Question

The resistance $R=\frac{V}{I}$, where $V=(50\pm 2)V$ and $I=(20\pm 0.2)A$. The percentage error in resistance is $\text{'}x\text{'}\%$. The value of $\text{'}x\text{'}$ to the nearest integer is _________.

Answer 1

Step 1: Given data

Voltage $\left(V\right)=(50\pm 2)V$

Current $\left(I\right)=(20\pm 0.2)A$

The percentage error in resistance is $\frac{∆R}{R}\times 100$ = $\text{'}x\text{'}\%$

Step 2: Formula used

From Ohm's law

It is understood that the resistance $R=\frac{V}{I}$. It means the resistance depends upon the voltage and current. Therefore,

The percentage error in resistance will be equal to the combined error of voltage and the current. So,

The maximum percentage error can be computed using the following formula:

$\left(\frac{∆R}{R}\right)\times 100=\left(\frac{∆V}{V}\right)\times 100+\left(\frac{∆I}{I}\right)\times 100$

Where $∆R=$Change in resistance, $∆V=$Change in voltage, $∆I=$Change in current

Step 3: Compute the value of $x$

From comparing the given equation of voltage $\left(V\right)=(50\pm 2)volt$ with $V\pm ∆V$

Thus, $V=50V,∆V=2V$

Comparing the given equation of current current $\left(I\right)=(20\pm 0.2)A$ with $I\pm ∆I$

Thus, $I=20A,∆I=0.2A$

Substitute the known values in the formula,

$$\begin{gathered} x=\left(\frac{2}{50}\right)\times 100+\left(\frac{0.2}{20}\right)\times 100 \\ x=4+1 \\ x=5\% \end{gathered}$$

Hence, the value of $x$ will be $5\%$.