The electric current passing through a resistor is measured as (5.0 \pm 0.1)\text{ A}, and… — JAMB UTME

The electric current passing through a resistor is measured as $(5.0 \pm 0.1)\text{ A}$ , and the resistance is measured as $(10.0 \pm 0.3)\text{ }\Omega$ . What is the maximum percentage error in the calculated electrical power dissipated by the resistor?

$7.0\%$ Answer
B
$5.0\%$
C
$4.0\%$
D
$12.0\%$

Answer

The maximum percentage error in the calculated power is

7.0\%

Electrical power is calculated using the formula

P = I^2 R

. In error analysis, the maximum fractional error for a quantity

y = a^n b^m

\frac{\Delta y}{y} = n\frac{\Delta a}{a} + m\frac{\Delta b}{b}

. Therefore, the maximum percentage error in

P

is equal to

2 \times (\text{percentage error in } I) + (\text{percentage error in } R) = 2(2.0\%) + 3.0\% = 7.0\%

Step-by-Step Solution

Calculate the percentage error in the measured current

I

\frac{\Delta I}{I} \times 100\% = \frac{0.1}{5.0} \times 100\% = 2.0\%

Percentage error is the ratio of absolute uncertainty to the measured value expressed as a percentage.

Calculate the percentage error in the measured resistance

R

\frac{\Delta R}{R} \times 100\% = \frac{0.3}{10.0} \times 100\% = 3.0\%

Applying the same percentage error formula to the resistance measurement.

Apply the error propagation formula for power

P = I^2 R

\frac{\Delta P}{P} \times 100\% = 2\left(\frac{\Delta I}{I} \times 100\%\right) + \left(\frac{\Delta R}{R} \times 100\%\right) = 2(2.0\%) + 3.0\% = 7.0\%

When physical quantities are raised to a power and multiplied, their fractional/percentage errors are multiplied by the exponent and summed to get the maximum error.

Key Concept

Propagation of errors in derived quantities involving powers and products

Alternative Method

Calculate absolute values of minimum and maximum possible power, then compute percentage variation relative to nominal power.

Estimated Time:1m 30s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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