The Alternating Current Motor

The first electric dynamo was constructed in 1831 by Michael Faraday. An electric dynamo is, of course, a device which transforms mechanical energy into electrical energy. An electric motor, on the other hand, is a device which transforms electrical energy into mechanical energy. In other words, an electric motor is an electric dynamo run in reverse. It took a surprisingly long time for scientists in the nineteenth century to realize this. In fact, the message only really sank home after a fortuitous accident during the 1873 Vienna World Exposition. A large hall was filled with modern gadgets. One of these gadgets, a steam engine driven dynamo, was producing electric power when a workman unwittingly connected the output leads from another dynamo to the energized circuit. Almost immediately, the latter dynamo started to whirl around at great speed. The dynamo was, in effect, transformed into an electric motor.

An AC electric motor consists of the same basic elements as an AC electric generator: i.e., a multi-turn coil which is free to rotate in a constant magnetic field. Furthermore, the rotating coil is connected to the external circuit in just the same manner as in an AC generator: i.e., via two slip-rings attached to metal brushes. Suppose that an external voltage source of emf $V$ is connected across the motor. It is assumed that $V$ is an alternating emf, so that

$$V = V_{\rm max} \,\sin (2\pi\, f\, t),$$ (219)

where $V_{\rm max}$ is the peak voltage, and $f$ is the alternation frequency. Such an emf could be obtained from an AC generator, or, more simply, from the domestic mains supply. For the case of the mains, $V_{\rm max} = 110$V and $f=60$Hz in the U.S. and Canada, whereas $V_{\rm max} = 220$V and $f=50$Hz in Europe and Asia. The external emf drives an alternating current

$$I = I_{\rm max} \, \sin (2\pi\, f\, t)$$ (220)

around the external circuit, and through the motor. As this current flows around the coil, the magnetic field exerts a torque on the coil, which causes it to rotate. The motor eventually attains a steady-state in which the rotation frequency of the coil matches the alternation frequency of the external emf. In other words, the steady-state rotation frequency of the coil is $f$. Now a coil rotating in a magnetic field generates an emf ${\cal E}$. It is easily demonstrated that this emf acts to oppose the circulation of the current around the coil: i.e., the induced emf acts in the opposite direction to the external emf. For an $N$-turn coil of cross-sectional area $A,$ rotating with frequency $f$ in a magnetic field $B$, the back-emf ${\cal E}$ is given by

$${\cal E} = {\cal E}_{\rm max}\,\sin (2\pi\, f\, t),$$ (221)

where

$${\cal E}_{\rm max} = 2\pi\,N\,B\,A\,f,$$ (222)

and use has been made of the results of Sect. 9.6.

Figure 43: Circuit diagram for an AC motor connected to an external AC emf source.

Figure 43 shows the circuit in question. A circle with a wavy line inside is the conventional way of indicating an AC voltage source. The motor is modeled as a resistor $R$, which represents the internal resistance of the motor, in series with the back-emf ${\cal E}$. Of course, the back-emf acts in the opposite direction to the external emf $V$. Application of Ohm's law around the circuit gives

$$V = I\, R + {\cal E},$$ (223)

or

$$V_{\rm max}\, \sin (2\pi\, f \,t) = I_{\rm max} \,R\,\sin (2\pi\, f\, t) + {\cal E}_{\rm max}\, \sin (2\pi\, f\, t),$$ (224)

which reduces to

$$V_{\rm max} = I_{\rm max} \,R + {\cal E}_{\rm max}.$$ (225)

The rate $P$ at which the motor gains electrical energy from the external circuit is given by

$$P = {\cal E}\,I = P_{\rm max}\, \sin^2(2\pi\, f\, t),$$ (226)

where

$$P_{\rm max} = {\cal E}_{\rm max}\, I_{\rm max} = \frac{ {\cal E}_{\rm max}\, (V_{\rm max} - {\cal E}_{\rm max})}{R}.$$ (227)

By conservation of energy, $P$ is also the rate at which the motor performs mechanical work. Note that the rate at which the motor does mechanical work is not constant in time, but, instead, pulsates at the rotation frequency of the coil. It is possible to construct a motor which performs work at a more uniform rate by employing more than one coil rotating about the same axis.

As long as $V_{\rm max} > {\cal E}_{\rm max}$, the rate at which the motor performs mechanical work is positive (i.e., the motor does useful work). However, if $V_{\rm max} < {\cal E}_{\rm max}$ then the rate at which the motor performs work becomes negative. This means that we must do mechanical work on the motor in order to keep it rotating, which is another way of saying that the motor does not do useful work. Clearly, in order for an AC motor to do useful work, the external emf $V$ must be able to overcome the back-emf ${\cal E}$ induced in the motor (i.e., $V_{\rm max} > {\cal E}_{\rm max}$).