A coil of resistance `20Omega` and inductance `0.5H` is switched to `DC200V` supply. Calculate the rate of increase of current
a. at the instant of closing the switch and
b. after one time constant.
c. Find the steady state current in the circuit.

Explanation

Here, `R = 20 Omega, L = 0.5 H, E = 200 V`
As `I = I_(0) (1 - e^((-R)/(L) t))`
`:. (dI)/(dt) = I_(0) [0 - ((-R)/(L)) e^((-R)/(L))]`
`(dI)/(dt) = (I_(0) R)/(L) e^(-(R//L)t) = (E)/(L) e^((-R)/(L) t)`
(i) When the switch is just closed, t = 0
`(dI)/(dt) = (E)/(L) xx 1 = (200)/(0.5) xx 1 = 400 As^(-1)`
(ii) At `t = (L)/(R )`,
`(dI)/(dt) = (E)/(L) e^(-1) = (200)/(0.5) xx (1)/(2.718) = 147.17 As^(-1)`
(iii) Steady value of current
`I_(0) = (E)/(R ) = (200)/(20) = 10 A`