Jitender Pal
Sol. We know i = i0 ( 1- e^–t/r)
(a) 90/100 i base 0 = i base 0 ( 1-e^-t/r)
⇒ 0.9 = 1 – e^-t/r
⇒ e^-t/r = 0.1
Taking ℓn from both sides
ℓn e^–t/r = ℓn 0.1 ⇒ –t = –2.3 ⇒ t/r = 2.3
(b) 99/100 i base 0 = i base 0 (1 – e^-t/r)
⇒ e^–t/r = 0.01
ℓne^–t/r = ℓn 0.01
or, –t/r = –4.6 or t/r = 4.6
(c) 99.9/100 i base 0 = i base 0 (1-e^-t/r)
e^–t/r = 0.001
⇒ lne^–t/r = ln 0.001 ⇒ e^–t/r = –6.9 ⇒ t/r = 6.9.
