A 45-mH ideal inductor is connected in series with a 60 ohm resistor through an ideal 15- V-DC power supply and an open switch. If the switch is closed at t 0, what is the current 7.0 ms later?
The time constant T is T L/R. So for your problem T 45x10^-3H/60 Ohm 7.5x10^-4 s Now the current in the curcuit after a very long time is simply I0 V/R 15/60 0.25 A. The current as a function of time is given by I(t) I0*(1 - e^(-t/T)) set t 7x10^-3 and solve for I I 0.25 A* (1 - e^(-7x10^-3/(7.5x10^-4))) 0.2499 A the current has been flowing for almost 10 time constants which means it is almost at full value.
