The voltage across a lamp is `(6.0 pm 0.2)` volt and current that passes through it is `(2.0 pm 0.1)` ampere. The power of the lamp in watt is :

Options

A. `(6pm0.1)`
B. `(12pm0.1)`
C. `(24 pm 0.1)`
D. `(4 pm 0.1)`

Answer

B

Explanation

`"Power "=P=VxxI`
`=6xx2`
`= 12 W`
`DeltaV=0.2`
`DeltaI=0.1`
From the principle of error we get
`(DeltaP)/(P) = pm ((DeltaV)/(V)+(DeltaI)/(I))`
`therefore (DeltaP)/(12)=pm ((0.2)/(6)+(0.1)/(2))`
`(DeltaP)/(12)=pm((0.2+0.3)/(6))`
`=pm (0.5)/(6)=pm((1)/(12))`
`therefore DeltaP=pm1`
`therefore " Power "theta=(12pm0.1)" watt"`