Question

The circuit shown below has a 100 Ohm resistor and 50 Volt-sec/amp (a unit of inductance called the Henry) inductor connected to a 12V battery. It is a freshman physics tradition that inductors are very poorly The switch is closed at t = 0 seconds. What is the magnitude of the current in amps at t = 1 second?drawn.

Answer 1

V = i R + L di/dt

i will be of form i = I (1-e^kt)
where I is the current after a long time determined only by the battery and the resistance because the inductor has no resistance once steady stage is reached.
at t = infinity, i =V/R
so I = V/R
and at t = 0 , V = L di/dt and di/dt=-kI
so V/L = -k V/R and therefore k = -R/L
so
i = V/R (1-e^-Rt/L)
i = (12/100)(1-e^-100t/50)
= .12(1 - e^-2t)
if t = 1 then
i = .12 (1- e^-2 )
e^-2 = .1353
so
i = .12 (1 - .1353)
= .104 amps




di/dt = -Ike^-kt
V = i R + L Ik(-e^-kt)
V = IR(1-e^kt) - LIk e^-kt
V = I R -I (R + Lk)e^-kt
at t = 0, V = I R
so R + Lk = 0

I = V/R =12/100 = .12 amps

Answer 2

forget the last few lines there, you do not need it