Question

The resistance of a metal is given by \[R = \dfrac{V}{I}\], where $V$ is potential difference and $I$ is the current. In a circuit, the potential difference across resistance is $V = (8 \pm 0.5)V$ and current in resistance, $I = (4 \pm 0.2)A$. What is the value of resistance with its percentage error?
$
  (A){\text{ }}(2 \pm 5.6\% ) \\
  (B){\text{ }}(2 \pm 0.7\% ) \\
  (C){\text{ }}(2 \pm 35\% ) \\
  (D){\text{ }}(2 \pm 11.25\% ) \\
 $

Answer 1

Hint: In this question use the concept of the Ohm’s Law, find the resistance according to the given formula in the question. Then go for finding the percentage error part in resistance. At last take together the actual value of resistance and the percentage value in resistance.

Complete step by step solution:
As we know that Ohm’s Law says that the current through a particular conductor between two points is directly proportional to the voltage across the two points. Here the proportionality constant is said as resistance.
As in the question, the formula for resistance is given as \[R = \dfrac{V}{I}\]. Where, $V$ is potential difference and $I$ is the current.
According to the question we can see that the value of $V$ and $I$ are given as $V = (8 \pm 0.5)V$ and $I = (4 \pm 0.2)A$ respectively.
Here, let we first calculate the value of resistance using the formula as given \[R = \dfrac{V}{I}\].
Taking $V = 8V$ and $I = 4A$ (as given in the question (here we are excluding the $ \pm $ part because that is considered as error)
So, $R = \dfrac{8}{4}\Omega = 2\Omega $
Now, percentage error formula for resistance is given by:
$
  \dfrac{{\Delta R}}{R} \times 100 = \left( {\dfrac{{\Delta V}}{V} \times 100} \right) + \left( {\dfrac{{\Delta I}}{I} \times 100} \right) \\
   \Rightarrow \dfrac{{\Delta R}}{R} \times 100 = \left( {\dfrac{{0.5}}{8} \times 100} \right) + \left( {\dfrac{{0.2}}{4} \times 100} \right) \\
   \Rightarrow \dfrac{{\Delta R}}{R} \times 100 = 6.25 + 5 \\
   \Rightarrow \dfrac{{\Delta R}}{R} \times 100 = 11.25\% \\
 $
So, the percentage error in resistance is given by $11.25\% $.
Now considering the actual resistance with percentage error in resistance, we get
$R = (2 \pm 11.25\% )\Omega $

So, option $(D)$ is the correct option.

Note: As we know that the selection of the actual value of resistance is the main point in the provided question. Percentage error calculation should be done carefully by selecting the part of error from the provided values.