A resistance of `20 Omega` is connected to a source of alternating current rated `110 V, 50 Hz`. Find (a) the `rms` current, (b) the maxium instantaneous current in the resistor and the time taken by the current to change from its maximum value to the rms value.

Explanation

(a) `I_(r.m.s) = (E_(r.m.s))/(R ) = (110)/(20) = 5.5 A`
(b) `I_(0) = sqrt2 I_(r.m.s) = 1.414 xx 5.50 = 7.8 A`
Let the ac be represented by `i= i_(0) cos omega t`
At `t = 0, i= i_(0) cos 0^(@) = i_(0)` (max)
At t = t, let `i= i_(v) = (i_(0))/(sqrt2) = i_(0) cos omega t`
`:. cos omega t = (1)/(sqrt 2) = cos (pi)/(4)`
`omega t (pi)/(4)` or `2 pi v t = (pi)/(4)` ,
`t = (1)/(8 v) = (1)/(8 xx 50) = 2.5 xx 10^(-3) s`