Why do La and J not affect steady-state speed?

di/dt and dw/dt are zero, so they only affect transients

Armature current in terms of speed:

Ia = (TL + B w)/Kb = 125 + 0.0125 w

2073 Chaitra · 6 marks

Speed of separately excited DC motor (given J, B)

Short answer

In steady state, Ia = (100 + 0.01 omega)/0.8 and 220 = 0.8 omega + 0.5 Ia. Solving gives omega = 195.35 rad/s, about 1865 rpm, with armature current 127.4 A.

A separately excited dc motor with the following parameters: Ra = 0.5 Ω, La = 0.003H, and Kb = 0.8 V/rad/sec, is driving a load of J = 0.0167 kg-m², B1 = 0.01 N-m/rad/sec with a load torque of 100 N-m. Its armature is connected to a dc supply voltage of 220 V and is given the rated field current. Find the speed of the motor.

In steady state the armature current and speed are constant, so $L_a\, di/dt = 0$ and $J\, d\omega/dt = 0$ . $L_a$ and $J$ only affect the transient.

Steady-state equations

\begin{aligned} V &= K_b\omega + R_a I_a \\ K_b I_a &= B\omega + T_L \end{aligned}

From the torque equation:

I_a = \frac{T_L + B\omega}{K_b} = \frac{100 + 0.01\omega}{0.8} = 125 + 0.0125\,\omega

Substituting into the voltage equation:

\begin{aligned} 220 &= 0.8\,\omega + 0.5(125 + 0.0125\,\omega) \\ 220 &= 0.80625\,\omega + 62.5 \\ \omega &= \frac{157.5}{0.80625} = 195.35\ \text{rad/s} \end{aligned}

N = \frac{60\,\omega}{2\pi} = \frac{60\times195.35}{2\pi} = 1865.4\ \text{rpm}

Armature current: $I_a = 125 + 0.0125\times195.35 = 127.44$ A.

Answer: Motor speed = 195.35 rad/s ≈ 1865 rpm (armature current 127.4 A).

✓ Answer checked by Claude Opus 5.5 (high effort)

Chopper ON time for separately excited DC motor · Utilization of Electrical Energy, chapter 3
Plugging of 220 V separately excited motor (100 A) · Utilization of Electrical Energy, chapter 3
Plugging of 200 V, 1000 rpm separately excited motor · Utilization of Electrical Energy, chapter 3
DC chopper: input power, speed, torque, speed range · Utilization of Electrical Energy, chapter 3 · asked 2×
DC shunt motor speed at reduced voltage and starting resistance · Utilization of Electrical Energy, chapter 3
Single-phase fully controlled converter DC motor speed · Utilization of Electrical Energy, chapter 3
25 kW separately excited DC machine as motor or generator · Electrical Machines I, chapter 3
Separately excited elevator motor with field flux reduced · Electrical Machines I, chapter 4
Armature-controlled DC motor with gear and load · Control Systems, chapter 2
Shunt motor driving pump: torque proportional to speed squared · Electrical Machines, chapter 4 · asked 2×

All Utilization of Electrical Energy chapter 3 questions →

Speed of separately excited DC motor (given J, B)

Steady-state equations

Related questions