Prof said one thing, book says another. R/L source-free circuit, pic included

[mr_coffee]

Prof said one thing, book says another. R/L source-free circuit, pic included :)

Hello everyone. This time after reviewing my professors work, and comparing it to the books solution manual i bought, they don't seem to match. The way the professor explained it to me made me think she was right. She said since it wants u to find Isw @ 5x10^-6 seconds, the current is diminshed enough that u can say its gone. Since IL and Isw are in opposite directions, Isw is just -IL. But the book said f that.
Here is my work and the problems directions:
After being in the configuration shown for hours, the switch in the circuit is closed at t = 0. At t = 5micro seconds, calculate (a) IL and (b) Isw
http://img236.imageshack.us/img236/9536/lastscan0zc.jpg

 

[nrqed]

Hello everyone. This time after reviewing my professors work, and comparing it to the books solution manual i bought, they don't seem to match. The way the professor explained it to me made me think she was right. She said since it wants u to find Isw @ 5x10^-6 seconds, the current is diminshed enough that u can say its gone.

The current where? This does not make sense to me because the time is not much larger than the time constant! So no approximation of that sort can be made

Since IL and Isw are in opposite directions, Isw is just -IL. But the book said f that.
Here is my work and the problems directions:
After being in the configuration shown for hours, the switch in the circuit is closed at t = 0. At t = 5micro seconds, calculate (a) IL and (b) Isw

In a circuit like this, where the middle branch had no resistance (when the switch is closed), you can treat the two loops (left and right) separately and combine the two currents in the middle branch, where the switch is (that's a special case!). So the effect of the left loop (with the 9 volts and the 1 kilo-ohms) is to produced a current downward of 9/1000 = 9 mA downward. Then you can calculate separately the current produced by the right branch (with the inductor and the other 1 kilo-ohm. Upon combining the two currents in the middle branch, you find the current in the switch.

 

[mr_coffee]

That does make sense, I think her assumption was wrong because like you said, the time isn't very long at all, so there could still be current flowing through the inductor. Thanks again for the explanation!