---
title: "Speed of separately excited DC motor (given J, B)"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:separately-excited-motor-speed-load-100nm"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/separately-excited-motor-speed-load-100nm
site: BCT Engineering
---

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

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2073 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/separately-excited-motor-speed-load-100nm
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

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.

## 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.

## Answer

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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why do La and J not affect steady-state speed?
   A. They are zero
   B. di/dt and dw/dt are zero, so they only affect transients
   C. They cancel Ra
   D. They are in the field
   Answer: B. Constant current and speed give zero derivatives. (Key point: Steady state)

2. Armature current in terms of speed:
   A. Ia = (TL + B w)/Kb = 125 + 0.0125 w
   B. Ia = TL/B
   C. Ia = TL Kb
   D. Ia = (TL - B w)/Kb
   Answer: A. Kb Ia = B w + TL. (Key point: Armature current)

3. The resulting speed is:
   A. ~195 rad/s (1865 rpm)
   B. ~275 rad/s
   C. ~100 rad/s
   D. ~220 rad/s
   Answer: A. 220 = 0.80625 w + 62.5, w = 195.35 rad/s. (Key point: Speed)

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