Three resistors having values `R_1,R_2 and R_3` are connected in series to a battery.Suppose `R_1` carries a current of 2.0 A,`R_2` has resistance of 3.0 ohms, and `R_3` dissipates. 6.0 watts of power. Then the voltage across `R_3` is

Options

A. 1 V
B. 2 V
C. 3 V
D. 4 V

Answer

C

Explanation

Given `I_(1)=2Am R_(2)= 3 Omga, P_(3)=6W`
We know in series current remains constant which means current in `R_(3) is I_(3)=2A`.
Power `P=VI`
`rArr V=(P)/(I)=(6W)/(2A)=3V`