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Problem Description

I am trying to find the current through the inductor in this circuit. When I approach it, I get a differential equation that does not seem like something I've learned how to solve. Am I making a mistake somewhere? Is there another approach that gets me a simpler differential equation to solve?

Here are my steps:

enter image description here

I use current distribution. Step2

Then I get this. It is a differential equation where the coefficient of y' is y.

Edit: Ignore the text in parenthesis: the voltage source turns on at t = 0

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  • \$\begingroup\$ Which "voltage source turns on at t=0?" If the current source goes to 0 at t=0, then it has infinite impedance and is "open." But I don't see any voltage source. Perhaps I'm missing something? O(r is it just that the inductor is the only source at t>=0?) \$\endgroup\$ – jonk Jun 1 '19 at 23:11
  • \$\begingroup\$ Sorry that was a typo in the problem. The current source turns off a 0s from 4mA. \$\endgroup\$ – mk3009hppw Jun 1 '19 at 23:20
  • \$\begingroup\$ First, your first equation only has one \$R\$, not \$R_1\$ and \$R_2\$. Second, if you stand by it, how did you arrive at it? \$\endgroup\$ – TimWescott Jun 1 '19 at 23:59
  • \$\begingroup\$ TimWescott: The current through R1 is given. so it doesn't matter. I am simply using the current division rule for I1 at the junction between the inductor and R2. \$\endgroup\$ – mk3009hppw Jun 2 '19 at 0:36
  • \$\begingroup\$ Your question is a mess, making it hard for anyone to understand and respond to. Please remove the voltage source comment from the image, put your equations in the right order so we can see what the "steps" are and where they came from, clarify "current distribution" and clarify "y". \$\endgroup\$ – Heath Raftery Jun 2 '19 at 0:45
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I think you started by using the current divider formula to find the current in the inductor, but no part of it is correct. To use that formula you would need to find the impedance of each parallel path as well as the total impedance. It doesn't look like you've done that, and it's an unnecessarily complicated way to go anyway.

Instead, consider the two scenarios: t<0 and t>0.

For t<0 the current is constant. For constant current the impedance of an inductor is zero. So all the current (4mA) is flowing through the inductor. That scenario is now done.

For t>0 the current source goes to zero. With no current in the current source branch, the circuit reduces to just two elements: L1 and R2. You then just need the inductor formula:

\$v_{L1} = L1 * \frac{di_{L1}}{dt}\$

And the resistor formula:

\$v_{R2} = R2 * i_{R2}\$

Since the voltage across the two elements must sum to zero, and the current flowing through them is the same, the two equations combine:

\$L1 * \frac{di_{L1}}{dt} = - R2 * i_{L1}\$

Which is a differential equation that probably looks a lot more familiar.

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From the other comments and answers, I think you already well-understand that at \$t=0^-\$, \$I_\text{L}=+4\:\text{mA}\$: top of your schematic to bottom in the direction of conventional current. When the current source goes to zero, \$R_1\$ has no current in it and is effectively cut out of the circuit. So all that remains is your inductor and \$R_2\$. In keeping with your own references, I'll call \$L_1\$ just \$L\$, for short. And I'll call \$R_2\$ just \$R\$, for short. I'll also call your top node, the one shared by \$R_1\$, \$R_2\$, and \$L_1\$, just simply \$V\$.

Using nodal analysis (KCL) and keeping in mind that \$I_\text{L}=\frac{1}{L}\int V\:\text{d}t\$. you find that:

$$\frac{1}{L}\int V\:\text{d} t+\frac{V}{R}=0\:\text{A}$$

Taking the differental with respect to time, this is simply:

$$\begin{align*}\frac{V}{L}+\frac{1}{R}\frac{\text{d} V}{\text{d} t}&=0\:\frac{\text{A}}{\text{s}}\\\\\therefore\quad \frac{\text{d} V}{\text{d} t}+\frac{R}{L}\,V&=0\:\frac{\text{A}}{\text{s}}\end{align*}$$

This last equation is in standard form for first order ordinary linear differential equations. If you set \$\tau=\frac{L}{R}\$, then it can be easily solved using an integrating factor, \$\mu=e^{\int\frac{R}{L}\:\text{d}t}=e^{^{\frac{t}{\tau}}}\$. So:

$$V=\frac{1}{\mu}\int0\cdot\mu\:\:\text{d}t=A\:e^{^{\frac{-t}{\tau}}}$$

where \$A\$ is the constant of integration that needs to be set based upon the initial conditions.

In your case, you know that \$I_\text{L}=4\:\text{mA}\$ at \$t=0\$. The sum of the two currents, that in \$L\$ and in \$R\$, must sum to zero. Therefore, \$V_{t=0}=-4\:\text{mA}\cdot 4\:\text{k}\Omega=-16\:\text{V}\$. Clearly then, \$A=-16\:\text{V}\$. From this, find: \$V=-16\:\text{V}\cdot e^{^\frac{-t}{\tau}}\$. So, for the resistor, \$I_\text{R}=\frac{V}{R}=-4\:\text{mA}\cdot e^{^\frac{-t}{\tau}}\$.

The current in the inductor is of opposite sign to the current in the resistor, so:

$$I_\text{L}=4\:\text{mA}\cdot e^{^\frac{-t}{\tau}}$$

With \$\tau=250\:\text{ps}\$, the current rapidly declines towards zero. Reaching zero, for all intents and purposes, by \$4-5\tau\$ or by \$t=1\:\text{ns}-1.25\:\text{ns}\$.

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guidance only:

Your equations are crap. I1=the current of R2 + the current of L1. At t=0 this becomes to zero. Before t=0 it's 4mA and some part of it surely goes through L1. Deduce how much it is, because that's the starting current of L1 in the beginning of the transient.

After t=0 I1=0, but L1's current doesn't stop immediately, it continues through R2 and decays. The key of the right equation is that the voltage over L1 is the same as the voltage over R2.

This is the end of the guidance.

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  • \$\begingroup\$ Thats exactly what I did to make my equation. R is R2 in my equation. I am using the current division rule to write the current through the inductor in terms of the rate of change of the current through the inductor. \$\endgroup\$ – mk3009hppw Jun 2 '19 at 0:38
  • \$\begingroup\$ @mk3009hppw you should learn that the voltage over an ideal inductor is L(di/dt) without multiplying it with anything. You should learn that voltage over a resistance is Current x R. You should write an equation where L1 and R2 have same current and voltage except the current directions are up-down and down-up. \$\endgroup\$ – user287001 Jun 2 '19 at 0:53
  • \$\begingroup\$ @mk3009hppw (continued) and the sameness is true after t=0 because then nothing comes nor goes through R1 \$\endgroup\$ – user287001 Jun 2 '19 at 1:01

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