No one needs to be reminded of the importance of electric motors in contemporary society. Their operation depends on magnetic forces on current-carrying conductors. As an example, let’s look at a simple type of direct-current motor, shown in Fig. 28—33. The center part is the armature, or rotor; it is a cylinder of soft steel that rotates about its axis (perpendicular to the plane of the figure).
Embedded in slots in the rotor surface (parallel to its axis) are insulated copper conductors. Current is led into and out of these conductors through graphite brushes making contact with a segmented cylinder called the commutator, shown in principle in Fig. 28—34. The commutator is an automatic switching arrangement that uses stationary sliding contacts called brushes to maintain the currents in the conductors in the directions shown in Fig. 28—33 as the rotor turns. The current in the field coils F and F’ sets up a magnetic field in the motor frame and in the gap between the pole pieces P and P’ and the rotor. (In some small motors the magnetic field is supplied by permanent magnets instead of electromagnets.) Some of the magnetic field lines are shown as blue lines. With the directions of field and rotor currents shown, the side thrust on each conductor in the rotor is such as to produce a counterclockwise torque on the rotor.
A motor converts electrical energy to mechanical energy or work, and it requires electrical energy input. If the potential difference between its terminals is Vab and the current is I, then the power input is P = Vab I. Even if the motor coils have negligible resistance, there must be a potential difference between the terminals if P is to be different from zero. This potential difference results principally from magnetic forces exerted on the currents in the conductors of the rotor as they rotate through the magnetic field. The associated electromotive force 6 is called an induced emf, or a back emf, referring to the fact that its sense is opposite to that of the current. In Section 30 we will study induced emf’s resulting from motion of conductors in magnetic fields.
28—33 Schematic diagram of a dc motor. The armature,
or rotor, rotates on a shaft along its axis.
28—34 The segments of this commutator are insulated from
one another. Conductors on the rotor carry currents in the directions shown.
(a) The brushes are aligned with the commutator segments for the loop shown,
which carries current in the directions indicated. (b) The rotor has turned
so that the brushes are aligned with the segments for the next conducting
loop; it is connected into the circuit, and the previous loop is disconnected.
In a series motor the rotor and the field windings are connected in series; in a shunt motor they are connected in parallel. In a series motor with internal resistance r, V is greater than Ε, and the difference is the drop Ir across the internal resistance; that is,
Vab = Ε + Ir. (eqn. 28—28)
Because the magnetic-field force is proportional to velocity, Ε is not constant but is proportional to the speed of rotation of the rotor.
A dc motor is analogous to a battery being charged, which we discussed in sub-section 26—4. In a battery, electrical energy is converted to chemical rather than mechanical energy.
EXAMPLE 28—11
A series dc motor: A dc motor with its rotor and field coils connected in series has an internal resistance of 2.00 Ω. When running at full load on a 120-V line, it draws a current of 4.00 A.
a) What is the emf in the rotor? b) What is the power delivered to the motor? c) What is the rate of dissipation of energy in the resistance of the motor? d) What is the mechanical power developed? e)What is the efficiency of the motor? f)What happens if the machine that the motor is driving jams and the rotor suddenly stops turning?
SOLUTION
a) From Eq. (28—28),
Vab = Ε Ir,
120 V = Ε + (4.0 A)(2.0 Ω),
Ε = 112V.
b) The power input P from the source is
P = Vab I = (120 V)(4.0 A) = 480 W.
c) The power P dissipated in the resistance r is
P = I2r = (4.0 A)2- (2.0 Ω) = 32 W
d) The mechanical power output is the electrical power input minus the rate of dissipation of energy in the motor’s resistance (assuming that there are no other power losses):
p = 480W — 32 W = 448 W
e) The efficiency e is the ratio of mechanical power output to electrical power input:
e = 480W/480W = 0.93 = 93%.
f) With the rotor stalled, the back emf Ε (which is proportional to rotor speed) goes to zero. From Eq. (28—28) the current becomes
I = Vab/r = 120V/2.0 = 60A
and the power dissipation P in the resistance r becomes
P =I2r =(60A)2(2 Ω ) =7200W.
If this massive overload doesn’t blow a fuse or trip a circuit breaker, the coils will quickly melt. When the motor is first turned on, there’s a momentary surge of current until the motor picks up speed. This surge causes greater-than-usual voltage drops in the power lines supplying the current. Similar effects are responsible for the momentary dimming of lights in a house when an air-conditioner or dishwasher motor starts.