Question

The heat generated in circuit $q=I^2$ Rt where I is current, R is resistance and t is time. If the percentage error in measuring I, R and t are $2\%,1$ respectively, then the maximum error in measuring heat will be

• A. $2\%$
• B. $3\%$
• C. $4\%$
• D. $6\%$

Answer 1

The correct option is D

$6\%$

Heat generated,$H=I^{2}Rt$

The error can be expressed as :

$\frac{\Delta H}{H}=2\times \frac{\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t}$

$So,\text{the required error}=2\times 2\%+1\%+1\%=6\%$