1. The problem statement, all variables and given/known data For the circuit in the attached file, find io (current) for t>0 2. Relevant equations 3. The attempt at a solution I have included an attachment which shows the solution. However, I will copy the solution here as well for reference: When t<0, the switch is closed, the inductor acts as a short circuit to dc. The question I have in this step is what happens to the resistor that is in series with the inductor when the inductor becomes a short circuit? The resulting circuit is one with the 24V source, 4 ohm resistor and the 4 ohm and 8 ohm resistors in parallel. The 4//8=8/3 ohms. i1=24/(8/3+4)=3.6 A We obtain i(t) from t1, using current division, by writing, i(t)= 3.6*4/(4+8) = 1.2A, t<0 Since the current through the inductor cannot change instantaneously, i(0)=i(0-)=1.2A When t=0, the switch is open and the voltage source is disconnected. Now we have a circuit with 4 ohm in series with the 4H inductor and the 8 ohm resistor. Req=4+8=12 ohms T(tow)=L/R=4/12=1/2 seconds io(t)=io(0)e^(-t/tow) io(t)=1.2e^-3t The solution manual shows that io(0)=1.4118 A. I am not sure how they obtained it. I will post the solution for reference as well. Can someone tell me where I went wrong?