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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

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

1 Answer

$\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 $d$


answered Jun 19, 2013 by meena.p
edited Jun 20, 2013 by meena.p

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