The heat dissipated in a resistance can be determined by measuring resistance current and time. If the maximum error in the above quantities are `1%, 2% and 1%` respectively. Then calculate the maximum error in the determination of the dissipated heat?

Explanation

Heat developed
`H=I^(2)Rt`
`DeltaI=2% DeltaR=1% Deltat=1%`
`(DeltaH)/(H)=(2DeltaI)/(I)+(DeltaR)/(R )+(Deltat)/(t)`
`(DeltaH)/(H)xx100=2(DeltaI)/(I)xx100+(DeltaR)/(R )xx100+(Deltat)/(t)xx100`
`=2xx2%+1%+1%`
`=4%+1%+1%6%`
The maximum error in heat `=6%`