Question

In the circuit as shown in figure, the energy supplied by the battery is calculated using a stop watch with an error of $\pm 2\%$ in time measurement. The used resistor having tolerance $\pm 6\%$ and the ammeter error is $\pm 5\%$. The switch S is closed at $t=0$ and opened at time $t$. What is the percentage of error in energy used in the circuit ?

• A. $18$ %
• B. $13$ %
• C. $16$ %
• D. $10$ %

Answer 1

Answer: (A)

$18$ %

Energy used in the circuit at time $t$ is $E=i^{2}Rt$
Take $ln$ and then differentiate:
We get the relative error, $\frac{\Delta E}{E}=2\frac{\Delta i}{i}+\frac{\Delta R}{R}+\frac{\Delta t}{t}$

The percentage error in energy $=\frac{\Delta E}{E}\times 100=(2\frac{\Delta i}{i}\times 100)+(\frac{\Delta R}{R}\times 100)+(\frac{\Delta t}{t}\times 100)$
$=(2\times 5)+6+2=18$%