Because of overloading, the power of a heater connected

Because of overloading, the power of a heater connected to a network of voltage $U_{1} = 220\ \text{V}$ drops from $P_{1} = 1000\ \text{W}$ to $P_{2} = 800\ \text{W}$. Find the drop in the network voltage, assuming the resistance $R$ stays constant.

Step 1: The relation between power and voltage

With constant resistance, $P = \frac{U^{2}}{R}$.

Step 2: Find $R$ from the initial state

\[ R = \frac{U_{1}^{2}}{P_{1}} = \frac{220^{2}}{1000} = \frac{48400}{1000} = 48.4\ \Omega. \]

Step 3: Differentiate and invert

\[ \frac{dP}{dU} = \frac{2U}{R} \;\Longrightarrow\; \Delta P \approx \frac{2U_{1}}{R}\,\Delta U \;\Longrightarrow\; \Delta U \approx \frac{R\,\Delta P}{2U_{1}}. \]

This is the differential used backwards: the observed change in the effect $P$ is converted into an estimate of the change in the cause $U$.

Step 4: Substitute the numbers

\[ \Delta P = P_{2}-P_{1} = 800-1000 = -200\ \text{W}, \]

\[ \Delta U \approx \frac{48.4\cdot(-200)}{2\cdot 220} = \frac{-9680}{440} = -22\ \text{V}. \]

The minus sign says the voltage fell.

Step 5: Verification — and the size of the error

Here the exact answer is available, since $U = \sqrt{RP}$: \[U_{2} = \sqrt{48.4\cdot 800} = \sqrt{38720} = 220\sqrt{0.8} \approx 196.7740\ \text{V}\]

\[\Delta U_{\text{exact}} = 196.7740-220 = -23.2260\ \text{V}\]

\[\text{Error of the linear estimate: } 1.226\ \text{V}, \text{ i.e. about } 5.3\% \text{ of the true drop}\]