Problem — under-powered heater

Problem

A heater rated $P_\text{nom} = 2000\ \text{W}$ at $V_\text{nom} = 230\ \text{V}$ is powered by mistake at $V = 115\ \text{V}$. Calculate the power actually dissipated. For what physical reason does it scale with the square of the voltage?

Solution

Resistance of the heater — this is a construction property, practically constant. I derive it from the rated data: $R = \frac{V_\text{nom}^2}{P_\text{nom}} = \frac{(230\ \text{V})^2}{2000\ \text{W}} = \frac{52900}{2000} \approx 26{,}5\ \Omega$

Power at halved voltage — with the same $R$ but $V = 115\ \text{V}$: $P = \frac{V^2}{R} = \frac{(115)^2}{26{,}5} = \frac{13225}{26{,}5} \approx 500\ \text{W}$

Shortcut — since the voltage is halved ($115 = 230/2$) and $R$ is fixed: $P = \frac{V^2}{R} \propto V^2 \quad\Rightarrow\quad P = \left(\frac{1}{2}\right)^2 P_\text{nom} = \frac{1}{4}\cdot 2000 = 500\ \text{W}$

Why the square — at fixed resistance, lowering the voltage reduces the power through two effects that multiply together: (1) the current drops in proportion to $V$, since $i = V/R$; (2) the power $P = V\,i$ is the product voltage $\times$ current. Combining, $P = V\cdot(V/R) = V^2/R$: halving $V$ halves $i$, and the product of the two drops to a quarter.

$\ev{P \approx 500\ \text{W} \quad (\text{a quarter of the rated value})}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Potenza · Effetto Joule