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A de machne is connected to a switching power pole. That power pole is controlled by PWM control circuit with an equavalent gain of K PWM =24. The de machune has a voliage constant KE =0.8 V. If the control voltage is v꜀=5 V, what would be the no load speed of the machine?

a. 96 rad/s
b. 300 rad/s
c. 150 rad/s
d. 75 rad/s

1 Answer

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Final answer:

The no-load speed of the DC machine is calculated using the control voltage and the gain of the PWM circuit. With a control voltage of 5 V and a gain of 24, the effective voltage is 120 V. Dividing this by the voltage constant 0.8 V gives a no-load speed of 150 rad/s.

Step-by-step explanation:

When a DC machine is connected to a switching power pole controlled by a PWM control circuit with an equivalent gain of KPWM = 24, and the machine has a voltage constant of KE = 0.8 V, the no-load speed of the machine can be calculated using the control voltage vc and the gain.

To find the no-load speed, first, calculate the effective voltage applied to the motor (Veffective) using the control voltage and the gain: Veffective = vc × KPWM. Then, use the voltage constant to find the no-load speed (ωno-load): ωno-load = Veffective / KE.

Given vc = 5 V:

  • Veffective = 5 V × 24 = 120 V,
  • ωno-load = 120 V / 0.8 V = 150 rad/s.

Therefore, the no-load speed of the machine is 150 rad/s, which corresponds to option (c).

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