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A series-wound motor in operation

The total resistance ( R f + R a ) of a series-wound dc motor is 2.0 Ω ( [link] ). When connected to a 120-V source ( ε S ), the motor draws 10 A while running at constant angular velocity. (a) What is the back emf induced in the rotating coil, ε i ? (b) What is the mechanical power output of the motor? (c) How much power is dissipated in the resistance of the coils? (d) What is the power output of the 120-V source? (e) Suppose the load on the motor increases, causing it to slow down to the point where it draws 20 A. Answer parts (a) through (d) for this situation.

Schematic shows the circuit of a series-wound direct current motor. It consists of two emfs and two resistors.
Circuit representation of a series-wound direct current motor.

Strategy

The back emf is calculated based on the difference between the supplied voltage and the loss from the current through the resistance. The power from each device is calculated from one of the power formulas based on the given information.

Solution

  1. The back emf is
    ε i = ε s I ( R f + R a ) = 120 V ( 10 A ) ( 2.0 Ω ) = 100 V .
  2. Since the potential across the armature is 100 V when the current through it is 10 A, the power output of the motor is
    P m = ε i I = ( 100 V ) ( 10 A ) = 1.0 × 10 3 W .
  3. A 10-A current flows through coils whose combined resistance is 2.0 Ω , so the power dissipated in the coils is
    P R = I 2 R = ( 10 A ) 2 ( 2.0 Ω ) = 2.0 × 10 2 W .
  4. Since 10 A is drawn from the 120-V source, its power output is
    P s = ε s I = ( 120 V ) ( 10 A ) = 1.2 × 10 3 W .
  5. Repeating the same calculations with I = 20 A , we find
    ε i = 80 V , P m = 1.6 × 1 0 3 W , P R = 8.0 × 10 2 W , and P s = 2.4 × 10 3 W .

    The motor is turning more slowly in this case, so its power output and the power of the source are larger.

Significance

Notice that we have an energy balance in part (d): 1.2 × 10 3 W = 1.0 × 10 3 W + 2.0 × 10 2 W .

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Summary

  • An electric generator rotates a coil in a magnetic field, inducing an emf given as a function of time by ε = N B A ω sin ( ω t ) where A is the area of an N -turn coil rotated at a constant angular velocity ω in a uniform magnetic field B .
  • The peak emf of a generator is ε 0 = N B A ω .
  • Any rotating coil produces an induced emf. In motors, this is called back emf because it opposes the emf input to the motor.

Problems

Design a current loop that, when rotated in a uniform magnetic field of strength 0.10 T, will produce an emf ε = ε 0 sin ω t , where ε 0 = 110 V and ω = 120 π rad/s .

three turns with an area of 1 m 2

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A flat, square coil of 20 turns that has sides of length 15.0 cm is rotating in a magnetic field of strength 0.050 T. If the maximum emf produced in the coil is 30.0 mV, what is the angular velocity of the coil?

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A 50-turn rectangular coil with dimensions 0.15 m × 0.40 m rotates in a uniform magnetic field of magnitude 0.75 T at 3600 rev/min. (a) Determine the emf induced in the coil as a function of time. (b) If the coil is connected to a 1000 - Ω resistor, what is the power as a function of time required to keep the coil turning at 3600 rpm? (c) Answer part (b) if the coil is connected to a 2000- Ω resistor.

a. ω = 120 π rad/s, ε = 850 sin 120 π t V;
b. P = 720 sin 2 120 π t W;
c. P = 360 sin 2 120 π t W

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The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V. (a) What is the frequency of the output voltage? (b) What is the strength of the magnetic field in which the coil is turning?

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A flip coil is a relatively simple device used to measure a magnetic field. It consists of a circular coil of N turns wound with fine conducting wire. The coil is attached to a ballistic galvanometer, a device that measures the total charge that passes through it. The coil is placed in a magnetic field B such that its face is perpendicular to the field. It is then flipped through 180 ° , and the total charge Q that flows through the galvanometer is measured. (a) If the total resistance of the coil and galvanometer is R , what is the relationship between B and Q ? Because the coil is very small, you can assume that B is uniform over it. (b) How can you determine whether or not the magnetic field is perpendicular to the face of the coil?

a. B is proportional to Q ; b. If the coin turns easily, the magnetic field is perpendicular. If the coin is at an equilibrium position, it is parallel.

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The flip coil of the preceding problem has a radius of 3.0 cm and is wound with 40 turns of copper wire. The total resistance of the coil and ballistic galvanometer is 0.20 Ω. When the coil is flipped through 180 ° in a magnetic field B , a change of 0.090 C flows through the ballistic galvanometer. (a) Assuming that B and the face of the coil are initially perpendicular, what is the magnetic field? (b) If the coil is flipped through 90 ° , what is the reading of the galvanometer?

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A 120-V, series-wound motor has a field resistance of 80 Ω and an armature resistance of 10 Ω . When it is operating at full speed, a back emf of 75 V is generated. (a) What is the initial current drawn by the motor? When the motor is operating at full speed, where are (b) the current drawn by the motor, (c) the power output of the source, (d) the power output of the motor, and (e) the power dissipated in the two resistances?

a. 1.33 A; b. 0.50 A; c. 60 W; d. 22.5 W; e. 2.5W

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A small series-wound dc motor is operated from a 12-V car battery. Under a normal load, the motor draws 4.0 A, and when the armature is clamped so that it cannot turn, the motor draws 24 A. What is the back emf when the motor is operating normally?

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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