GTU BE03009021 Electrical Machines I Summer 2026 Solved Paper & Solutions

A shaded pole motor is a small, cheap, self-starting single-phase induction motor.

Construction

• The stator has salient (projecting) poles, each carrying the main winding fed from the single-phase supply.
• Each pole is split by a slot. The smaller part carries a thick copper ring called the shading ring (shading coil).
• The rotor is a squirrel cage rotor.

Working

1. The alternating current sets up an alternating flux in the pole.
2. The changing flux induces a current in the shading ring. By Lenz's law, this current opposes the change of flux, so the flux in the shaded part lags the flux in the unshaded part.
3. The two fluxes are apart in space and in time, so they make a weak moving field, from the unshaded part to the shaded part.
4. This field induces currents in the rotor bars and gives a small starting torque. The rotor turns from the unshaded part towards the shaded part.

Features: very low starting torque, low efficiency and fixed direction. Used in table fans, hair dryers, exhaust fans and record players (up to about 40 W).

The circle diagram is the locus of the stator current of an induction motor as the load changes. It is drawn from the results of the no-load test and the blocked rotor test, and it gives the full performance of the motor without loading it.

Data needed

• No-load test: $V$, $I_0$ and $P_0$, so $\cos\phi_0 = \dfrac{P_0}{\sqrt{3}\,V I_0}$.
• Blocked rotor test: $V_{sc}$, $I_{sc}$ and $P_{sc}$, so $\cos\phi_{sc} = \dfrac{P_{sc}}{\sqrt{3}\,V_{sc} I_{sc}}$. The current at rated voltage is $I_{SN} = I_{sc}\dfrac{V}{V_{sc}}$.
• The stator resistance per phase, measured by a DC test.

Construction

1. Choose a current scale. Draw the voltage $V$ along the vertical axis from O.
2. Draw $OO' = I_0$ at angle $\phi_0$ behind $V$.
3. Draw $OA = I_{SN}$ at angle $\phi_{sc}$ behind $V$.
4. Join $O'A$. This is the output line.
5. Draw a horizontal line through $O'$. Draw the perpendicular bisector of $O'A$. It cuts the horizontal line at the centre C.
6. With centre C and radius $CO'$, draw a semicircle through $O'$ and $A$.
7. From A, drop a vertical $AG$ to the horizontal line through $O'$. Divide it at E so that $\dfrac{AE}{EG} = \dfrac{\text{rotor Cu loss}}{\text{stator Cu loss}}$ at short circuit. Join $O'E$. This is the torque line.

Full load performance

Find the full load output in watts, change it to a length with the power scale, and locate the point P on the circle so that the vertical distance from P to the output line equals this length. Drop a vertical from P to the X axis.

All lengths are read with the power scale:

• Input $= PT$, output $= PQ$
• Rotor Cu loss $= QR$, stator Cu loss $= RS$, constant (iron and friction) losses $= ST$
• Rotor input $= PR$, which also gives the torque in synchronous watts

The maximum output, maximum torque and starting torque can also be found from tangents and the point A.

Crawling: the motor runs steadily at a speed much lower than normal, about one seventh of the synchronous speed. The air-gap flux contains harmonics. The 7th harmonic field produces a torque dip near $N_s/7$. If the load torque is more than the motor torque there, the motor stays at that speed. Remedy: chorded, distributed windings.

Cogging: the motor refuses to start at all. When the number of stator slots equals the number of rotor slots (or a multiple of it), the teeth face each other, and the strong magnetic pull locks the rotor. Remedy: unequal slot numbers and skewed rotor slots, which also reduce crawling.

Let the primary be connected to a sinusoidal supply of frequency $f$. The flux in the core changes sinusoidally:

So the emf lags the flux by $90^\circ$.

Dividing, $\dfrac{E_2}{E_1} = \dfrac{N_2}{N_1} = K$, the transformation ratio. The emf per turn, $4.44 f \Phi_m$, is the same in both windings.

Clock notation

In a vector group, the HV line voltage phasor is placed at 12 o'clock. The LV phasor is then placed at the hour that shows its phase shift. Each hour is $30^\circ$. Hour 11 means the LV leads by $30^\circ$, hour 1 means it lags by $30^\circ$, and hour 6 means $180^\circ$. The first capital letter gives the HV connection and the small letter gives the LV connection.

