I want to check my answers for this problem: Part c) below doesn't agree with the book. And even though part a) agrees, I'm not sure about my method. I've attached a figure for reference. For the attached figure, the switch is closed fo t<0, and steady-state conditions are established. The switch is thrown open at t = 0. a) Find the initial voltage across L just after t = 0. Which end of the coil is at higher potential: A or B? c) How long after t = 0 does the current in the 6kOhm resistor have the value 2.00 mA. For part a), I used Kirchhoff's loop rule to get the currents at steady state. 6kOhmI1 = 18V I1 = 3mA 2kOhmI2 - 0.4H(di2/dt) = 18V ==> di2/dt = 0 at steady state. I2 = 9ma Itotal = 12mA Once the switch is thrown open I have an RL series circuit. In a series circuit I should be the same across both resistors so: I1 goes to zero, so I total in the circuit is 9mA. 6kOhmI + 2kOhmI - 0.4H(di/dt) = 0 0.4(di/dt) = 8kOhmI = 8kOhm(9mA) = 72Volts; B has the higher potential. If my reasoning above is correct then for part c) I = Iinitial(e^(-t/T)) where T = L/R = .4/8000 = 50 microseconds 2mA = 9mA(e^(-t/50us) ln(1/6) = -t/50us t = 50ln(1/6) = 75.2 microseconds. The answer is the book is 75.2us Thanks in advance for the help.