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Question

Physics Question on Resistance

The resistance R=VIR = \frac{V}{I} where V=(200±5)VV = (200 \pm 5) \, \text{V} and I=(20±0.2)AI = (20 \pm 0.2) \, \text{A}. The percentage error in the measurement of RR is:

A

3.5%

B

7%

C

3%

D

5.5%

Answer

3.5%

Explanation

Solution

Step 1: Express RR in Terms of VV and II

R=VIR = \frac{V}{I}

Step 2: Calculate the Percentage Error Using Error Analysis

The relative error in RR is given by:

ΔRR=ΔVV+ΔII\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}

Substitute the values:

ΔRR=5200+0.220\frac{\Delta R}{R} = \frac{5}{200} + \frac{0.2}{20}

Step 3: Simplify the Expression

ΔRR=5200+0.220=5200+2200=7200\frac{\Delta R}{R} = \frac{5}{200} + \frac{0.2}{20} = \frac{5}{200} + \frac{2}{200} = \frac{7}{200}

Step 4: Calculate the Percentage Error

Percentage Error=ΔRR×100=7200×100=3.5%\text{Percentage Error} = \frac{\Delta R}{R} \times 100 = \frac{7}{200} \times 100 = 3.5\%

So, the correct answer is: 3.5%