Browse Questions

# Heat generated in a circuit is given by $H=I^2Rt$, $I$ is current $R$ is resistance, $t$ is time. If percentage error in measuring $I$, $R$ and $t$ are $2 \%,1\%$ and $1\%$ respectively. what is the maximum error in measuring heat

$(a)\;2 \%\quad (b)\;3\% \quad (c)\;4 \%\quad (d)\;6 \%$

$\large\frac{\Delta H}{H}$$=2 \large\frac{\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t} \large\frac{\Delta H}{H}$$\times 100= 2 \times 2 \%+1 \%+1 \%$
$\qquad\qquad=6 \%$
Hence the correct option is $d$

edited Jun 20, 2013 by meena.p