The whole Electrical Machines formula sheet, to print

$E_{rms}$ in V, $f$ in Hz, $N$ = turns of that winding, ${\Phi }_m$ = peak core flux in Wb (${\Phi }_m=B_m{A}_i$, with $A_i$ the net iron cross-section). The derivation assumes the core flux is sinusoidal. It applies to the primary and secondary separately, each with its own $N$.

An ideal transformer has no winding resistance, no leakage flux, no core loss and an infinitely permeable core, so it needs no magnetising current. In a practical transformer, $V_1/V_2=a$ holds only approximately (closest at no load), and $I_2/I_1=a$ holds only when the no-load current $I_0$ is neglected.

$P_h$ = hysteresis loss (the Steinmetz exponent $x$ is empirical, commonly taken as 1.6). $P_e$ = eddy-current loss, where $t$ is the lamination thickness. With $B_m\propto V/f$: at constant $V/f$, $P_h\propto f$ and $P_e\propto f^2$. At constant $f$, $P_e\propto V^2$. At constant $V$, $P_e$ does not depend on $f$, and $P_h\propto V^x{f}^{1-x}$.

Here $a=N_1/N_2$. To refer a primary quantity to the secondary, use the inverse: $V_1^{\prime }=V_1/a$, $I_1^{\prime }=a{I}_1$, $Z_1^{\prime }=Z_1/a^2$. An impedance always grows when it is referred to the side with more turns.

$Z_{01}=\sqrt{R_{01}^2+X_{01}^2}$ and $Z_{02}=\sqrt{R_{02}^2+X_{02}^2}=Z_{01}/a^2$. In the approximate equivalent circuit the shunt branch ($R_c$ in parallel with $X_m$) is moved to the supply terminals, ahead of the series impedance. This is justified because the no-load current $I_0$ is a small fraction of rated current.

These values hold at the voltage and frequency of the test. Core loss is not linear in flux, so $R_c$ and $X_m$ are not constants at other voltages.

$R_{eq}$ and $X_{eq}$ are referred to the energised winding. Multiply or divide by $a^2$ before combining them with quantities on the other side.

If the SC test was not run at rated current, scale its reading by the square of the current ratio to get the full-load copper loss. $\%Z$ is the SC-test voltage at rated current, expressed as a percentage of rated voltage. It has the same value whichever side it is calculated on, provided $I$, $Z$ and $V$ all belong to that side.

$V_{2,nl}$ is the no-load secondary voltage and $V_{2,fl}$ the full-load (rated) secondary voltage. Some texts divide by $V_{2,nl}$ instead ("regulation down"). These notes use the rated full-load voltage as the base.

Use $+$ for a lagging and $-$ for a leading power factor. $I$ and $V$ are the rated current and voltage on the same side as $R_{eq}$, $X_{eq}$ (that is, $R_{01}$, $X_{01}$ with primary quantities, or $R_{02}$, $X_{02}$ with secondary quantities). ${\epsilon }_r=I{R}_{eq}/V\times 100$ and ${\epsilon }_x=I{X}_{eq}/V\times 100$ are the percentage resistance and reactance. The formula neglects the quadrature component of the drop.

Zero regulation is possible only at a leading power factor, where the resistive drop in phase with $V$ is cancelled by the component of the reactive drop. At any leading power factor more leading than this, the regulation is negative: the terminal voltage rises on load.

Obtained by setting $\frac{d}{d\varphi }\left(R_{eq}\cos \varphi +X_{eq}\sin \varphi \right)=0$. Maximum regulation occurs when the load power-factor angle equals the impedance angle of the series impedance, and its value is then the percentage impedance.

$S$ is the rated VA. $xS\cos \varphi =V_2{I}_2\cos \varphi$ is the output. $P_i$ is the constant core loss (from the OC test at rated voltage). $P_{cu,fl}=I_{2,rated}^2{R}_{02}$ is the full-load copper loss (from the SC test at rated current). The formula assumes the terminal voltage stays at its rated value.

Efficiency is highest at the load where the variable copper loss equals the constant core loss. That load is $x_{\eta max}S$ VA, or $I_2=I_{2,rated}\sqrt{P_i/P_{cu,fl}}$. The load fraction does not depend on power factor. The value of ${\eta }_{max}$ does: it is highest at unity power factor.

