Consider a motor application that requires a continuous torque *T *of 0.003 N-m at 5,000 rpm, with both factors referred to the motor shaft. The available power supply voltage is 12 Vdc with maximum allowable steady-state current *I *of 0.5 A.

First, calculate the required torque constant *Kt *using the torque and current. Thus, *Kt *= *T/I *= 0.003/0.5 = 0.006 N-m/A. Because *Kt *is expressed in the SI system of units, (N-m/A), the voltage constant *Ke *is numerically the same with units of V/rad/sec. Also, = 5,000 rpm X 6.28/60 = 523 rad/sec.

Next, calculate the required armature resistance, *R *= (V *Ke*)/I = (12 523 0.006)/0.5 = 17.7 Ω Finally, calculate the motor constant, *Km *= *Kt/R *^{0.5 }= 0.006/(17.7) ^{0.5 }= 0.0014 N-m/W ^{0.5 }.

Referring to the table, calculate *Km *for each of the motors listed:

- Motor 1;
*Km*= 0.001/(5)^{.5 }= 0.00045 N-m/W^{0.5 } - Motor 2;
*Km*= 0.005/(9.5)^{.5 }= 0.0016 N-m/W^{0.5 } - Motor 3;
*Km*= 0.012/(10)^{.5 }= 0.0038 N-m/W^{0.5 }

For the required *Km *of 0.0014 N-m/W ^{0.5 }, motor 1 does not meet the specification, motor 2 is adequate, and motor 3 has the best margin.