The heat generated in a circuit is given by `Q = I^(2) Rt` , where `I` is current , `R` is resistance , and `t` is time . If the percentage errors in measuring `I , R , and t are 2% , 1% , and 1%` , respectively , then the maximum error in measuring heat will be

Answer

`+- 6%`

Explanation

As `H = (I^2Rt)/(4.2)cal`
`:. (DeltaH)/(H) = +-((2DeltaI)/(I) +(DeltaR)/(R )+(Deltat)/(t))`
`=+-(2xx2% +1%+1%) =-6%`