If the measured values of the voltage across and the current through a resistor are (100 ± 5) V and (10 ± 0.2) A respectively, then the error in the determination of the resistance is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q81 – AP-EAMCET 2025

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Home Physics Units & MeasurementsQuestion #546

If the measured values of the voltage across and the current through a resistor are

(100 \pm 5) V

and

(10 \pm 0.2) A

respectively, then the error in the determination of the resistance is

Clue 1 of 3: Error Propagation

When physical quantities are combined through multiplication or division, their relative (or percentage) errors add up.

🧠 Full Solution Path

Step 1: Identify Given Values and Relationship✦ Active

The resistance $R$ is related to voltage $V$ and current $I$ by Ohm's Law. We are given the measured values with their absolute errors:

V = (100 \pm 5) V ⟹ V_{0} = 100 V, Δ V = 5 V

I = (10 \pm 0.2) A ⟹ I_{0} = 10 A, Δ I = 0.2 A

💡 Teacher's Secret Hint

Remember that $V_{0}$ and $I_{0}$ represent the nominal or measured values, and $Δ V$ and $Δ I$ are their respective absolute errors.

Step 2: Calculate Percentage Error in Voltage○ Expand

The percentage error in voltage is calculated as:

Percentage error in V = (\frac{Δ V}{V_{0}} \times 100) %

Percentage error in V = (\frac{5}{100} \times 100) % = 5 %

💡 Teacher's Secret Hint

Ensure you use the correct nominal value in the denominator for calculating percentage error.

Step 3: Calculate Percentage Error in Current○ Expand

Similarly, the percentage error in current is calculated as:

Percentage error in I = (\frac{Δ I}{I_{0}} \times 100) %

Percentage error in I = (\frac{0.2}{10} \times 100) % = (0.02 \times 100) % = 2 %

💡 Teacher's Secret Hint

Pay attention to decimal places during calculation to avoid small errors.

Step 4: Apply Error Propagation Rule for Resistance○ Expand

For a quantity $R$ that is a quotient of two other quantities, $R = \frac{V}{I}$ , the maximum fractional error in $R$ is the sum of the fractional errors in $V$ and $I$ . Therefore, the maximum percentage error in $R$ is the sum of the individual percentage errors:

Percentage error in R = (Percentage error in V) + (Percentage error in I)

Percentage error in R = 5 % + 2 % = 7 %

💡 Teacher's Secret Hint

Remember that for multiplication and division, relative (or percentage) errors add up. For addition and subtraction, absolute errors add up.

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Physics Question 81 – AP-EAMCET 2025

Clue 1 of 3: Error Propagation

🧠 Full Solution Path