The heat dissipated in a resistance can be obtained by the measurement of resistance, current and time. If the maximum percentage error in the measurement of these quantities are `1%,2% and 1% ` respectively. The maximum percentage error in the determination of the dissipated heat is -

Options

A. 0.04
B. 0.06
C. `4/3%`
D. `2%`

Answer

B

Explanation

The heat dissipatd in a resitance R carrying current I in the t is given by `H=I^(2)Rt`
Take logs and differentiate both sides we get
`(DeltaH)/H=(2DeltaI)/I+(DeltaR)/R+(Deltat)/t" ":.(DeltaH)/Hxx100=2(2%)+1%+1%=6%`