Question

If across a resistance, voltage applied is \((20 \pm 0.2)\,\text{V}\) and the current passing through it is \((10 \pm 0.1)\,\text{A}\). Find the maximum error in resistance.

• A. \(0.04\)
• B. \(0.05\)
• C. \(0.06\)
• D. \(0.07\)

Hint:

For multiplication or division of quantities: \[ \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} \] Always add fractional errors.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem requires finding the maximum absolute error in a calculated quantity.
Resistance is calculated by taking the ratio of Voltage to Current based on Ohm's Law.
When quantities are divided, their relative fractional errors are added.
Step 2: Key Formula or Approach:
From Ohm's Law, $R = \frac{V}{I}$.
The maximum fractional error is given by:
\[ \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \]
Therefore, maximum error $\Delta R$ is $\Delta R = R \left( \frac{\Delta V}{V} + \frac{\Delta I}{I} \right)$.
Step 3: Detailed Explanation:
Given data:
$V = 20 \text{ V}$, $\Delta V = 0.2 \text{ V}$
$I = 10 \text{ A}$, $\Delta I = 0.1 \text{ A}$
First, calculate the nominal value of $R$:
\[ R = \frac{V}{I} = \frac{20}{10} = 2 \ \Omega \]
Now, calculate the maximum fractional error:
\[ \frac{\Delta R}{R} = \frac{0.2}{20} + \frac{0.1}{10} \]
\[ \frac{\Delta R}{R} = 0.01 + 0.01 = 0.02 \]
Finally, find the absolute error $\Delta R$:
\[ \Delta R = R \times 0.02 = 2 \times 0.02 = 0.04 \ \Omega \]
Step 4: Final Answer:
The maximum error in resistance is 0.04.