This is the ratio of energy output to energy input over 24 hours, with $t_k$ in hours and $P_{out,k}=x_{k}S\cos {\varphi }_k$. The core loss is counted for all 24 h, including no-load periods, because it is present whenever the transformer is energised. Copper loss is counted only for the hours the transformer carries each load.

$Z_A$ and $Z_B$ are complex ohmic impedances referred to the same side. $z_A$ and $z_B$ are per-unit impedances, each on its own transformer's rating. Each unit's share is inversely proportional to its ohmic impedance. The share is proportional to its rating only when $z_A=z_B$. The unit with the lower per-unit impedance takes more than its proportional share and reaches full load first.

The common winding carries $I_{LV}-I_{HV}$ and the series winding carries $I_{HV}$. Only $S_{ind}$ passes through transformer action, so the windings and core need only be sized for $S_{ind}$.

This compares the same output and the same voltages, with the same current density and the same mean length of turn in both designs. The saving approaches 100% as $a\to 1$ and is small for large ratios. For example, $a=2$ saves 50% and $a=4$ saves only 25%.

Each winding keeps its rated voltage and current, so the inductively transformed power equals the two-winding rating $S_{2w}$. The auto-transformer's losses are the same as the two-winding transformer's, but its output is larger, so its efficiency is higher. Its per-unit series impedance on its own rating is $(1-1/a_{auto})$ times that of the two-winding transformer. This gives lower regulation and a higher short-circuit current.

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.

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

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

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

The first two terms are reluctance torques. They exist only if a self-inductance varies with $\theta$ (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.

• $P$ = number of poles
• $\varphi$ = 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 $P$

$I_a$ is the total armature current. With $\varphi$ in Wb and $I_a$ in A, $T$ is in N m. $P$, $Z$ and $A$ are the same as in the EMF equation.

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

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.

$R_a$ 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 $V_{\text{brush}}$ is treated as a constant voltage when one is given, and neglected otherwise. It is subtracted for a generator and added for a motor.

$I_L$ is the line (load or supply) current and $R_f$ is the total resistance of the shunt field circuit. In a series machine, $I_a=I_L=I_{\text{series field}}$.

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

This follows directly from $E_b=\frac{P\varphi ZN}{60A}$, with $N$ in rpm. For two operating points of the same machine, $\frac{N_2}{N_1}=\frac{E_{b2}}{E_{b1}}\cdot \frac{{\varphi }_1}{{\varphi }_2}$. Every speed-control method below can be reasoned from this equation.

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

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 $I_a\ne I_L$. It is used there with $I_a\approx I_L$.

${\theta }_m$ 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 $Z{I}_c/(2P)$ armature ampere-turns per pole are cross-magnetising.

$f$ = supply frequency (Hz), $P$ = number of poles. This is the speed of the rotating stator field. In rad/s: ${\omega }_s=\frac{2\pi n_s}{60}=\frac{4\pi f}{P}$.

$F_{max}$ is the peak MMF of one phase winding. Holds for a balanced three-phase winding carrying balanced three-phase currents. The resultant has constant amplitude and rotates at $n_s$.

Dimensionless. $s=1$ at standstill and $s\to 0$ near synchronous speed. $0<s<1$: motoring. $s<0$: the rotor is driven above $n_s$ (induction generator). $s>1$: the rotor turns against the field (braking, as in plugging).

Frequency of the currents induced in the rotor at slip $s$, where $f$ is the stator supply frequency. At standstill ($s=1$), $f_r=f$.

The fixed part $R_2^{\prime }$ dissipates rotor copper loss. The slip-dependent part $R_2^{\prime }\frac{1-s}{s}$ is the equivalent load resistance, and the power it absorbs is the gross mechanical power developed, per phase.

Per-phase rms rotor current, with $V_1$ the per-phase voltage. The stator current is the phasor sum of $I_2^{\prime }$ and the no-load (magnetizing) current.

$P_g=3(I_2^{\prime }{)}^2{R}_2^{\prime }/s$ is the total air-gap power, $P_{cu2}=s{P}_g$ the total rotor copper loss and $P_m=(1-s)P_g$ the gross mechanical power. The split gives rotor power quantities without computing currents. Since $P_m/P_g=1-s$, the motor's overall efficiency cannot exceed $1-s$.

