The heat generated in a circuit is given by $\mathrm{Q}=\mathrm{I}^{2} \mathrm{Rt},$ where I is current, $\mathrm{R}$ is resistance and tis time. If the percentage errors in measuring I. $\mathrm{R}$ and $\mathrm{t}$ are $2 \%, 1 \%$ and $1 \%$ respectively, then the maximum error in measuring…

Question

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

Aicharya · Accepted Answer

$\begin{aligned} \frac{\Delta Q}{Q} 	imes 100 &=\frac{2 \Delta I}{I} 	imes 100+\frac{\Delta R}{R} 	imes 100+\frac{\Delta t}{t} 	imes 100 \ &=2 	imes 2 \%+1 \%+1 \%=6 \% \end{aligned}$

The heat generated in a circuit is given by $\mathrm{Q}=\mathrm{I}^{2} \mathrm{Rt},$ where I is current,…

Solution