Worked exercise — electric motor under load

Problem

A motor with $R = 2\;\Omega$ is powered by $\mathcal{E} = 48\;\text{V}$. At full running speed it generates a back electromotive force $\mathcal{E}_c = 44\;\text{V}$. Find: the current drawn, the mechanical power delivered, the efficiency.

Solution

Current drawn. The back e.m.f. opposes the source, so: $i = \frac{\mathcal{E} - \mathcal{E}_c}{R} = \frac{48 - 44}{2} = \ev{2\;\text{A}}$

Total electrical power input: $P_\text{el} = \mathcal{E}\cdot i = 48\cdot 2 = 96\;\text{W}$

Power dissipated by the Joule effect: $P_J = R\,i^2 = 2\cdot 2^2 = 8\;\text{W}$

Mechanical power delivered: $P_\text{mech} = \mathcal{E}_c\cdot i = 44\cdot 2 = \ev{88\;\text{W}}$

Efficiency: $\eta = \frac{P_\text{mech}}{P_\text{el}} = \frac{88}{96} \approx \ev{91{,}7\%}$

Check: $8 + 88 = 96 = P_\text{el}$. ✓

Note: with the motor stalled $\mathcal{E}_c = 0$, the current would rise to $48/2 = 24\;\text{A}$ and the dissipated power to $R\cdot 24^2 = 1152\;\text{W}$: this is why electric motors are damaged if mechanically blocked while powered.

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Topics: Magnetismo Concepts: Dipolo magnetico · Forza elettromotrice · Effetto Joule