculated value of the current (the actual polarity of the currents relative to your assumption). The calculated value and polarity plug into the variable and its polarity which represents the assumption of direction that you made. Only together can they tell you the actual direction (relative to what you have drawn).

If you drew I3 to point down on the drawing, Iy CW and Iz CW, then your equation would be
I3 = Iz - Iy

Drawing I3 like this and writing out this equation is the same as saying:

"I assume I3 is in the same direction as Iz and opposite direction of Iy, as drawn (not in actuality)"
"I assume I3 is in the same direction as Iz, assuming that my direction assumption for Iz was correct."
"I hereby define down as the positive direction for I3 which just so happens to be the same definition that I have given Iz, both of which are the opposite definition given to Iy, which is defined as up being positive."

But then you might end up calculating that Iz = 3A, Iy = -1A.

That means that your assumption for Iz's direction was right, but your assumption for Iy was wrong. If you need to display Iy at this point you can draw a little schematic with the arrow in the proper direction for Iy and use +3A as the value. But if you need to use Iy for another calculation, then don't use the corrected value. Continue using what you originally defined and calculated with your conventions:

I3 = Iz - Iy = (3A) - (-1A) = 4A.

Since it is positive, that means your assumption of I3 (the down arrow) was correct

But if you got Iz = -1A and Iz = 3A

Then I3 = Iz - Iy = (-1A) - (3A) = -4A

Since I3 is negative, that means your down arrow for I3 was wrong and it actually flows in the opposite direction (up).

It's super mindless once you grasp it. You just blindly work through the math and stay consistent.
 
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Thread starter #30
Yes, but you would have known your assumption for Iz was wrong the second it popped out of the calculation as a negative number. if you need to display it as an answer then you can correct the direction at that point if required, but if you need to continue using the value in other calcuations in the mesh analysis then...


...you don't need to recalculate or re-draw if your assumption was wrong. Just continue working with all the values as is it by reversing direction in the drawing if your calculated current was negative (or voltage for that matter). This is probably best anyways since it will just cause confusion in your conventions if you make correct direction assumptions here and there. If you were consistent it should carry all the way through to the end. That's kind of why you treat it mechanically and blindly like a computer. Only whenever you actually require displaying a result do you need to correct for the bad assumption just so it makes sense to the person reading it.
Hmm, you're right, it was perhaps silly of me to have not immediately seen the error of my assumptions when the negative value came up (and not realized it sooner). Oh, well I think you may be right about not needlessly involving more work with all the re-calculating and re-drawings. My final values did end up positive for i3(t<0) and i3(t=infinity), as 2.67 A and 1.78 A, respectively.

Upon inserting these values into my general equation for the i3(t), I ended up with:
\[ i_3(t)\;=\;1.78\;+\;0.89\ast e^{-\frac t{0.0000006296}} \]

However, that tau value is very strange, as far as the RL circuit is concerned. Despite knowing that the time constant is L/R (or Req, based on the equivalent resistance as 1.59 ohms), did I perhaps incorrectly calculate the time constant for the circuit?
 

dknguyen

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#32
http://tinyurl.com/ya35oa7w

Bah. I am having trouble getting the timescale in nano seconds which is where tau suggests we need to be at to see the rise. Might have to increase the L just to be able to observe the time constant at work since there seems to be a us lower limit on the scope in this simulator.

BUt the simulator indicates that the I3 is 1.33A before the switch closes and 444.44mA in the far future.

I might give it a shot tommorrow by hand myself.
 
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Thread starter #33
Found a mistake in my last post and corrected it.

It's possible that the method I suggested to calculate tau is not valid. I don't see why it wouldn't be though. We might have to run this through a circuit simulator to check
https://www.falstad.com/circuit/
Ah, your current statements into post #29 now ring true with respect to the example you clarified for the CW orientation of both Iz and Iy. It certainly demonstrates the point of just maintaining consistency for the drawing, currents and loops setup that allow you to test whether your assumptions were found to be true. It's more thorough than what I expected, but it emphasizes your point of working through the calculations in a consistent manner.
Found a mistake in my last post and corrected it.

It's possible that the method I suggested to calculate tau is not valid. I don't see why it wouldn't be though. We might have to run this through a circuit simulator to check
https://www.falstad.com/circuit/
Hmm, well it's anyone guess as to whether the method taken was correct in determining tau. At least, when I tried to visualize opening the inductor and seeing how the resistors are oriented to find the equivalent resistance, that generated a lot of confusion.

Oh, well this is a circuit simulator that I have not used before. I'm not exactly sure what key variables am I supposed to be identifying for the circuit shown?
 

dknguyen

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#34
Ah, your current statements into post #29 now ring true with respect to the example you clarified for the CW orientation of both Iz and Iy. It certainly demonstrates the point of just maintaining consistency for the drawing, currents and loops setup that allow you to test whether your assumptions were found to be true. It's more thorough than what I expected, but it emphasizes your point of working through the calculations in a consistent manner.

Hmm, well it's anyone guess as to whether the method taken was correct in determining tau. At least, when I tried to visualize opening the inductor and seeing how the resistors are oriented to find the equivalent resistance, that generated a lot of confusion.

