A light bulb is rated at 100 W for a 220 V supply. Find
(a) the resistance of the bulb.
(b) the peak voltage of the source
(c ) the rms current through the bulb.

Explanation

Here, `P = 100 W, E_(v) = 220 V`,
`R = ? E_(0) = ? I_(v) = ?`
From, `P = (E_(v)^(2))/(R ) , R = (E_(v)^(2))/(R ) = (220 xx 220)/(100) = 484 Omega`
`E_(0) = sqrt2 E_(v) = 1.414 xx 220 = 311 V`
`I_(v) = (P)/(E_(v)) = (100)/(220) = 0.45 A`