Limiting error in calculated power

Power and its limiting error

Given: $V = 120$ V ($\pm 0.5\%$), $R = 50\ \Omega$ ($\pm 1\%$).

$$P = \frac{V^2}{R} = \frac{120^2}{50} = 288\ \text{W}$$

For a quantity $P = V^2 R^{-1}$, limiting (worst-case) relative errors add, each multiplied by its power:

$$\frac{\delta P}{P} = 2\,\frac{\delta V}{V} + \frac{\delta R}{R} = 2(0.5) + 1 = 2\%$$ $$\delta P = 0.02 \times 288 = 5.76\ \text{W}$$

Answer: $P = 288 \pm 5.76$ W (limiting error $\pm 2\%$).

If the errors are independent, the probable error is $\sqrt{(2\times 0.5)^2 + 1^2} = 1.41\%$, about $\pm 4.1$ W.

Systematic and random errors