Current was measured during a test as of 30.4 A, flowing in a resistor of 0.105 Ω. It was discovered later that the ammeter reading was low by 1.2 percent and the marked resistance was high by 0.3 percent. What is the true power as a percentage of the power that was originally calculated?

given  I_measured=30.4A

Resistance=0.105Ω 

ammeter reading was low by 1.2 percent

\(I_{true}=30.4+(30.4×\frac{1.2}{100} )\)

I_true=30.7648A

marked resistance was high by 0.3 percent

\(R_{true}=0.105-(0.105\times \frac{0.3}{100})\)

R_true=0.10468Ω

\(true \ \rm power=I^2_{true}R_{true}=(30.7648)^2\times 0.10468\)

\(true \ \rm power=99.0815\ \rm watt\)

\(measured \ \rm power =I^2_{measured}R_{measured}=(30.4)^2\times 0.105\)

\(measured \ \rm power =97.0368\ \rm watt\)

\(true \ \rm power\ \rm as\ \rm a \ \rm percentage\ \rm of \ \rm the \ \rm power\ \rm that \ \rm was \ \rm originally\ \rm calculated\ \rm = \frac{true \ \rm power}{measured power}\times 100\)

\(true \ \rm power\ \rm as\ \rm a \ \rm percentage\ \rm of \ \rm the \ \rm power\ \rm that \ \rm was \ \rm originally\ \rm calculated\ \rm=\frac{99.0815} {97.0368}\times 100=102.1\% \)

Option 3 is correct.