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 isA. 384 WB. 576 WC. 1500 WD. 1800 W

Two resistors of `6 Omega " and " 9 Omega` are connected in series to

1.	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 isA. 384 WB. 576 WC. 1500 WD. 1800 W
Answer» Correct Answer - A 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`

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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 isA. 384 WB. 576 WC. 1500 WD. 1800 W

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