---
title: "Equilibrium points and steady-state stability"
subject: "Utilization of Electrical Energy"
chapter: 2
key: "semester_7/utilization_of_electrical_energy:equilibrium-points-steady-state-stability"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/equilibrium-points-steady-state-stability
site: BCT Engineering
---

# Equilibrium points and steady-state stability

Utilization of Electrical Energy · Chapter 2: Electric Drive System

- Asked in: 2069 Chaitra
- 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/equilibrium-points-steady-state-stability
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-2

## Question

Obtain the equilibrium points and determine their steady-state stability when motor and load torque are Tm = −(1 + 2ωm) and TL = −3√ωm respectively.

## Short answer

Equilibrium at T_m = T_L: ω_m = 1 rad/s and ω_m =0.25 rad/s. Stability using dT_L/dω - dT_m/dω >0: at ω=1, diff = 0.5 (>0, stable); at ω=0.25, diff = -1 (<0, unstable).

## Answer

**Equilibrium points:** where motor torque equals load torque, $T_m = T_L$:

$$
\begin{aligned}
-(1 + 2\omega_m) &= -3\sqrt{\omega_m} \\
2\omega_m - 3\sqrt{\omega_m} + 1 &= 0
\end{aligned}
$$

Let $x = \sqrt{\omega_m}$: $2x^2 - 3x + 1 = 0 \Rightarrow (2x-1)(x-1) = 0$, so $x = 1$ or $x = 0.5$.

$$
\omega_m = 1\ \text{rad/s} \quad \text{and} \quad \omega_m = 0.25\ \text{rad/s}
$$

**Stability criterion.** An equilibrium point is steady-state stable if a small speed increase produces a net decelerating torque:

$$
\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} > 0
$$

Derivatives:

$$
\frac{dT_m}{d\omega_m} = -2, \qquad \frac{dT_L}{d\omega_m} = -\frac{3}{2\sqrt{\omega_m}}
$$

**At $\omega_m = 1$:**

$$
\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2} + 2 = 0.5 > 0 \;\Rightarrow\; \text{stable}
$$

**At $\omega_m = 0.25$:**

$$
\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2\times0.5} + 2 = -3 + 2 = -1 < 0 \;\Rightarrow\; \text{unstable}
$$

**Check at $\omega_m = 0.25$**: torque $T_m = T_L = -1.5$ N-m; at $\omega_m = 1$: $T_m = T_L = -3$ N-m.

| Equilibrium speed | Torque (N-m) | $dT_L/d\omega - dT_m/d\omega$ | Stability |
|---|---|---|---|
| 1 rad/s | -3 | +0.5 | Stable |
| 0.25 rad/s | -1.5 | -1 | Unstable |

**Answer:** equilibrium points at $\omega_m = 1$ rad/s (stable) and $\omega_m = 0.25$ rad/s (unstable).

## Check yourself

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

1. Equilibrium points are where:
   A. Tm = TL
   B. dTm/dw = 0
   C. Tm = 0
   D. TL = 0
   Answer: A. Motor torque equals load torque. (Key point: Equilibrium condition)

2. Setting x = sqrt(wm) gives 2x^2 - 3x + 1 = 0. Equilibrium speeds are:
   A. 1 and 0.25 rad/s
   B. 2 and 1 rad/s
   C. 1 and 0.5 rad/s
   D. 3 and 1 rad/s
   Answer: A. x = 1 or 0.5, so wm = 1 or 0.25 rad/s. (Key point: Solving)

3. A point is stable if:
   A. dTL/dw - dTm/dw > 0
   B. dTL/dw - dTm/dw < 0
   C. dTm/dw > 0
   D. Tm > TL
   Answer: A. A speed rise must produce net decelerating torque. (Key point: Stability criterion)

4. At wm = 0.25 rad/s the value is -1. The point is:
   A. Stable
   B. Unstable
   C. Marginal
   D. Not an equilibrium
   Answer: B. Negative value means unstable; wm = 1 (value +0.5) is stable. (Key point: Result)

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