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.

# An alternating voltage given by E = 140 sin 314t

**prasanna**#1

**prasanna**#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)