Question

A voltage across a lamp is $\left( {6.0 \pm 0.1} \right)V$ and the current flowing through it is $\left( {4.0 \pm 0.2} \right)A$. Find the power consumed with maximum permissible error in it.

Answer 1

Hint:The question is somewhat the same as simply applying the formula of power. Power is a product of voltage and the current. The only difference is that we are provided the voltage and current in terms of error. So, we need to find the power in terms of error. Errors are always addictive in nature.

Formula used:
Applying the formula $\dfrac{{\Delta P}}{P} = \dfrac{{\Delta V}}{V} + \dfrac{{\Delta I}}{I}$ where delta$\left( \Delta \right)$ represents the error.

Complete step by step answer:
We know that power is equal to the product of voltage and the current. $P = VI$
We are given the values in the form of errors. The facts about errors are:
Errors are always added and multiplicative in theory. The reason is because we take the logarithm first and then differentiate the equation to find the error. Hence, going back to the equation
$\dfrac{{\Delta P}}{P} = \dfrac{{\Delta V}}{V} + \dfrac{{\Delta I}}{I}$
Putting the values from the question:
$V = V + \Delta V$ where $\Delta V$ is the error
$\Rightarrow\dfrac{{\Delta P}}{P} = \dfrac{{0.1}}{{6.0}} + \dfrac{{0.2}}{{4.0}}$
Solving the above equation
$
\dfrac{{\Delta P}}{P} = \dfrac{{0.1 \times 4 + 0.2 \times 6}}{{6 \times 4}} \\
\therefore\dfrac{{\Delta P}}{P} = \dfrac{{0.4 + 1.2}}{{24}} = \dfrac{{1.6}}{{24}} \\ $
So, the power consumed is $(24 \pm 1.6)Watt$.

Note:Accuracy is the extent to which reported measurement approaches the true value of the quantity measured. The difference between the actual value and the measured value is known as error of measurement. There are three types of errors:
-Random errors
-Systematic errors
-Least count error