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 (c) the time taken by the current to change from its maximum value to the rms value.

Explanation

(a) The rms potential difference `= 110 V` and so the rms current `= (110 V)/(20 Omega) = 5.5 A`.
(b) The maximum instantaneous current
`= sqrt(2) (rms current)`
`= sqrt(2) xx 5.5 A = 7.8 A`.
(c) Let the current be `I = i_0 sin ometat`.
It `t_1 and t_2` be the time instants for consecutive appearances of the maximum value and the rms value of the current,
i_0 = i_0 sin omegat_1`
and `i_0/(sqrt) 2 = i_0 sin omegat_2`.
If `omegat_1 = (pi)/2 , (omegat_2 = (pi)/2 + (pi)/4`.
Hence, `t_2 - t_1 = (pi)/(4omega)`
`= (pi)/(4 xx 2piv) = 1/(8 xx 50) s = 2.5 ms`