---
title: "Output equation of DC machine"
subject: "Electric Machine Design"
chapter: 5
key: "semester_5/electric_machine_design:dc-output-equation"
source: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-output-equation
site: BCT Engineering
---

# Output equation of DC machine

Electric Machine Design · Chapter 5: DC Machine Design

- Asked in: 2075 Asoj, 4 marks; 2071 Chaitra, 8 marks; 2070 Asar, 8 marks (asked 3 times)
- Video: [Output Equation of DC Machine |  Electrical Machine Design #education  #viral #dcmachine](https://www.youtube.com/watch?v=xm9XCD-ksEs) — Prasant Decodes
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/dc-output-equation
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-5

## Question

Derive the output equation for the design of dc machine.

## Short answer

The output equation is Pa = C0 D^2 L n, where C0 = π^2 B_av ac × 10^-3. It is derived by expressing armature power as induced emf times armature current, substituting flux per pole from average gap density (PΦ = B_av π D L) and total ampere-conductors (I_z Z = ac π D).

## Answer

The **output equation** of a dc machine relates the power developed in the armature $P_a$ to the main dimensions ($D$, $L$), the specific loadings ($B_{av}$, $ac$) and the speed $n$.

#### Symbols
$P$ = poles, $\Phi$ = flux per pole, $Z$ = armature conductors, $a$ = parallel paths, $n$ = speed (rps), $I_a$ = armature current, $I_z = I_a/a$ = conductor current, $E$ = induced emf.

#### Derivation
Power developed in the armature:
$$
P_a = E I_a \times 10^{-3}\ \text{kW}
$$
Emf (with $n$ in rps):
$$
E = \frac{\Phi Z n P}{a}
$$
So
$$
P_a = \frac{\Phi ZnP}{a} \times I_a \times 10^{-3} = (P\Phi)(I_z Z)\, n \times 10^{-3}
$$

Total flux and specific magnetic loading:
$$
B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow P\Phi = B_{av}\pi DL
$$
Total ampere-conductors and specific electric loading:
$$
ac = \frac{I_zZ}{\pi D} \Rightarrow I_zZ = ac\,\pi D
$$
Substituting:
$$
\begin{aligned}
P_a &= B_{av}\pi DL \times ac\,\pi D \times n \times 10^{-3} \\
&= \pi^2 B_{av}\, ac\, D^2 L n \times 10^{-3} \\
&= C_0 D^2 L n
\end{aligned}
$$
where the **output coefficient** is
$$
C_0 = \pi^2 B_{av}\, ac \times 10^{-3}
$$

#### Relation between $P_a$ and rated output $P$
Assuming armature copper loss is about one-third of total losses (Sawhney):
$$
\text{Generator: } P_a = P\,\frac{1+2\eta}{3\eta}, \qquad \text{Motor: } P_a = P\,\frac{2+\eta}{3\eta}
$$
For small machines $P_a$ may also be taken as $E I_a$ directly when $E$ and $I_a$ are known.

#### Use
$$
D^2L = \frac{P_a}{C_0 n}
$$
$D$ and $L$ are then separated using the pole proportions (square pole face: $L = \psi\tau = \psi\pi D/P$) and the peripheral speed limit (about 30 m/s).

Example: $B_{av} = 0.5$ Wb/m², $ac = 22000$ A/m gives $C_0 = \pi^2 \times 0.5 \times 22 = 108.6$.

## Check yourself

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

1. Which expression gives the armature power in terms of total flux and ampere-conductors?
   A. Pa = (P Phi)(Iz Z) / n
   B. Pa = Phi Z n x 10^-3 only
   C. Pa = (P Phi)(Iz Z) n x 10^-3 kW
   D. Pa = (Phi/P)(Iz Z) n
   Answer: C. With E = Phi Z n P / a and Iz = Ia/a, E Ia = (P Phi)(Iz Z) n. (Key point: Power developed)

2. What does P Phi equal?
   A. Bav D L
   B. Bav pi D^2 L
   C. ac pi D
   D. Bav pi D L
   Answer: D. Bav = P Phi/(pi D L), the total flux over the armature surface. (Key point: Specific magnetic loading)

3. What is C0 in the dc output equation?
   A. C0 = pi Bav ac x 10^-3
   B. C0 = 11 Bav ac x 10^-3
   C. C0 = pi^2 Bav / ac
   D. C0 = pi^2 Bav ac x 10^-3
   Answer: D. Substituting Bav pi D L and ac pi D gives pi^2 Bav ac D^2 L n x 10^-3. (Key point: Output coefficient)

4. On what assumption is Pa related to rated output for a generator?
   A. Armature copper loss is about one-third of total losses
   B. Armature copper loss is half of total losses
   C. Iron loss equals copper loss
   D. No losses in the armature
   Answer: A. This gives Pa = P(1+2 eta)/(3 eta) for a generator. (Key point: Pa to output)

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