Question

(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.

Answer 1

(a) N//A
(b) `A = 200 cm^(2)`
`= 200 xx 10^(-4)m^(2) = 2 xx 10^(-3) m^(2)`
`n= 20`
`omega = 50 "Rad" s^(-1)`
`B = 3 xx 10^(-2)T`
Max. emf `E_(o) = 3 xx 10^(-2) T`
`= 20 xx 3 xx 10^(-2) xx 50 = 6000 xx 10^(-4) = 0.` Volt
Max. Current `I_(0) = (E_(0))/(R ) = (0.6)/(R )A`.