IN THE EE SYLLABUS

Losses and efficiency of electric machines

Losses in an electric machine split into two groups. Constant losses are the iron (core) loss, which is hysteresis plus eddy current, together with friction and windage and, in a shunt DC machine, the shunt field copper loss. Variable losses grow with load: the I²R copper loss, stray load loss, and in DC machines the brush contact loss. Efficiency is output/input. With output proportional to current, efficiency is maximum when the losses that scale as I² equal the constant losses. The topic covers classifying each loss, separating hysteresis from eddy current loss using core-loss data at two frequencies, how core loss scales with voltage and frequency, and computing efficiency, the load for maximum efficiency and all-day efficiency for transformers, DC machines and induction motors.

2 previous-year questions · 2 solved · 5 practice questions

What are losses, and why they matter

Every electric machine converts input power to output power, and some power is dissipated inside it as heat. These dissipations are called losses. The ratio of useful output power to input power is the machine's efficiency. This topic covers three things: classifying a loss correctly; separating hysteresis loss from eddy current loss using core-loss data at different frequencies; and computing efficiency, and the load at which efficiency is maximum, for transformers, DC machines and induction motors.

Conventions used throughout:

Efficiency is $η = P_{o u t} / P_{in}$ , expressed as a percentage.
For a motor, $P_{o u t}$ is the mechanical shaft power. For a generator or transformer, $P_{o u t}$ is the electrical power delivered to the load.
For three-phase machines, circuit quantities (voltage, current, impedance) are per phase unless stated otherwise. Powers and losses in efficiency calculations are three-phase totals, i.e. $3 \times$ the per-phase value.
All loss formulas assume the stated voltage, frequency and flux density unless a variation is asked for.

Classification of losses

Loss type	Where it occurs	Depends on	Constant or variable with load?
Copper ( $I^{2} R$ ) loss	Armature/stator/rotor windings, transformer windings, field windings	Winding current $I$ and winding resistance $R$	Variable (as $I^{2}$ ) for load-carrying windings. Shunt or separately excited field copper loss is constant when the field current is constant
Hysteresis loss	Magnetic core (laminations)	Peak flux density $B_{ma x}$ , frequency $f$ of flux reversal, core material	Constant for a transformer at fixed $V$ and $f$ . In a DC machine the armature core sees $f = P n /120$ ( $P$ poles, $n$ in rpm), so the loss is constant at fixed flux and speed
Eddy current loss	Magnetic core (laminations)	$B_{ma x}$ , $f$ , lamination thickness $t$ , core resistivity $ρ$	Constant under the same conditions as hysteresis loss
Mechanical (friction + windage) loss	Bearings, brushes, air resistance (rotating machines only)	Speed	Roughly constant at a fixed speed
Brush contact loss (DC machines)	Brush–commutator contact	Armature current $I_{a}$ (brush voltage drop $\times I_{a}$ )	Variable, proportional to $I_{a}$ (not $I^{2}$ )
Stray load loss	Leakage-flux effects and extra eddy currents in conductors and structural parts caused by load current	Load current (taken as proportional to $I^{2}$ )	Variable

Hysteresis loss and eddy current loss together make up the core loss or iron loss ( $P_{i}$ ).

In a transformer:

The open-circuit (no-load) test at rated voltage and frequency gives $P_{i}$ . The small no-load copper loss is neglected.
The short-circuit test at rated current gives the full-load copper loss $P_{c u, f l}$ . The test runs at a much reduced voltage, so the core loss is negligible.

In a DC machine, iron loss + friction and windage + shunt field copper loss form the constant loss. Armature copper loss, brush contact loss and stray load loss form the variable loss.

In an induction motor, stator iron loss + friction and windage form the constant loss. Stator and rotor copper losses plus stray load loss form the variable loss.

EXAM TIP

Watch the word "stray". For DC machines (for example, in no-load or Swinburne-type test calculations), "stray losses" is often used to mean the rotational losses: iron loss + friction + windage. Those are constant. "Stray load loss", on the other hand, is load-dependent and variable. Use the definition the problem gives. If a separate stray load loss figure is given, it goes on the variable side.

Hysteresis and eddy current loss — separating the two

HYSTERESIS LOSS (STEINMETZ'S LAW)

P_{h} = k_{h} B_{ma x}^{1.6} f V

$k_{h}$ 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 $B_{ma x}$ .

