Two resistors of `6 Omega " and " 9 Omega` are connected in series to a 120 V source. The power consumed by the `6 Omega` resistor is

Options

A. 384 W
B. 576 W
C. 1500 W
D. 1800 W

Answer

A

Explanation

V=120 V
`V=V_(1)+V_(2)`
`V_(1)=iR_(1)`
`V_(2)=iR_(2) implies (V_(1))/(R_(1)) =(V_(2))/(R_(2))`
From Eq. (i), `V=V_(1)+V_(2)=V_(1)+(R_(2))/(R_(1))V_(1)`
`V=V_(1)[(R_(1)+R_(2))/(R_(1))] implies V_(1)=(120xx6)/(15)=48 V`
Power in ` 6 Omega, P_(1)=(V_(1)^(2))/(R_(1))=(48xx48)/(6)implies P_(1)=384 W`