$P_g$ is the total three-phase air-gap power and ${\omega }_s=2\pi n_s/60$ the synchronous speed in rad/s. ${\omega }_r=(1-s){\omega }_s$ is the rotor speed in rad/s. Divide air-gap power by synchronous speed, or mechanical power by rotor speed, and never mix the two. Shaft torque is lower by the rotational-loss torque.

Approximate equivalent circuit. $V_1$ is the per-phase stator voltage. For the exact circuit use the Thevenin values $V_{th},R_{th},X_{th}$ in place of $V_1,R_1,X_1$.

Approximate equivalent circuit. Independent of the applied voltage and proportional to $R_2^{\prime }$. Adding rotor resistance moves the maximum-torque point towards standstill.

Approximate equivalent circuit. Independent of $R_2^{\prime }$: changing the rotor resistance moves where the maximum torque occurs, not its value. Proportional to $V_1^2$.

The general torque expression with $s=1$.

Valid when the stator resistance is neglected ($R_1\approx 0$), so that $s_{mT}=R_2^{\prime }/(X_1+X_2^{\prime })$. The same supply voltage and frequency apply to $T$ and $T_{max}$.

$V_{NL}$ and $I_{NL}$ are line values and $P_{NL}$ is the total three-phase input power. Per-phase values are taken for an equivalent star connection. The second relation uses the low power factor (impedance almost entirely reactive).

$V_{BR}$ and $I_{BR}$ are line values and $P_{BR}$ is the total three-phase input power. Results are per phase for an equivalent star connection. If the test is run at a reduced frequency $f_{BR}$, multiply the reactance by $f/f_{BR}$. The test cannot separate $X_1$ from $X_2^{\prime }$, so a split (often equal) must be assumed or given.

From $T\propto (I_2^{\prime }{)}^2{R}_2^{\prime }/s$ with $R_2^{\prime }$ constant and the magnetizing current neglected. $I_{sc}$ is the DOL (full-voltage) starting current, $I_{fl}$ the full-load current, $s_{fl}$ the full-load slip. $x=1$ for DOL, $x=1/\sqrt{3}$ for star-delta, and $x$ = tap ratio for an autotransformer.

Wound-rotor motor, approximate equivalent circuit. $R_{ext}^{\prime }$ is the added rotor resistance referred to the stator. This choice makes $s_{mT}=1$.

$n$ is the rotor speed in the forward direction. At standstill $s_f=s_b=1$ and $T_f=T_b$, so the net torque $T=0$.

$n_s$ in rpm, $f$ = supply or generated frequency in Hz, $P$ = number of poles. In steady-state operation the rotor and the rotating armature field both turn at this speed. Unlike an induction machine, there is no slip.

$k_w$ = winding factor ($\le 1$), $\Phi$ = fundamental flux per pole (Wb), $T$ = series turns per phase. All quantities are per-phase RMS values of the fundamental, for sinusoidally distributed flux.

$\alpha$ = the angle, in electrical degrees, by which the coil span falls short of full pitch ($180^{\circ }$). For a full-pitch coil, $\alpha =0$ and $k_p=1$.

$m$ = slots per pole per phase. $\beta$ = slot pitch in electrical degrees $=180^{\circ }/(\text{slots per pole})$.

All quantities are per-phase phasors, and $j$ is the imaginary unit. $I$ lags $V$ by the power-factor angle $\theta$ for a lagging load and leads it for a leading load. The equation is valid for a round rotor, where a single $X_s$ applies whatever the rotor position.

Motor convention: $I$ is the current drawn from the supply and $E$ is the back emf. In motoring operation $E$ lags $V$ by the load angle $\delta$.

$E$ = the terminal voltage on no load, with the field current and speed unchanged, when that field current gives rated $V$ at the specified load. $V$ = rated terminal voltage. Both are per phase. Regulation is positive for lagging-pf loads (the terminal voltage falls on loading) and can be zero or negative for leading-pf loads.

Here $X_d$ is the saturated direct-axis (synchronous) reactance in per unit, taken from the OCC at rated voltage. A higher SCR means a smaller synchronous reactance, a larger maximum power ($EV/X_s$), better steady-state stability and a smaller voltage change with load, at the cost of a larger machine.

Per-phase power in W, with per-phase $E$ and $V$. Multiply by 3 for the total 3-phase power. Valid for round-rotor machines only. For a generator, $\delta >0$ when $E$ leads $V$.