EDDY CURRENT LOSS

P_{e} = k_{e} B_{ma x}^{2} f^{2} t^{2} V

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

TOTAL CORE LOSS VS. FREQUENCY

P_{i} = P_{h} + P_{e} = A f + B f^{2} (at constant B_{ma x})

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

To separate $P_{h}$ and $P_{e}$ experimentally, excite the core at the same peak flux density $B_{ma x}$ but at two or more frequencies, and measure the total core loss $P_{i}$ at each. At fixed $B_{ma x}$ , $P_{h} \propto f$ and $P_{e} \propto f^{2}$ . So a plot of $P_{i} / f$ against $f$ is a straight line, $P_{i} / f = A + B f$ : the intercept $A$ gives the hysteresis coefficient and the slope $B$ gives the eddy current coefficient. Solving the same relation as simultaneous equations, without plotting, gives the identical answer.

Write $P_{i} = k_{1} f + k_{2} f^{2}$ for each test frequency, with $B_{ma x}$ fixed.
Substitute the known $(f, P_{i})$ pairs to get simultaneous equations in $k_{1}$ and $k_{2}$ .
Solve for $k_{1}$ (hysteresis loss per hertz) and $k_{2}$ (eddy loss per hertz squared).
Hysteresis loss at any frequency $f$ is $k_{1} f$ and eddy loss is $k_{2} f^{2}$ . As a check, their sum must reproduce each given total core loss.

WHY B_MAX FOLLOWS V/F (SINUSOIDAL SUPPLY)

V \approx E = 4.44 f N Φ_{ma x} = 4.44 f N B_{ma x} A_{c} \Rightarrow B_{ma x} \propto \frac{V}{f}

$N$ 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 $V / f$ ), $B_{ma x}$ stays fixed.

CORE-LOSS SCALING WITH SUPPLY VOLTAGE AND FREQUENCY (SAME WINDING)

P_{e} \propto B_{ma x}^{2} f^{2} \propto V^{2}, P_{h} \propto B_{ma x}^{1.6} f \propto V^{1.6} f^{- 0.6}

Follows from $B_{ma x} \propto V / f$ . 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).

WORKED EXAMPLE

A transformer designed for 50 Hz has hysteresis loss 600 W and eddy current loss 400 W at rated voltage. It is operated at the same voltage but at 60 Hz. Take the Steinmetz exponent as 1.6. Find the new core loss.

Solution

At constant $V$ : $B_{ma x} \propto 1/ f$ , so $B_{60} / B_{50} = 50/60 = 0.8333$ .
Eddy loss $\propto B_{ma x}^{2} f^{2} \propto V^{2}$ , so it is unchanged: $P_{e} = 400$ W.
Hysteresis loss ratio $= (0.8333)^{1.6} \times (60/50) = 0.7470 \times 1.2 = 0.8964$ .
New hysteresis loss $= 600 \times 0.8964 \approx 537.8$ W.
New core loss $= 537.8 + 400 \approx 937.8$ W.

Answer: About 937.8 W (hysteresis ≈ 537.8 W, eddy unchanged at 400 W).

EXAM TIP

If a question changes both $V$ and $f$ while keeping $V / f$ constant, $B_{ma x}$ stays fixed automatically. Then $P_{h} \propto f$ and $P_{e} \propto f^{2}$ apply directly.

Efficiency and the condition for maximum efficiency

EFFICIENCY (GENERAL)

η = \frac{P _{o u t}}{P _{in}} \times 100 = \frac{P _{o u t}}{P _{o u t} + \sum P _{l osses}} \times 100 = \frac{P _{in} - \sum P _{l osses}}{P _{in}} \times 100

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

CONDITION FOR MAXIMUM EFFICIENCY

η = η_{ma x} when (loss \propto I^{2}) = (constant loss)

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 $d η / d I = 0$ 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 $η_{ma x}$ does depend on it, and is highest at unity power factor.

LOAD FRACTION (AND CURRENT) FOR MAXIMUM EFFICIENCY

x = \frac{P _{i}}{P _{c u, f l}}, I_{η ma x} = I_{f l} \frac{P _{i}}{P _{c u, f l}}

$x$ is the fraction of rated (full) load, $P_{i}$ the constant loss, and $P_{c u, f l}$ 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 $x S$ (kVA).