The windings $A_1A_2$ and $a_1a_2$ are on the same limb, so their voltages are in phase. The same is true of $B_1B_2$, $b_1b_2$ and of $C_1C_2$, $c_1c_2$.

(a) Dz6: delta HV, zigzag LV, $180^\circ$ shift

Connections:

• HV delta: $A_2$ to $B_1$, $B_2$ to $C_1$, $C_2$ to $A_1$. Line terminals at $A_2$, $B_2$, $C_2$.
• LV zigzag: each LV winding is split into two halves, and each leg uses halves from two different limbs, joined in opposition. Leg a runs from the star point n through the half on limb b (forward), then the half on limb a (reversed), to terminal a. Legs b and c follow the same pattern with limbs (c, b) and (a, c).

Phasor sum: $V_{na} = k\,(E_b - E_a) = -k\,(E_a - E_b)$. With the HV delta joined as above, the HV phasor of line A points along $E_a - E_b$, so the LV phasor points the opposite way. The shift is $180^\circ$ (hour 6).

(b) Yz11: star HV, zigzag LV, LV leads by $30^\circ$

Connections:

• HV star: $A_1$, $B_1$, $C_1$ joined to the neutral. Line terminals at $A_2$, $B_2$, $C_2$.
• LV zigzag: leg a runs from the star point n through the half on limb b (reversed), then the half on limb a (forward), to terminal a. Legs b and c follow the same pattern with limbs (c, b) and (a, c).

Phasor sum: $V_a = k\,(E_a - E_b)$. With $E_a$ at $90^\circ$ and $E_b$ at $-30^\circ$, the sum points at $120^\circ$, which is $30^\circ$ ahead of $E_A$.

(c) Yd11: star HV, delta LV, LV leads by $30^\circ$

Connections:

• HV star: $A_1$, $B_1$, $C_1$ joined to the neutral. Line terminals at $A_2$, $B_2$, $C_2$.
• LV delta: $a_2$ to $b_1$, $b_2$ to $c_1$, $c_2$ to $a_1$. Line terminals at $a_2$, $b_2$, $c_2$.

Phasor geometry: the three LV winding voltages form the sides of a triangle. The line from the centre of this triangle to $a_2$ points $30^\circ$ ahead of $E_A$, so the LV line voltage leads the HV line voltage by $30^\circ$ (hour 11).

Summary

The stator of a single-phase induction motor carries a single winding fed by AC, so it produces a pulsating field, not a rotating one. The field only grows and shrinks along one axis.

By the double revolving field theory, this pulsating field equals two fields of half the size, rotating at synchronous speed in opposite directions. At standstill, each field produces an equal torque on the rotor, in opposite directions, so the net starting torque is zero.

So the motor does not start by itself. Once pushed either way, it keeps running in that direction. An auxiliary winding or a shading ring is used to start it.

$V_2$ and $\cos\phi$ are fixed, so $\eta$ is maximum when the denominator is minimum.

Given: 100 kVA, 1100/440 V, single phase. O.C. test: iron loss $P_i = 500$ W. S.C. test (on the HV side, secondary shorted): $V_{sc} = 40$ V, $P_{sc} = 850$ W at full load current, so the full load copper loss is $P_{cu} = 850$ W. Power factor $0.8$ lagging.

Full load efficiency

Regulation

Check with per cent values: $\%R = \dfrac{850}{100000} \times 100 = 0.85\%$ and $\%Z = \dfrac{40}{1100} \times 100 = 3.64\%$, so $\%X = \sqrt{3.64^2 - 0.85^2} = 3.54\%$. Regulation $= 0.85 \times 0.8 + 3.54 \times 0.6 = 2.80\%$.

A Buchholz relay is a gas-operated relay that protects an oil-filled transformer against internal faults. It is used on transformers with a conservator.

Position: it is fitted in the pipe between the main tank and the conservator, so all gas from the tank must pass through it.

Construction: a chamber filled with oil, with an upper float connected to an alarm circuit, and a lower float (or flap) connected to the trip circuit of the circuit breaker. Each float carries a mercury switch.

Working:

1. Minor fault (insulation failure between turns, core heating, a loose joint): the oil slowly breaks down and forms gas. The gas rises and collects at the top of the relay, pushing the oil level down. The upper float tilts and rings the alarm.
2. Major fault (short circuit between windings or to earth): a large volume of gas forms suddenly and pushes a surge of oil towards the conservator. This tilts the lower float, which trips the circuit breaker and disconnects the transformer.