Per phase. $E$ lies along the q-axis. $I_d$ and $I_q$ are the components of $I$ perpendicular to and along $E$.

Per phase. $\theta$ = power-factor angle (lagging). Method: form $E^{\prime }=V+jI{X}_q$, whose angle is $\delta$, then add $I_d(X_d-X_q)$ along the same direction to get $E$. For a leading pf, replace $\theta$ with $-\theta$.

Per phase. The first term is the excitation power: it has the same form as for a round rotor, with $X_d$ in place of $X_s$. The second term is the reluctance power. It arises only from saliency ($X_d\ne X_q$) and is present even with zero excitation ($E=0$). With $E=0$ it is the only term, and its maximum is at $\delta =45^{\circ }$. With $E>0$ and $X_d>X_q$, maximum power occurs at some $\delta <90^{\circ }$ that depends on the excitation.

Per phase. The rotor is driven slightly away from synchronous speed with the field winding open, and a reduced balanced 3-phase voltage is applied to the armature. The armature voltage and current swing slowly. $V_{max}$ and $I_{min}$ occur when the armature mmf lies along the d-axis, and $V_{min}$ and $I_{max}$ when it lies along the q-axis.

Per phase, cylindrical rotor, $R_a\approx 0$, in W per electrical radian at operating angle $\delta$. It is the restoring power per unit change in $\delta$ when synchronism is briefly disturbed. Stable operation needs $P_{sy}>0$, which is why a cylindrical rotor is limited to $\delta <90^{\circ }$.

$k_h$ is a material constant and $V$ is the core volume. The exponent (the Steinmetz coefficient) depends on the core material, and 1.6 is the classical value. A problem will state the exponent to use, or will say only that $P_h\propto f$ at constant $B_{max}$.

$t$ is the lamination thickness, and $k_e\propto 1/\rho$ (core resistivity). This form assumes sinusoidal flux and thin laminations. At the same $B_{max}$, $f$ and core volume, halving $t$ cuts $P_e$ to one quarter. This is why cores are laminated rather than solid.

Valid only when $B_{max}$ is held constant across the tests. A core-loss separation test is designed exactly this way.

$N$ is the number of turns of the winding and $A_c$ the net core area. The winding resistance and leakage drops are neglected, so $V\approx E$. If $V$ and $f$ are changed in the same ratio (constant $V/f$), $B_{max}$ stays fixed.

Follows from $B_{max}\propto V/f$. At constant voltage, the eddy current loss does not depend on frequency, and the hysteresis loss falls as frequency rises (with a Steinmetz exponent of 1.6).

Applies to any machine: motor, generator or transformer. Express $P_{out}$, $P_{in}$ and every loss in the same unit (W or kW). The last form is convenient for a motor, when the electrical input is known.

Assumes the output is proportional to the load current (fixed terminal voltage and power factor), a constant loss that does not depend on load, and a variable loss that scales as $I^2$. Setting $d\eta /dI=0$ then gives the condition shown. Any loss that is proportional to $I$ (such as the DC brush-drop loss) drops out of the condition. For a transformer, the load (kVA) at which efficiency is maximum does not depend on power factor. The value of ${\eta }_{max}$ does depend on it, and is highest at unity power factor.

$x$ is the fraction of rated (full) load, $P_i$ the constant loss, and $P_{cu,fl}$ the full-load variable ($I^2$) loss. Valid when the constant loss stays fixed and the variable loss scales as $x^2$. For a transformer, the load at maximum efficiency is $xS$ (kVA).

Output energy over input energy across 24 hours. $t_k$ is the time in hours spent at load fraction $x_k$, with output $P_{out,k}$. The core loss applies for all 24 h while the transformer is energised, even at no load. The copper loss applies only while the transformer carries load.

$s=(n_s-n)/n_s$ is the per-unit slip, with $n_s=120f/P$ rpm. $P_{airgap}$ is the power crossing the air gap: stator input minus stator copper and stator iron losses. The ratio holds for three-phase totals, or per phase if used consistently. So $P_{rotor,cu}=s{P}_{airgap}$ and $P_{mech,gross}=(1-s)P_{airgap}$. Shaft output = $P_{mech,gross}$ − friction and windage − stray load loss. Because of this chain, efficiency can never exceed $(1-s)$, and in practice is below it.