Question

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 $2\%,1\%and1\%$, respectively, then the maximum error in measuring heat will be

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

Answer 1

Answer: (D)

$6\%$

$Q=I^{2}Rt$

$\Rightarrow \frac{\Delta Q}{Q}=\frac{2\Delta I}{I}+\frac{\Delta R}{R}+\frac{\Delta t}{t}$

$\Rightarrow \frac{\Delta Q}{Q}\times 100=2\left(2\right)+1+1$

$\Rightarrow \frac{\Delta Q}{Q}\times 100=6$%