Question 6.6.1: The parameter values for a certain motor are KT = Kb = 0.05 ......
The parameter values for a certain motor are
K_{T}=K_{b}=0.05\,\mathrm{N}\cdot\mathrm{m/A}c=10^{-4}\,\mathrm{N}\cdot\mathrm{m}\cdot\mathrm{s/rad}\qquad R_{a}=0.5\,\Omega
The manufacturer’s data states that the motor’s maximum speed is 3000 rpm, and the maximum armature current it can withstand without demagnetizing is 30 A.
Compute the no-load speed, the no-load current, and the stall torque. Determine whether the motor can be used with an applied voltage of \ v_{a} = 10 V.
For \ v_{a} = 10 V, (6.6.5) and (6.6.6) give
i_{a}={\frac{c\,V_{a}+K_{b}T_{L}}{c R_{a}+K_{b}K_{T}}} (6.6.5)
\omega={\frac{K_{T}V_{a}-R_{a}T_{L}}{c R_{a}+K_{b}K_{T}}} (6.6.6)
i_{a}=0.392+19.61T_{L}\,\mathrm{A}\qquad\omega=196.1-196.1T_{L} rad/s
The no-load speed is found from the second equation with {{T}}_{L} = 0. It is 196.1 rad/s, or 1872 rpm, which is less than the maximum speed of 3000 rpm. The corresponding no-load current is i_{a}=0.392\;A, which is less than the maximum allowable current of 30 A. The no-load current is required to provide a motor torque K_{T}i_{a} to cancel the damping torque cω.
The stall torque is found by setting ω = 0. It is TL = 1 N · m. The corresponding stall current is i_{a} = 20 A, which is less than the maximum allowable current.