The collected gas can be tested to find the type of fault. The relay gives early warning of slow faults.

Methods of speed control

The speed of an induction motor is $N = N_s(1 - s) = \dfrac{120f}{P}(1 - s)$. So it can be changed by changing $f$, $P$ or $s$.

From the stator side:

1. Supply voltage control
2. Supply frequency control (V/f control)
3. Changing the number of poles (pole changing)
4. Adding resistance or reactance in the stator circuit

From the rotor side (slip ring motors only):

5. Rotor resistance control
6. Injecting an emf into the rotor (slip power recovery, Kramer and Scherbius drives)
7. Cascade (concatenation) of two motors

V/f control in detail

Principle: the synchronous speed is $N_s = \dfrac{120f}{P}$, so changing the supply frequency changes the speed smoothly over a wide range.

Why the voltage must change with frequency: the stator voltage is $V \approx 4.44\,f N \Phi K_w$, so the air-gap flux is

• If $f$ is lowered at a fixed $V$, the flux rises, the core saturates, and the magnetising current and losses rise.
• If $f$ is raised at a fixed $V$, the flux falls and the torque falls.

So the voltage is changed in proportion to the frequency, keeping $V/f$ constant. The flux stays at its rated value, and the maximum torque stays the same at every speed.

How it is done: a variable frequency drive (VFD). A rectifier changes the AC supply to DC, and a PWM inverter changes the DC back into AC of the wanted voltage and frequency.

Torque-speed curves: each frequency gives a curve of the same shape and the same maximum torque, shifted along the speed axis. This gives a constant torque region below the rated speed. Above the rated frequency, the voltage cannot rise further, so the flux and the torque fall (constant power or field weakening region).

Advantages: smooth, wide and efficient speed control, low starting current, high starting torque and soft starting.

Uses: pumps, fans, conveyors, lifts, electric vehicles and machine tools.

Double field revolving theory: a pulsating (alternating) field of maximum value $\Phi_m$ can be replaced by two fields, each of value $\Phi_m / 2$, rotating at synchronous speed in opposite directions.

At any instant, the sum of the two rotating fields gives the actual pulsating field. When both point in the same direction, the sum is $\Phi_m$. When they point in opposite directions, the sum is zero.

In a single-phase induction motor, each field makes its own torque. At standstill, the forward and backward torques are equal and opposite, so there is no starting torque. Once the rotor turns in one direction, the forward torque becomes larger and the motor keeps running.

An energy efficient (premium efficiency) induction motor is designed to have lower losses than a standard motor of the same rating, so it uses less electricity for the same output. It meets the higher efficiency classes, such as IE3 or IE4.

How the losses are reduced: more and better copper in the windings, thinner laminations of better steel, a longer core, a better fan and better bearings, and a smaller air gap.

It is best for motors that run for long hours, such as pumps, fans and compressors.

The Scott connection changes a three-phase supply into a two-phase supply (or the reverse), using two single-phase transformers. The two-phase voltages are equal in size and $90^\circ$ apart.

Connection

• The main transformer has $N_1$ primary turns with a centre tap O. Its ends are joined to lines R and Y.
• The teaser transformer has $\dfrac{\sqrt{3}}{2}N_1 = 0.866\,N_1$ primary turns. One end is joined to O and the other to line B.
• Both secondaries have the same turns $N_2$ and give the two phases.

Phasor analysis

Let $V_L$ be the line voltage. Points R, Y and B form an equilateral triangle, and O is the mid-point of RY.

• The main primary has $V_{RY} = V_L$ across it.
• The line BO is the height of the triangle, so it is at $90^\circ$ to RY and its length is

The teaser has $0.866\,N_1$ turns, so both transformers have the same volts per turn. The secondary voltages are

They are equal in size and $90^\circ$ apart, which is a balanced two-phase supply. If the two-phase load is balanced, the three line currents are also balanced.

Applications

1. Electric arc furnaces and induction furnaces that need a two-phase supply.
2. Railway traction: two single-phase loads (two sections of track) can be fed from a three-phase grid while keeping the grid fairly balanced.
3. Two-phase motors and servo systems, from a three-phase supply.
4. Two-phase to three-phase conversion, by using the connection in reverse.
5. Phase angle measurement and testing, where a $90^\circ$ supply is needed.