In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t
`i = 20 sin( 30t-(pi)/(4))` In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

Options

A. `(50)/(sqrt(2))` , 0
B. 50 , 0
C. 50 , 10
D. `(1000)/(sqrt(2))` , 10

Answer

D

Explanation

(d) `V= 100 sin 20t`
`I = 20 sin (30 t - (pi)/(4))`
Hence phase difference between voltage and current is `Phi = pi//4`.
`V_(0)= 100 implies V_("rms") = 100//sqrt(2)`
`I_(0)= 20 implies I_("rms") = 20//sqrt(2)`
Average power consumed can be written as follows :
`P_("avg")= V_("rms")I_("rms") cosPhi`
`= (100)/(sqrt(2))xx(20)/(sqrt(2))xx(1)/(sqrt(2))= (1000)/(sqrt(2))` W
Wattless current `=I_("rms") sinPhi`
=`(I_(0)sinPhi)/(sqrt(2))= (20)/(sqrt(2))xx(1)/(sqrt(2))= 10A.`
Hence option (d) is correct.