Worked exercise — resistor used as a heater

Problem

A resistor $R_x$ is connected to a generator $\mathcal{E} = 200\;\text{V}$ with internal resistance $r = 4\;\Omega$. The resistor must bring $m = 2\;\text{kg}$ of water from $20\,^\circ$C to $100\,^\circ$C in $t = 5\;\text{min} = 300\;\text{s}$. Specific heat of water $c = 4180\;\text{J}/(\text{kg}\cdot\text{K})$. Find the value of $R_x$.

Real generator ($\mathcal{E}$, internal resistance $r$) powering the heater $R_x$.

Solution

Required thermal energy. To heat the water by $\Delta T = 80\;\text{K}$: $\Delta E = m\,c\,\Delta T = 2 \times 4180 \times 80 = 668\,800\;\text{J}$

Power to be dissipated on $R_x$: $P_{R_x} = \frac{\Delta E}{t} = \frac{668\,800}{300} \approx 2229\;\text{W}$

Circuit balance ($P_{\text{gen}} = P_r + P_{R_x}$, i.e. the power delivered by the e.m.f. is dissipated in the internal resistance plus the heater): $\mathcal{E}\,i = r\,i^2 + P_{R_x} \;\Rightarrow\; 200\,i = 4\,i^2 + 2229$ that is, the quadratic equation $4\,i^2 - 200\,i + 2229 = 0$

Discriminant and solutions: $\Delta = 40000 - 35664 = 4336$ $i = \frac{200 \pm \sqrt{4336}}{8} \approx \frac{200 \pm 65{,}9}{8}$ The physically sensible solution is $i \approx 16{,}7\;\text{A}$ (the other, $i \approx 33{,}2\;\text{A}$, would dissipate too much power in the internal resistance $r$).

Resistance of the heater: $R_x = \frac{P_{R_x}}{i^2} = \frac{2229}{16{,}7^2} \approx 8\;\Omega$

$\ev{R_x \approx 8\;\Omega}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Potenza · Effetto Joule Competenze: Conservazione dell’energia