Alex needs to determine the power dissipated in a resistor. She measures the current through the resistor to be I=10.0±0.6 A, and its resistance to be R=45.0±2.0 Ω. Assuming her calculations are correct, Alex should find that the uncertainty in power dissipated in the resistor due only to the uncertainty in the current is:

Given that,

Current measure is

i=10±0.6 Amps

And also,

R=45.0±2.0 Ω

Power dissipated by

P=i²R

Then

P = (10±0.6) ² (45.0±2.0)

P=10²*45

P=450Watts

Now, calculating the uncertainty

∆P=|P| • √ (2 (∆i/i) ² + (∆R/R) ²)

∆P=450√ (2 * (0.6/10) ² + (2/45) ²)

∆P=450√ (0.0072+0.001975)

∆P=450√0.009175

∆P=43.1

The uncertainty in power is 43.1

Then,

P=450 ± 43.1 Watts