Mechatronic Systems: practice problems and exam questions

DC motor: the armature is a resistance R_a in series with the back-EMF E, so V = E + I_a·R_a. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·I_a, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·I_a = T·ω; the input power is V·I_a, and the copper loss I_a²R_a and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − R_a·T/k², from the no-load speed V/k to zero at the stall torque k·V/R_a. At standstill E = 0, so the starting current V/R_a is large unless a resistor limits it.

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.