In the circuit shows in Fig. , `E = 10 V`, `R_(1) = 5 Omega`, `R_(2) = 10 Omega`, and `L = 5 H`. For the two separate conditions, (i) switch `S` is just closed and (ii) switch `S` is closed for a long time, calculate
a. current `i_(1)` through `R_(1)`,
b. current `i_(2)` through `R_(2)`,
c. current `i` through the switchs,
d. the potential difference across `R_(2)`,
e. the potential difference across `L`,
f. `di_(2)//dt`.

Explanation

(I) At `t = 0`, inductor acts as open circuit:

a. `i_(l) = (E)/(R_(1)) = 2A` b. `i_(2) = 0A`
c. `i = i_(1) + i_(2) + 0 = 2 A` d. `V` across resistor `R_(2) = 0`
e. `V` across `L = 10 V`
f. `L(di_(2))/(dt) = 10 rArr (di_(2))/(dt) = (10)/(5) = 2 A//s`
(II) At `t = oo`, inductor acts as a conducing wire:

a. `i_(1) = (E)/(R_(1)) = 2 A`
b. `i_(2) = (E)/(R_(2)) = (10)/(10) = 1 A`
c. `i_("switch") = i_(1) + i_(2) = 3 a`
d. `V_(R2) = 10 V`
e. `V_(L) 0 V`
f. Current in `R_(2)` is not changing with time, so
`(di_(2))/(dt) = 0`