Hello eveeryone! exam time monday! Just doing some last checks to make sure i'm understanding this correctly. Here is the directions and circuit: THe circuit shown below has beeen in the form shown for a very long time. THe switchopens at t =0. FInd iR at t equal to (a)0-, (b) 0+, (c) infinity, and (d) 1.5 I'm showing u my work for (a) and (b). FOr par (a) WHat i'm confused on is, U see that wire that says at time t =0, its going to switch open. At time t < 0, does that mean there is just a wire there? Does that mean all the current is going to go through that wire and say screw all the other compoents like the 60, and 40 ohm resistors? THe answer in the back of the book is iR = 0; But is it 0 because all the current will go into that wire with no resistance? I remember the professor said, if the inductor is shorted(in this case it would be because the circuit has been sitting there for a long time) then anything in parellel with that shorted inductor is also shorted out? For part (b) iR = 10mA Is this iR 10mA because when that wire is opened that means the conductor isn't shorted but acts as a huge resistor. Not wanting any current to go through it, so it makes all the current go through the 60 ohm resistor? For partt (d) 1.5ms I need to find x = L/Req, to find Req, ur suppose to "look" into the inductor and take out all power supplies, in this case I think you would be left with 60 and 40 in parellel, so i got Req = 4mA, L = .1H I know i'm going to use the equation: iL(t) = io*e^t/x, where x = L/R, L being the indcutor, R being the equivlent resistance, but what io do i use? io is the initial current i though, usually i use the io i found at time i(0-) but in this case its 0! The book is getting an answer of 5.34mA. Thanks!