Question

Question 1 0/10 pts A permanent magnet dc motor is to be started from rest. This motor is driving a load of inertia J = 0.045 kg-m2, and a load torque of T = 2.1 Nm. The motor current must not exceed 15A First, find the expression of voltage va as functions of time which must be applied to bring this motor to a steady state speed of 300 rad/s as quickly as possible. Then, at time t= 1.4 seconds, what is the voltage va in Volts? R = 0.35 Ohms ke=0.5 V/(rad/s) kr=0.5 Nm/A Jm=0.025 kg-m2 You Answered 156.2 Correct Answer 59.25 margin of error +/- 2%

Answer 1

1 given data's JL = 0.045kg.m TL = 2.1 Nm the motor current must not exceeed Is A. Ra: 0.354, KE -0.5 rados) , kq=0.5 Nm/A Jm=0.025kg - my Steady state speed = 300 rad's at t= 1.4 sec write the expression of Torque to bring steady state Tem = Ky imax (0.5)(15) = 7.5Nem : Ten-T- Tace = 7.55 -2.1 = 5.4 Nom wam 3oornd/sec Tacc 1. ) (+ Tace Ł JL+Im 5.4 t 0.045 + 0.025 = 77148t talt) - Keln) ) : 0.5) 77.14244) = 38.591%t)

Rala= 0.35 15 = 5.25V Valt) = Calt) + Raia = (38.571 € +5.25) v 5 38.571*(1-4) +5.25 Valt) = 59.2494 v
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