This is more of a conceptual issue. In this problem, the circuit is in steady state when t is approaching 0, correct? So assuming that, the inductor acts as a short circuit and receives all of the 100 mA current. But because the current is constant, that means the voltage across the inductor (which is equal to v(t)) is 0 V since i(t) = L*(di/dt). That doesn't seem to make any sense, though, because then the voltage would just be zero or infinite over the entire time. Clearly there's something I'm not grasping here. Am I wrong in saying that the current is constant when trying to find initial and final conditions?
What does di/dt mean? If di/dt = 0, does it always mean that i = 0? Since you stated that there is no current in the resistor, what is the voltage drop across the resistor when i = constant? At t = 0, when the current source changes from 100mA to 0mA, what happens to the current in the inductor?
1. di/dt is the derivative (or rate of change) of the current. I'm thinking of the formula for voltage over an inductor, v(t) = L*(di/dt). In that case, if di/dt is 0 (which happens when the current is constant), then v(t) is also 0. 2. No. But as far as I can tell, it does always mean that v(t) over an inductor is zero when a constant current is applied. 3. When i = constant? It'd just be iR (i*20 in this problem). But if there's no current across the resistor it's of course 0. 4. Nothing. Current can't change instantaneously in an inductor.
So where does the current go since it can no longer go through the source? Think of a switch is opened to disable the source.
Correct. WHEN the current THROUGH the inductor is not changing, the voltage ACROSS the inductor will be 0V. Where do you get either of those conclusions from? Namely, where does the "infinite" part of it come from and where does the "entire time" part come from. You only have that the voltage across the inductor WHEN the current THROUGH the inductor is not changing is 0V. No, but you are trying to apply the conclusion from a specific point in time to every point in time.
Okay. When I say it's infinite I'm referring to when the line is vertical and the derivative is undefined- when the current changes instantaneously. And I guess saying "entire time" was pretty unclear- I meant at any instance over the course of the given time interval. Okay, so it looks like my misunderstanding lies in interpreting the behavior of the voltage and current at t=0 and t=0.2. I know that v(t) will of course be the same over the 20 ohm resistor and the inductor since they are in parallel, but I don't understand how to establish the initial conditions at t = 0+, and I feel like that has something to do with the behavior of the current due to the inductor. Are you saying then that in the moment that the current changes, it cannot go over the inductor and goes entirely over the branch with the resistor?
The graph is for the current through the current supply. There is nothing that says that the current through a current supply can't change instantly when it is turned on and off. The voltage across the inductor will be zero when the current through the inductor is zero. The current through the inductor is NOT always the same as the current through the current source. You find the conditions at t=0- and then apply the changes, noting that things that cannot change instantaneously must be the same at t=0+ as they were at t=0- Current does NOT go "over" things! It goes THROUGH them! And since the current in an inductor cannot change instantaneously, any instantaneous change in current coming into or out of a node to which an inductor is connected must be due to the other branches connected to the node.
Right. But it will be until t = 0 because the circuit will have reached steady state. Yeah. So iL(0-) = iL(0+) = 100mA. Whoops. Duly noted. So that means v(0+) = iL(0+)*R = 100mA*20Ω = 2 V?
You are making progress, but not quite there. You need to always keep in mind that voltage and current are SIGNED quantitites and that the sign matters. Your v(t) has a specified polarity. What is the direction of current through the resistor that is consistent with this polarity? What is the direction of the current that is actually flowing through the resistor at t=0+? You have v(t) = R*i_R(t) You have i_L(0+) = 100mA and the implied direction is downward through the inductor (and this shouldn't be implied -- you would put an arrow on the diagram to indicate the symbolic direction of i_L). So what is i_R(0+) in terms of i_L(0+)?
I thought all the signage was consistent in this problem...? The current from the current source will flow clockwise and down through the two branches, which is consistent with the polarity of the voltage, so it's all positive...right? I don't know why it wouldn't be. And I thought that based on what you had said, at t = 0, i_R = i_S. Because no current can flow through the inductor.
