A dc machine has the following parameters: kE-kr = 0.8 Vs (voltage and torque co

Question

A dc machine has the following parameters: kE-kr = 0.8 Vs (voltage and torque constants) R, = 0.0 (armature resistance ignored) a) For an applied terminal voltage of 120 V what would be the speed? i) 120.0.8 ii) 120 0.82 120 0.8 ii) 120.0.8 iv) b) If the machine is to run at -120 rad/s, what would be the applied voltage? 120 0.8) 120 0.8 iii) 0.8.(-120) iv) -120-0.8 The machine is to be run as a generator delivering 500 W of electrical power while running at 150 rad/s. What would be the current being delivered? c) 150.0.8 0ii) 150.500 i 0.8 ii) -150-500-0.8 iv) 10-0.8 500 0.8.150

Explanation / Answer

a) Ke=Kt=08

Vt=120

Ea=Vt=120

Ea=Ka*Wm

120=.8*Wm

Wm=120/.8

b) Wm=-120 rad/sec

Ea=k*Wm

Ea=0.8*(-120)

Ea=Vt=.8*(-120)

c) P=500w

Wm=150 rad/sec

P=Ea*Wm=Te*

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