Numford
Error analysis

Error in electric power

P = V²/R. P is the power, V the potential difference and R the resistance.

Error in a formula

Product, quotient, power. The percentage errors add. Each counts as many times as its power. Sum or difference. The absolute errors add. A percentage needs the value to become an absolute error. Both in one formula. Worked out with the values, as the sum of slope times error. This is the worst case. The answer is always the maximum error.
V
R

Maximum error in P

4%

  • V22 × 1%2%
  • R2%2%
  • Totaladd them4%

Product, quotient or power: the percentage errors add, each counted |power| times.

Worked example

4% maximum error in P

ΔP/P × 100 = 2 × ΔV/V × 100 + ΔR/R × 100

Product, quotient or power: the percentage errors add, each counted |power| times. With P = V I the errors of V and I simply add, giving the same answer for V ± 1% and I ± 2% as for R ± 2% here.

Related

Other formulas

Measured the same thing several times? Use repeated readings, or find the least count of your instrument.

Questions

Error in electric power: quick answers

What is the formula for the error in electric power?

P is the power, V the potential difference and R the resistance. With P = V²/R, the maximum error is given by ΔP/P × 100 = 2 × ΔV/V × 100 + ΔR/R × 100. Product, quotient or power: the percentage errors add, each counted |power| times.

How do I work out the error with real numbers?

Take V with 1% error and R with 2% error. The maximum percentage error in P is 2 × 1% + 2% = 4%.

Why are the errors added and never subtracted?

The error shown is the largest possible one. In the worst case every measurement is off in the direction that pushes the result the same way, so the errors pile up. They are added even when the quantity is divided or subtracted.

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