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This is a homework problem for my circuit analysis class. I can't figure out what the heck I did wrong. I know the solution to the relevant ODE is of the form I(t)=k1+k2*e^(-t/(R/L)), so I just need the values of the constants. Since the switch is being opened, I(infinity)=0 which means that K1=0 and K2=I(0+). Since this is an RL circuit, we should have that I_L(0-)=I_L(0+), right? I found I_L(0-) by drawing the circuit with the switched closed and the inductor as a short circuit, then I used loop analysis to get I. I know that part is right, because I plugged that version of the circuit into a simulator and got the same answer. Unless I'm misunderstanding when I(0-)=I(0+) (and if I am, please tell me), that's the value of K2. K1 is zero because the current will quickly go to zero after the switch is opened, making I(infinity)=0. L is given as 2H and I found R_th by drawing the circuit for t=infinity, removing the sources, and getting the equivalent resistance, which is how you get the Thevenin resistance value when there are only independent sources present, yes? I tried it with R_th=4, 12, and 4||12=3 as well as 4+12=16, but all of those answers were marked wrong as well. Can someone please help?

  • 1
    \$\begingroup\$ I'll look at this in a dozen hours or so. It looks like you put in some work and is like to help if I can. Just need to sleep now and then do some work for others tomorrow. (I'm not retired.) But I'll see if I can help out, later. \$\endgroup\$
    – jonk
    Commented Sep 8, 2020 at 14:29
  • \$\begingroup\$ I guess I need to assume where the switch was. If so, I'd guess it's located where the blue colored "t=0" is located. Is there a reason why you didn't bother yourself to use the schematic editor that is available to you here? \$\endgroup\$
    – jonk
    Commented Sep 9, 2020 at 3:22

1 Answer 1


I hope that you can readily follow the schematics and their transitions, below:


simulate this circuit – Schematic created using CircuitLab

Prior to the switch being opened at \$t=0^+\$, \$L_1\$ has a fixed current in it and therefore has zero voltage across it. I've also then swapped positions of \$V_2\$ and \$R_2\$ (which is perfectly valid for a series pair like this) and so, on the right, you see the equivalent circuit to analyze in order to work out the steady state current in \$L_1\$ at \$t=0^-\$. This is, clearly, just \$750 \:\text{mA}\$, and in the direction shown.

Upon opening the switch, \$V_1\$ and \$R_1\$ are no longer relevant. So they can be removed, entirely. What remains, at \$t=0^+\$, is then shown on the left side of the bottom row. It's equivalent circuit is then shown on the right. (I've chosen a convenient place for ground on the left.) You should be able to see this transformation and agree with it, as well.

This is really a very simple remaining circuit to analyze. The nodal equation is:

$$\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t=\frac{6\:\text{V}}{R_\text{SUM}}$$

Now, we take the derivative of the above in order to help simplify the solution and arrive at:

$$\begin{align*} \frac{\text{d}}{\text{d}t}\left[\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t\right.&=\left.\frac{6\:\text{V}}{R_\text{SUM}}\right]\\\\ \frac{\text{d}}{\text{d}t}\left[\frac{V_x}{R_\text{SUM}}+\frac1{L_1}\int V_x\:\text{d}t\right]&=\frac{\text{d}}{\text{d}t}\left[\frac{6\:\text{V}}{R_\text{SUM}}\right]\\\\ \frac{1}{R_\text{SUM}}\frac{\text{d}}{\text{d}t}V_x+\frac1{L_1}V_x&= 0\:\text{A} \end{align*}$$

The solution to the above (integrating factor or whatever else you may prefer) yields:

$$V_x=C_0\cdot e^{\left[-\frac{R_\text{SUM}}{L_1}\cdot t\right]}$$

We know that \$V_x=18\:\text{V}\$ at \$t=0^+\$ (you should easily see why.) Therefore, \$C_0=18\:\text{V}\$. At this point you have everything you need. You are trying to compute:

$$I\left(t=70\:\text{ms}\right)=\frac{+6\:\text{V}-18\:\text{V}\cdot e^{\left[-\frac{R_\text{SUM}=16\:\Omega}{L_1=2\:\text{H}}\cdot \,t=70\:\text{ms}\right]}}{R_\text{SUM}=16\:\Omega}$$

Which should be very easy to compute.


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