
Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Question
Chapter 47, Problem 1SQ
To determine
Explain the origin of DC power for the DC motor.
Expert Solution & Answer

Explanation of Solution
DC motors are highly used in industries due to its versatility property. DC motors provide torque at a low speed. Even at the nameplate speed or above the nameplate speed, the motor can be operated constantly. DC power supply is required to operate the DC motor.
Due to the wide use of solid state rectifiers, a DC power plant and the distribution system is not required to develop the DC power. In AC distribution system, AC power can be rectified to DC power by using the solid state electronic devices at the motor site.
Conclusion:
Thus, the origin of DC power for the DC motor is explained.
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Students have asked these similar questions
Assume a JFET device with VGS(0) = -1.3 and ipss = 20 mA. Design a self-biased
(Fig. 2) JFET common-source amplifier with the gain of -2 and a DC biasing that
allows the largest swing in ip. Note that you can choose Vcc to arrive at a desired
RD to meet the gain requirement. Since you are designing for a given gain, you
may have to check to see if JFET is biased correctly. (Hint: First find Rs for correct
VGs and then use the gain to compute RD. Finally, use RD and Rs to determine
Vcc). Assume that the amplifier is to interface a source that expects a load of 50
. Also, assume that the amplifier circuit is AC coupled at both ends with 3 dB
corner frequency of 15 kHz.
Rearrange the circuit in step 1 to implement a common-drain amplifier. Do note
that the output capacitor (C2) must be redesigned as the output impedance of
common-drain is different from that of common-source amplifier. What is the
actual gain? What is the input impedance?
Assume a JFET device with VGS(0) = -1.3 and ipss = 20 mA. Design a self-biased
(Fig. 2) JFET common-source amplifier with the gain of -2 and a DC biasing that
allows the largest swing in ip. Note that you can choose Vcc to arrive at a desired
RD to meet the gain requirement. Since you are designing for a given gain, you
may have to check to see if JFET is biased correctly. (Hint: First find Rs for correct
VGs and then use the gain to compute RD. Finally, use RD and Rs to determine
Vec). Assume that the amplifier is to interface a source that expects a load of 50
2. Also, assume that the amplifier circuit is AC coupled at both ends with 3 dB
corner frequency of 15 kHz.
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