EFFICIENCY AT A FRACTION X OF FULL LOAD (TRANSFORMER)

η (x) = \frac{x S cos ϕ}{x S cos ϕ + P _{i} + x ^{2} P _{c u, f l}} \times 100

$S$ is the rated apparent power in VA (with the losses in W), and $cos ϕ$ is the load power factor. Use this form for efficiency at, say, half load or three-quarter load rather than full load.

ALL-DAY (ENERGY) EFFICIENCY OF A TRANSFORMER

η_{a l l - d a y} = \frac{\sum _{k} P _{o u t, k} t _{k}}{\sum _{k} P _{o u t, k} t _{k} + 24 P _{i} + \sum _{k} x _{k}^{2} P _{c u, f l} t _{k}} \times 100

Output energy over input energy across 24 hours. $t_{k}$ is the time in hours spent at load fraction $x_{k}$ , with output $P_{o u t, k}$ . 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.

WORKED EXAMPLE

A 20 kVA transformer has full-load copper loss of 500 W and iron loss of 320 W. Find (a) the efficiency at half load, unity power factor, and (b) the load (in kVA) at which efficiency is maximum.

Solution

At half load, $x = 0.5$ . Output $= x S cos ϕ = 0.5 \times 20000 \times 1 = 10000$ W.
Copper loss at half load $= x^{2} P_{c u, f l} = 0.25 \times 500 = 125$ W. Iron loss stays at $320$ W (constant).
Input $= 10000 + 125 + 320 = 10445$ W.
$η = 10000/10445 \times 100 \approx 95.74%$ .
Part (b): $x = P_{i} / P_{c u, f l} = 320/500 = 0.64 = 0.8$ .
Load at maximum efficiency $= x S = 0.8 \times 20 = 16$ kVA, at any power factor.

Answer: (a) η ≈ 95.74% at half load; (b) maximum efficiency occurs at 16 kVA (80% of full load).

EXAM TIP

For a DC or induction machine, replace $x S cos ϕ$ with the actual output at that load. When the problem is given in amperes, use the armature or line current $I$ in place of $x$ . The maximum-efficiency condition is the same: the loss that scales as $I^{2}$ equals the constant loss.

Machine-specific notes on losses and efficiency

Machine	Main constant losses	Main variable losses	Point to remember
Transformer	Core (hysteresis + eddy) loss, measured by the OC test at rated voltage	Copper loss ( $I^{2} R$ in both windings), measured by the SC test at rated current	No friction or windage loss, because nothing rotates
DC machine	Iron loss + friction and windage + shunt field copper loss (constant when the field current is constant)	Armature copper loss ( $\propto I_{a}^{2}$ ) + brush contact loss ( $\propto I_{a}$ ) + stray load loss	Maximum efficiency when armature $I_{a}^{2} R_{a}$ loss (plus any $I^{2}$ stray load loss) equals the constant losses
Induction motor	Stator iron loss + friction and windage. Rotor iron loss is usually neglected because the rotor frequency $s f$ is low at normal slip	Stator and rotor copper loss + stray load loss	Rotor copper loss = slip × air-gap power

INDUCTION MOTOR POWER SPLIT

P_{ai r g a p} : P_{r o t or, c u} : P_{m ec h, g r oss} = 1 : s : (1 - s)

$s = (n_{s} - n) / n_{s}$ is the per-unit slip, with $n_{s} = 120 f / P$ rpm. $P_{ai r g a p}$ 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 $P_{r o t or, c u} = s P_{ai r g a p}$ and $P_{m ec h, g r oss} = (1 - s) P_{ai r g a p}$ . Shaft output = $P_{m ec h, g r oss}$ − friction and windage − stray load loss. Because of this chain, efficiency can never exceed $(1 - s)$ , and in practice is below it.

EXAM TIP

When losses are given as percentages of rated output or input, convert them to watts first using the rating. Then apply the same efficiency and maximum-efficiency formulas. Do not average percentages directly.

PREVIOUS-YEAR QUESTIONS

2 questions from the EE papers

Try each question first, then open the answer. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE EE 2024

GATE 2024 EE Q.17

GATE 2024 · EEQ.17 · 1 mark · MCQ

A three phase, 50 Hz, 6 pole induction motor runs at 960 rpm. The stator copper loss, core loss, and the rotational loss of the motor can be neglected. The percentage efficiency of the motor is

(A)
92
(B)
94
(C)
96
(D)
98

Answer and solution

GATE 2024 EE Q.17: answer and solution

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GATE 2024 EE Q.55

GATE 2024 · EEQ.55 · 2 marks · NAT

A 5 kW, 220 V DC shunt motor has 0.5 Ω armature resistance including brushes. The motor draws a no-load current of 3 A. The field current is constant at 1 A. Assuming that the core and rotational losses are constant and independent of the load, the current (in amperes) drawn by the motor while delivering the rated load, for the best possible efficiency, is _______ (rounded off to 2 decimal places).

