At time t=0, a potential difference of 475 V is applied to a coil with an inductance L=0.642 H and a resistance R=36 ohms. How long does it take to reach 20 percent of maximum energy in the inductor?

Answer 1

V = L di/dt + i R
i is of form i = (V/R)(1 - e^-kt)
di/dt = (V/R)k e^-kt
so
V = (L V/R )k e^-kt +R(V/R)(1-e^-kt)
1 = (L/R)k e^-kt + 1 - e^-kt
0 = (L/R)k -1
so
k = R/L
so
i = (V/R) (1 - e^-(R/L) t )
U = energy stored in magnetic field of inductor = (1/2) L i^2
U = (1/2) L V^2/R^2 (1-e^-kt)^2
this is maximum when i is maximum which is as t--> oo
U max = (1/2)L(V/R)^2
.2 Umax = 0.1 L(V/R)^2
so when is that true?
0.1 L(V/R)^2 = .5 L(V/R)^2 (1-e^-kt)^2
.2 = (1-e^-kt)^2
1 - e^-kt = sqrt .2 = .447
e^-kt = 1-.447 = .552786
1/e^kt = .552786
e^kt = 1.81
kt = ln 1.81 = .593
t = .593/k
but k = R/L = 36/.642
so
t = .593 (.642/36) = .0106 s

Answer 2

Thanks!

Answer 3

You are welcome but ALWAYS check my arithmetic :)