An alternating voltage given by E = 140 sin 314t

cbse
alternating-currents

#1

An alternating voltage given by E = 140 sin 314t is connected across a pure resistor of 50 Ω. Find
(i) the frequency of the source.
(ii) the rms current through the resistor.


#2

(i)As, given E = 140 sin 314 t
On comparing with E = ${ E }{ 0 }$ sin ω t , we have,
ω = 314, ${ E }
{ 0 }$ = 140V
ω = 2 $\pi$ v
v = ω / 2 $\pi$ = 314 / 2 x 3.14
= 31400/ 2 x 314 = 100/2 = 50Hz
(ii)