Answer and solution

GATE 2024 EE Q.55: answer and solution

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PRACTICE

5 questions to test yourself

PRACTICE 12 marks · NAT

In a core-loss separation test on a transformer, the total core loss was measured at two frequencies with $B_{ma x}$ held constant: 1000 W at 50 Hz and 380 W at 25 Hz. Using $P_{h} = k_{1} f$ and $P_{e} = k_{2} f^{2}$ , find the hysteresis loss (in W) at 50 Hz.

Answer and solution

Answer: $520$ (accepted range $\pm 5$ )

Set up two equations: $50 k_{1} + 2500 k_{2} = 1000$ and $25 k_{1} + 625 k_{2} = 380$ .
Double the second equation: $50 k_{1} + 1250 k_{2} = 760$ . Subtract it from the first: $1250 k_{2} = 240$ , so $k_{2} = 0.192$ .
Then $k_{1} = (380 - 625 \times 0.192) /25 = (380 - 120) /25 = 10.4$ .
Hysteresis loss at 50 Hz $= k_{1} f = 10.4 \times 50 = 520$ W.
Check: eddy loss $= 0.192 \times 2500 = 480$ W, and $520 + 480 = 1000$ W.

PRACTICE 21 mark · MCQ

For a transformer supplying a load at constant voltage and power factor, whose copper loss varies as the square of the load current and whose core loss is constant, efficiency is maximum when:

(A)
The copper (variable) loss equals the constant loss
(B)
The copper loss is zero
(C)
The iron loss is zero
(D)
The copper loss is twice the iron loss

Answer and solution

Answer: (A)

$η = x S cos ϕ / (x S cos ϕ + P_{i} + x^{2} P_{c u, f l})$ . Setting $d η / d x = 0$ gives $P_{i} = x^{2} P_{c u, f l}$ .
So efficiency is maximum when the copper (variable) loss at that load equals the constant (core) loss.
Neither loss is zero at that point, and there is no factor of 2.

PRACTICE 32 marks · NAT

A transformer is rated 10 kVA and supplies full load at 0.8 power factor lagging. At full load, the copper loss is 400 W and the iron loss is 225 W. Find the full-load efficiency (in %).

Answer and solution

Answer: $92.75$ (accepted range $\pm 0.3$ )

Output power $= S cos ϕ = 10000 \times 0.8 = 8000$ W.
Total losses at full load $= P_{c u} + P_{i} = 400 + 225 = 625$ W.
Input power $= 8000 + 625 = 8625$ W.
$η = 8000/8625 \times 100 \approx 92.75%$ .

PRACTICE 42 marks · NAT

For the same transformer (full-load copper loss = 400 W, iron loss = 225 W), find the fraction of full load $x$ (as a decimal) at which efficiency is maximum.

Answer and solution

Answer: $0.75$ (accepted range $\pm 0.02$ )

Condition for maximum efficiency: $x^{2} P_{c u, f l} = P_{i}$ .
So $x = P_{i} / P_{c u, f l} = 225/400 = 0.5625$ .
$x = 0.75$ : maximum efficiency occurs at 75% of full load.

PRACTICE 51 mark · MSQ

A transformer is connected to a supply of rated voltage and rated frequency, and its load is increased from no-load to full-load. Which of the following losses stay essentially constant (independent of load current) over this range?

(A)
Hysteresis loss
(B)
Winding copper loss
(C)
Eddy current loss
(D)
Stray load loss

Answer and solution

Answer: (A), (C)

$B_{ma x} \propto V / f$ , and both are fixed by the supply. The primary leakage-impedance drop is small, so $B_{ma x}$ is essentially constant with load.
Hysteresis loss depends on $B_{ma x}$ and $f$ , so it stays essentially constant.
Eddy current loss depends on $B_{ma x}$ and $f$ (and on lamination thickness and resistivity, which are fixed), so it also stays essentially constant.
Winding copper loss is $I^{2} R$ . It rises with load current, so it is variable.
Stray load loss comes from leakage flux produced by the load current, so it also rises with load and is variable.

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Types of losses and efficiency calculations of electric machines

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