Unformatted text preview: t = 0, just after the switch is closed. Now, just after the switch is closed, the inductor prevents the rapid buildup of current in its branch, so at that time, i 2 = 0 and i 1 = i . Thus i = i , so i 1 = ie − Rt/L and i 2 = i − i 1 = i ± 1 − e − Rt/L ² . (b) When i 2 = i 1 , e − Rt/L = 1 − e − Rt/L , so e − Rt/L = 1 2 . Take the natural logarithm of both sides and use ln(1 / 2) = − ln 2 to obtain ( Rt/L ) = ln 2 or t = L R ln 2 ....
View
Full
Document
This test prep was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.
 Spring '08
 Reich
 Current

Click to edit the document details