Problem — heating resistor

Problem

A resistor for heating water is connected to the mains at r.m.s. voltage $V = 230\ \text{V}$ and dissipates $P = 1500\ \text{W}$. Calculate: (a) the resistance; (b) the current drawn; (c) the energy consumed in one hour, expressed in kilowatt-hours.

Resistor $R$ supplied by the mains $V$, with an ammeter in series.

Solution

(a) Resistance — from $P = V^2/R$: $R = \frac{V^2}{P} = \frac{(230\ \text{V})^2}{1500\ \text{W}} = \frac{52900}{1500} \approx 35{,}3\ \Omega$

(b) Current — from $P = V\,i$: $i = \frac{P}{V} = \frac{1500\ \text{W}}{230\ \text{V}} \approx 6{,}5\ \text{A}$ (check: $i = V/R = 230/35{,}3 \approx 6{,}5\ \text{A}$ ✓)

(c) Energy in one hour — with $t = 1\ \text{h}$: $E = P\,t = 1500\ \text{W} \cdot 1\ \text{h} = 1500\ \text{Wh} = 1{,}5\ \text{kWh}$

$\ev{R \approx 35{,}3\ \Omega, \quad i \approx 6{,}5\ \text{A}, \quad E = 1{,}5\ \text{kWh}}$

Links

Topics: Circuiti elettrici Concepts: Effetto Joule · Potenza