Oh, well this is a circuit simulator that I have not used before. I'm not exactly sure what key variables am I supposed to be identifying for the circuit shown?
Well, I set up the circuit already and linked you to it. MIght require some tinkering though.
https://www.falstad.com/circuit/circuitjs.html?cct=$+1+0.000005+45.7144713268909+31+5+50 r+112+80+112+176+0+2 r+112+176+112+272+0+6 r+224+176+224+288+0+3 r+352+64+352+304+0+4 v+224+176+224+128+0+0+40+12+0+0+0.5 w+112+80+224+128+0 w+224+128+352+64+0 s+384+64+464+64+0+1+false w+352+64+384+64+0 i+480+240+464+64+0+2 w+480+240+352+304+0 w+112+272+224+336+0 w+224+336+352+304+0 l+112+176+224+176+0+0.000001+6.666666666666667 370+224+336+224+288+1+0 o+14+256+0+4099+5+1.6+0+2+14+3
 
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dknguyen

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#35
The simulation behaves completely differently when the switch is before and closed. There's a very rapid step change in I3 so maybe we're missing something.
 
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Thread starter #36
Oh, okay, before I had a problem with my computer opening it up, I can see the condition now for where the switch is open. You are right about the strange change in I3 as the switch closes. I'm trying to interpret the cause of this, but even I'm uncertain as to what exactly it could be.
 

dknguyen

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#37
The time constant is for the current in the inductor so that's what you have to look at, not I3. I observed it took 310ns to complete 37% of the change between the two steady states. That might mean that you can't stitch things together quite the way you want to for I3 since that is for the resistor, but tau is for the inductor...

But tau was calculated to be 630ns which still doesn't match...

I also realized that what we did with the "AC analysis" where we turned of all the sources and calculated the LR constant was the same as if you turned off all the sources, and calculated the thevnin resistance by looking out of the terminals of the inductor (treating it as the load for the thevnin circuit). In other words, it still seems to me that the tau calculation should have worked.

https://www.falstad.com/circuit/circuitjs.html?cct=$+1+1e-8+0.2954511527092107+31+5+50 r+80+80+80+176+0+2 r+80+176+80+272+0+6 r+224+176+224+288+0+3 r+352+64+352+304+0+4 v+224+176+224+128+0+0+40+12+0+0+0.5 w+80+80+224+128+0 w+224+128+352+64+0 s+384+64+464+64+0+0+false w+352+64+384+64+0 i+480+240+464+64+0+2 w+480+240+352+304+0 w+80+272+224+336+0 w+224+336+352+304+0 l+112+176+224+176+0+0.000001+6.223166169895595 370+224+336+224+288+1+0 370+80+176+112+176+1+0 o+15+1+0+12293+0.0001+6.666666666666481+0+2+15+3
 
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Thread starter #38
The time constant is for the current in the inductor so that's what you have to look at, not I3. I observed it took 310ns to complete 37% of the change between the two steady states.

You have to turn on the max scale option right clicking on the plot and then fiddle with the step time under OPtions->Other Options, and then again with the time scale slide bar found by right clicking the plot- > properties.

But tau was calculated to be 630ns. Hmmm...

https://www.falstad.com/circuit/circuitjs.html?cct=$+1+1e-8+0.2954511527092107+31+5+50 r+80+80+80+176+0+2 r+80+176+80+272+0+6 r+224+176+224+288+0+3 r+352+64+352+304+0+4 v+224+176+224+128+0+0+40+12+0+0+0.5 w+80+80+224+128+0 w+224+128+352+64+0 s+384+64+464+64+0+0+false w+352+64+384+64+0 i+480+240+464+64+0+2 w+480+240+352+304+0 w+80+272+224+336+0 w+224+336+352+304+0 l+112+176+224+176+0+0.000001+6.223166169895595 370+224+336+224+288+1+0 370+80+176+112+176+1+0 o+15+1+0+12293+0.0001+6.666666666666481+0+2+15+3

I also realized that what we did with the "AC analysis" where we turned of all the sources and calculated the LR constant was the same as if you turned off all the sources, and calculated the thevnin resistance by looking out of the terminals of the inductor (treating it as the load for the thevnin circuit). In other words, it still seems to me that the tau calculation should have worked.
Oh, okay, thanks for the heads up there. So the max scale option and the step time both help in identifying the time constant.

But actually, now that you mentioned the 630ns, that was associated with the equivalent resistance we found earlier, I think I may have realized a past error made in calculating Req. Usually superposition is used to find equivalent resistance for t = infinity (switch closes) right? I want to believe that we did find that correctly (based on posts #1 - #3), but the immensely small number of tau kind of stems some doubt?
 

dknguyen

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#39
THe time constant is correct. Me using 37% of the change as the measuring point in my previous post was wrong. I was supposed to use 63.7% of the change. I get 650ns in the simulation which is close enough the calculated 630ns considering I'm basically eyeballing the graph with cursors.

I'll try the mesh analysis later this week.
 
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dknguyen

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Most Helpful Member
#40
I think the concept is also right. Here's my math but I goofed a bit at the end of the first half

You can see in my work that I have that before the switch closes, I3 = Ic - Ib = 2 - (-0.666) = 2.666A. But it would be 1.33A (what the simulation says) if the signs were reversed so I probably missed something somewhere.

(EDIT: Yup, I forgot to carry over a negative sign right as I wrote Ic = 2. So if I make that correction, then I3 = Ic - Ib = (-2) - (-0.666) = -1.33A. So with that correction, the direction and magnitude now agree with the simulation for before the switch closes.)

My larger concern is the second half where T = infinite.
You can see that I ended up with the change in current being 444.44mA, but that's the change in current, whereas the simulation says I3 actually ends off at 444.44mA. So it seems something is not quite right in our interpretation of things.

Now that I think about it, you don't need to use superposition at all. I can't remember why I ever wanted to use superposition to begin with. You just calculate I3 before the switch closes (which we both did).Then you calculate I3 for after the switch closes (with both sources active and connected) in one go.
 

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