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EXAMPLE 21-8 Back emf in a motor. The armature windings of a dc motor have a resistance of 5.0 N. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed. APPROACH As the motor is just starting up, it is turning very slowly, so there is negligible back emf. The only voltage is the 120-V line. The current is given by Ohm's law with R = 5.0 N. At full speed, we must include as emfs both the 120-V applied emf and the opposing back emf. SOLUTION (a) At start up, the current is controlled by the 120 V applied to the coil's 5.0-N resistance. By Ohm's law, V 120 V I = 24 A. R 5.0 Ω (b) When the motor is at full speed, the back emf must be included in the equivalent circuit shown in Fig. 21–19. In this case, Ohm's law (or Kirchhoff's rule) gives 120 V – 108 V = 1(5.0 N). Therefore 12 V I 2.4 A. 5.0 Ω NOTE This result shows that the current can be very high when a motor first starts up. This is why the lights in your house may dim when the motor of the refrigerator (or other large motor) starts up. The large initial refrigerator current causes the voltage to the lights to drop because the house wiring has resistance and there is some voltage drop across it when large currents are drawn.

EXAMPLE 21-8 Back emf in a motor. The armature windings of a dc motor have a resistance of 5.0 N. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed. APPROACH As the motor is just starting up, it is turning very slowly, so there is negligible back emf. The only voltage is the 120-V line. The current is given by Ohm's law with R = 5.0 N. At full speed, we must include as emfs both the 120-V applied emf and the opposing back emf. SOLUTION (a) At start up, the current is controlled by the 120 V applied to the coil's 5.0-N resistance. By Ohm's law, V 120 V I = 24 A. R 5.0 Ω (b) When the motor is at full speed, the back emf must be included in the equivalent circuit shown in Fig. 21–19. In this case, Ohm's law (or Kirchhoff's rule) gives 120 V – 108 V = 1(5.0 N). Therefore 12 V I 2.4 A. 5.0 Ω NOTE This result shows that the current can be very high when a motor first starts up. This is why the lights in your house may dim when the motor of the refrigerator (or other large motor) starts up. The large initial refrigerator current causes the voltage to the lights to drop because the house wiring has resistance and there is some voltage drop across it when large currents are drawn.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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(II) What will be the current in the motor of Example 21–8
if the load causes it to run at half speed?

EXAMPLE 21-8 Back emf in a motor. The armature windings of a dc motor have a resistance of 5.0 N. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed. APPROACH As the motor is just starting up, it is turning very slowly, so there is negligible back emf. The only voltage is the 120-V line. The current is given by Ohm's law with R = 5.0 N. At full speed, we must include as emfs both the 120-V applied emf and the opposing back emf. SOLUTION (a) At start up, the current is controlled by the 120 V applied to the coil's 5.0-N resistance. By Ohm's law, V 120 V I = 24 A. R 5.0 Ω (b) When the motor is at full speed, the back emf must be included in the equivalent circuit shown in Fig. 21–19. In this case, Ohm's law (or Kirchhoff's rule) gives 120 V – 108 V = 1(5.0 N). Therefore 12 V I 2.4 A. 5.0 Ω NOTE This result shows that the current can be very high when a motor first starts up. This is why the lights in your house may dim when the motor of the refrigerator (or other large motor) starts up. The large initial refrigerator current causes the voltage to the lights to drop because the house wiring has resistance and there is some voltage drop across it when large currents are drawn.
Transcribed Image Text:EXAMPLE 21-8 Back emf in a motor. The armature windings of a dc motor have a resistance of 5.0 N. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed. APPROACH As the motor is just starting up, it is turning very slowly, so there is negligible back emf. The only voltage is the 120-V line. The current is given by Ohm's law with R = 5.0 N. At full speed, we must include as emfs both the 120-V applied emf and the opposing back emf. SOLUTION (a) At start up, the current is controlled by the 120 V applied to the coil's 5.0-N resistance. By Ohm's law, V 120 V I = 24 A. R 5.0 Ω (b) When the motor is at full speed, the back emf must be included in the equivalent circuit shown in Fig. 21–19. In this case, Ohm's law (or Kirchhoff's rule) gives 120 V – 108 V = 1(5.0 N). Therefore 12 V I 2.4 A. 5.0 Ω NOTE This result shows that the current can be very high when a motor first starts up. This is why the lights in your house may dim when the motor of the refrigerator (or other large motor) starts up. The large initial refrigerator current causes the voltage to the lights to drop because the house wiring has resistance and there is some voltage drop across it when large currents are drawn.
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