DC machines formulas

Energy conversion and DC machines. 16 formulas for GATE EE, IN Electrical Machines, each with when to use it. Each is explained in the topic notes.

All 16 DC machines formulas

FORCE, CONSTANT-CURRENT FORM

F = \frac{\partial W _{f}^{'} ( i , x )}{\partial x}_{i = const}

Coenergy is written as a function of $i$ and $x$ , and the partial derivative is taken with $i$ held fixed. This form is convenient when the excitation current is known.

FORCE, CONSTANT-FLUX-LINKAGE FORM

F = - \frac{\partial W _{f} ( λ , x )}{\partial x}_{λ = const}

Field energy is written as a function of $λ$ and $x$ , and the partial derivative is taken with $λ$ held fixed. Note the minus sign. This gives the same force as the constant-current form.

TORQUE

T = \frac{\partial W _{f}^{'} ( i , θ )}{\partial θ}_{i = const} = - \frac{\partial W _{f} ( λ , θ )}{\partial θ}_{λ = const}

This is the rotational analogue of the force expressions. $θ$ is the mechanical angle in radians and $T$ is in N m.

LINEAR (UNSATURATED) SINGLY EXCITED CASE

W_{f}^{'} = \frac{1}{2} L (x) i^{2}, W_{f} = \frac{λ ^{2}}{2 L ( x )} \Rightarrow F = \frac{1}{2} i^{2} \frac{d L}{d x}

This applies when the system is magnetically linear, so that $λ = L (x) i$ . The force acts in the direction that increases the inductance, i.e. that reduces the reluctance of the magnetic path.

DOUBLY EXCITED SYSTEM (MAGNETICALLY LINEAR)

T = \frac{1}{2} i_{1}^{2} \frac{d L _{11}}{d θ} + \frac{1}{2} i_{2}^{2} \frac{d L _{22}}{d θ} + i_{1} i_{2} \frac{d L _{12}}{d θ}

The first two terms are reluctance torques. They exist only if a self-inductance varies with $θ$ (saliency). The third term is the mutual (alignment) torque, which needs both windings excited. In a machine with a uniform air gap only the mutual term survives.

GENERATED EMF

E = \frac{P ϕZ N}{60 A}

$P$ = number of poles
$ϕ$ = flux per pole (Wb)
$Z$ = total number of armature conductors
$N$ = speed (rpm)
$A$ = number of parallel paths: $A = P$ for a simplex lap winding, and $A = 2$ for a simplex wave winding whatever the value of $P$

ELECTROMAGNETIC TORQUE

T = \frac{P ϕZ I _{a}}{2 π A}

$I_{a}$ is the total armature current. With $ϕ$ in Wb and $I_{a}$ in A, $T$ is in N m. $P$ , $Z$ and $A$ are the same as in the EMF equation.

COMPACT FORM

E = k ϕ ω, T = k ϕ I_{a}, k = \frac{P Z}{2 π A}

$ω = 2 π N /60$ is the mechanical angular speed in rad/s. $k$ is a fixed constant of the winding, so both $E$ and $T$ are proportional to $ϕ$ .

ELECTROMAGNETIC POWER AND TORQUE

E I_{a} = T ω \Rightarrow T = \frac{E I _{a}}{ω} = \frac{60 E I _{a}}{2 π N}

The electromagnetic (developed) power $E I_{a}$ equals the electromagnetic torque times the speed, in both modes. Shaft power differs from it by the rotational (core and mechanical) losses.

TERMINAL VOLTAGE EQUATIONS

Generator: V = E - I_{a} R_{a} - V_{brush}, Motor: V = E_{b} + I_{a} R_{a} + V_{brush}

$R_{a}$ is the resistance of the armature circuit. It includes the series-field (and interpole) winding resistance where those windings carry $I_{a}$ , as in series and long-shunt compound machines. The brush contact drop $V_{brush}$ is treated as a constant voltage when one is given, and neglected otherwise. It is subtracted for a generator and added for a motor.

CURRENTS IN A SHUNT MACHINE

Shunt generator: I_{a} = I_{L} + I_{f}, Shunt motor: I_{a} = I_{L} - I_{f}, I_{f} = \frac{V}{R _{f}}

$I_{L}$ is the line (load or supply) current and $R_{f}$ is the total resistance of the shunt field circuit. In a series machine, $I_{a} = I_{L} = I_{series field}$ .

VOLTAGE REGULATION (GENERATOR) AND SPEED REGULATION (MOTOR)

% VR = \frac{V _{n l} - V _{f l}}{V _{f l}} \times 100, % SR = \frac{N _{n l} - N _{f l}}{N _{f l}} \times 100

Both are referred to the full-load value. A flat-compounded generator has zero voltage regulation.

SPEED EQUATION

N = \frac{E _{b}}{k ^{'} ϕ} = \frac{V - I _{a} R _{a}}{k ^{'} ϕ}, k^{'} = \frac{P Z}{60 A}

This follows directly from $E_{b} = \frac{P ϕZ N}{60 A}$ , with $N$ in rpm. For two operating points of the same machine, $\frac{N _{2}}{N _{1}} = \frac{E _{b 2}}{E _{b 1}} \cdot \frac{ϕ _{1}}{ϕ _{2}}$ . Every speed-control method below can be reasoned from this equation.

EFFICIENCY

η = \frac{Output}{Input} = \frac{Output}{Output + Losses}

For a generator, the output is electrical terminal power and the input is mechanical shaft power. For a motor it is the reverse.

CONDITION FOR MAXIMUM EFFICIENCY

η_{m a x} when I^{2} R = P_{const} \Rightarrow I_{η m a x} = \frac{P _{const}}{R}

Efficiency is maximum when the variable copper loss equals the constant losses.

Exact when the terminal voltage is constant and the same current $I$ flows as output current and through $R$ . Example: a separately excited generator, with $I = I_{a}$ and $R = R_{a}$ .
Approximate for shunt machines, where $I_{a} \neq = I_{L}$ . It is used there with $I_{a} \approx I_{L}$ .

ARMATURE-REACTION AMPERE-TURNS PER POLE (BRUSH SHIFT)

A T_{d} = \frac{Z I _{c} θ _{m}}{360}, A T_{c} = Z I_{c} (\frac{1}{2 P} - \frac{θ _{m}}{360}), I_{c} = \frac{I _{a}}{A}

$θ_{m}$ is the brush shift from the geometric neutral axis in mechanical degrees, and $I_{c}$ is the current in each conductor. With no brush shift, $A T_{d} = 0$ and all $Z I_{c} / (2 P)$ armature ampere-turns per pole are cross-magnetising.

Where this was asked: 2018 (2), 2020 (1), 2022 (1), 2023 (3), 2025 (1). See the 8 previous-year questions on DC machines. This shows what was asked, not what will be asked.

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DC machines: the notes and 8 solved previous-year questions