Question 8: Units , Dimensions and Error Analysis

Question

In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A , respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA , respectively. The estimated error in the resistance measurement is  $\_\_\_\_$ $\Omega$

Accepted Answer

According to Ohm's law, the resistance R is given by the ratio of potential difference V to the current I :
$$ \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}} $$
The reading of voltmeter is $\mathrm{V}=10 \mathrm{~V}$ and that of ammeter is $\mathrm{I}=5 \mathrm{~A}$
Then resistance is $\mathrm{R}=\frac{10}{5}=2 \Omega$
When a quantity is determined by the division of two other quantities $\left(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}}ight)$, the relative error in the result is the sum of the relative errors of the individual quantities.
The relative error is,
$$ \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{V}}{\mathrm{~V}}+\frac{\Delta \mathrm{I}}{\mathrm{I}} $$
Where $\Delta \mathrm{R}$ is the absolute error in resistance, $\Delta \mathrm{V}$ is the error in voltage, and $\Delta \mathrm{I}$ is the error in current.
The error in a measurement is taken as the least count of the instrument used.
So, error in voltmeter is $\Delta \mathrm{V}=500 \mathrm{mV}=0.5 \mathrm{~V}$
And the error in ammeter is $\Delta \mathrm{I}=200 \mathrm{~mA}=0.2 \mathrm{~A}$
Substituting values into the error equation
$$ \frac{\Delta \mathrm{R}}{2}=\frac{0.5}{10}+\frac{0.2}{5} $$
$$\Rightarrow $$ $\frac{\Delta \mathrm{R}}{2}=0.05+0.04=0.09$
$$\Rightarrow $$ $$ \Delta \mathrm{R}=0.09 	imes 2=0.18 \Omega $$
Therefore, the estimated error in the resistance measurement is $0.18 \Omega$.
Hence, the correct option is (D).

Answer

0.25

Answer

2

Answer

2.5

Answer

0.18

Units , Dimensions and Error Analysis

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