We are now at t=0+, the current source is no longer connected and the current through the inductor is 100mA. You just said that the current will now flow through the resistor and has a voltage drop of 2V. Is this current flowing in the direction that you have assigned it from when the current source was connected? I'm confused, what is i_S?
If I were to tell you that v(t) happened to be 10V at t=100s, what is the magnitude and the direction of the current flowing through the resistor? Top-to-bottom or bottom-to-top? If you say that v(0+) is 2V, what direction is the current flowing through the resistor? No. When the switch is open, what is the required relationship between i_L (the current flowing DOWN through the inductor) and i_R (the current flowing DOWN through the resistor)? If nothing else, go back to the KCL equation at the top node: i_S = i_R + i_L if the current source is turned off, then i_S=0. So what is i_R in terms of i_L?
The current source is no longer connected?! Now I'm lost. Why would that be the case? The current supplied by the source.
Hold on, I think I see what you're saying. When the current source is turned off, it is essentially not a part of the circuit since a non-operational current source acts as an open circuit. So the inductor and resistor are in series, and the 100 mA current that was flowing down through the inductor now flows UP through the resistor, creating a voltage of -2 V. Right?
Well, I'm going to need 2 equations- one for 0 < t < 0.2 and one for t > 0.2. And eventually I want the voltage, but I guess if I find the current first that'll be trivial. So I want to find the final condition i(∞) when the current source is off. That would be 0, would it not? The inductor will continuously decrease voltage in the circuit, which means current also decreases given that the resistance remains constant. In that case, v(t) = 0 + (-2 V - 0)*e^(-(R/L)*t) v(t) = -2*e^(-4000*t) V for 0 < t < 0.2 ms
It looks good so far. If you want to plot the result using 20 points, you will need to step t by 0.01 ms.
You are definitely getting there. I do recommend that you think about the chain of events a bit differently. The way you expressed it makes it seem like the inductor, for some unknown reason, is going to decrease the voltage and that that is why the current in the resistor is going to decrease. It is better to think of them as close to cause and effect as you can. You have an inductor that has a constant current flowing in it and, hence, zero voltage across it meaning that zero current is flowing in the attached parallel resistor. Once the current source is removed (or turned off, same thing) the current that was flowing in the inductor must remain continuous. At this point the only path available for the inductor current is the resistor. Therefore the inductor produces (or, more correctly, induces) a voltage across it's terminals that is sufficient to drive the inductor's present current through the available resistor. In order to keep the current flowing in the same direction, the voltage that is induced is negative compared to the current flow. As a result, that voltage not only causes a current to flow in the resistor, it also causes the current in the inductor to start changing because v(t)=Ldi/dt and, since this voltage is negative relative to the current, the current starts decreasing in magnitude. There are other ways to look at it, particularly from a power/energy viewpoint. It would be good practice to see if you can write an explanation that is based principally from that perspective. The other thing that I highly recommend you start doing is to track and use the proper units. All the time. Everywhere. Religiously. With a little practice it will become second nature and you will be amazed how much your grades will improve as you start catching your own mistakes early on. You aren't doing too bad, so it won't take much effort to really make the leap. You answer is v(t) = -2*e^(-4000*t) V But how is anyone using this result supposed to know if 't' is in seconds, milliseconds, hours, or what? You time constant was L/R = 5mH/20Ω = 0.25ms I would recommend you just use that directly in your answer v(t) = -2*e^(-t/0.25ms) V or v(t) = -2*e^(-t/250μs) V Though I think the first one makes more sense in this case. You now can immediately see that your voltage should basically reach zero in just a bit over 1ms since that would be four time constants. Not nearly so obvious from your expression. I would also tend to write v(t) = -2V * e^(-t/0.25ms) because I prefer to keep units associated with coefficients as much as possible. My contention is that this makes error tracking and checking easier and more reliable. But that is a stylistic choice on my part and putting it at the end, as you did, as its merits, too.