A resistance of `40 Omega` is connected to an source of 220 v, 50 Hz. Find the (i) rms current (ii) maximum instantaneous current in resistor.

Explanation

Here, `R = 40 Omegam E_(v) = 220 V, v = 50 Hz`,
(i) `I_(v) = (E_(v))/( R) = (220)/(40) = 5.5 A`
`I_(v) = sqrt 2 I_(v) = 1.414 xx 5.5 = 7.78 A`
If alternating current is given by `I = I_(0) sin omega t`, then
`I_(0) = I_(0) sin omega t_(1) = 1` or `omega t_(1) = (pi)/(2)`
and `I_(v) = (I_(0))/(sqrt 2) = I_(0) sin omega t_(2)`, which implies
`omega t_(2) (pi)/(2) + (pi)/(4)`
`:. omega (t_(2) - t_(1)) = (pi)/(2) + (pi)/(4) - (pi)/(2) = (pi)/(4)`
`t_(2) - t_(1) = (pi)/(4 omega) = (pi)/(4 xx 2 pi v) = (1)/(8 v) = (1)/(8 xx 50)`
`= 2.5 xx 10^(-3) s = 2.5 ms`