A resistor and an inductor are connected in series to an AC source of voltage 150 sin`(100 pit +pi) ` volt. If the current in the circuit is 5 sin `(100 pit + (2pi)/(3))` ampere then the average power dissipated and the resistance of the resistor are respectively

Options

A. `187.5 W 30 Omega`
B. `187.5 W 15 Omega`
C. `375 W 30 Omega`
D. `375 W 15 Omega`

Answer

B

Explanation

Given peak voltage `V_(0)= 150 v `
peak current `I_(0)= 5 `
Phase difference between voltage voltage and current
`phi = pi - (2pi)/(3) = (pi)/(3)= 60^(@)`
Now average power in R-L circuit ,
` P_(av)= V_("rms"). I_("rms"). Cos phi`
`=(V_(0))/(sqrt(2)).I_(0)/(sqrt(2))cos60^(@)= (150)/(sqrt(2))xx(5)/(sqrt(2))xx(1)/(2)`
`P_(av)= 187.5 W `
Impendence Z `= (V__(0))/(I_(0))= (150)/(5)`
Z= 30 `Omega`
`(R)/(Z)= cos phi`
`(R)/(30)= cos 60^(@) implies R=30xx(1)/(2)`
`R = 15 Omega`