DC Motor

Back emf is the emf induced in the armature conductors of a running DC motor. When the armature rotates in the magnetic field, its conductors cut the flux and an emf is induced (generator action). By Lenz's law this emf opposes the applied voltage, so it is called back emf or counter emf, $E_b$.

$$E_b = \frac{P \phi Z N}{60 A}, \qquad V = E_b + I_a R_a$$

   +----Ra----+
   |   Ia ->  |
   V         (Eb)  opposes V
   |          |
   +----------+

So the armature current is

$$I_a = \frac{V - E_b}{R_a}$$

Roles of back emf

1. Limits armature current: $R_a$ is very small (e.g. 0.5 Ω). Without $E_b$, a 220 V motor would draw $220/0.5 = 440$ A. With $E_b = 210$ V, $I_a = (220 - 210)/0.5 = 20$ A only. Back emf acts like a self-adjusting resistance.

2. Makes the motor self-regulating (adjusts current to load):

   • Load increases → speed falls slightly → $E_b$ falls → $I_a$ rises → torque rises to meet the load.
   • Load decreases → speed rises → $E_b$ rises → $I_a$ falls → torque falls. The motor automatically draws just the current needed by the load.

3. Makes energy conversion possible: multiplying $V = E_b + I_a R_a$ by $I_a$:

   $$V I_a = E_b I_a + I_a^2 R_a$$

   $E_b I_a$ is the gross mechanical power developed. Without back emf there would be no conversion; all input would be wasted as heat. Maximum mechanical power occurs when $E_b = V/2$.

4. Determines speed: $N \propto \dfrac{E_b}{\phi} = \dfrac{V - I_a R_a}{\phi}$. This is the basis of speed control by flux or armature voltage.

5. Need for a starter: at starting $N = 0$ so $E_b = 0$, and current would be very large. That is why a starter resistance is put in series until $E_b$ builds up.

A starter is a device that puts a resistance in series with the armature of a DC motor while starting, and cuts it out step by step as the motor speeds up. It also provides protection against no-voltage and overload.

Necessity of a starter

In a DC motor, $I_a = \dfrac{V - E_b}{R_a}$. At the instant of starting, speed is zero, so back emf $E_b = 0$:

$$I_{a,start} = \frac{V}{R_a}$$

$R_a$ is very small, so this current is huge. For example, a 220 V motor with $R_a = 0.5\ \Omega$ would draw $220/0.5 = 440$ A, perhaps 15-20 times full-load current. This would:

• burn the armature winding and damage insulation,
• cause heavy sparking at the commutator and brushes,
• give a sudden large torque that can damage the shaft and load,
• cause a big voltage dip on the supply.

So an external resistance $R$ is added at start: $I_{start} = V/(R_a + R)$, and it is cut out as $E_b$ builds up. Small motors (below about 1 hp) can be started directly because their $R_a$ is higher and inertia low.

Three-point starter

It has three terminals: L (line), F (field) and A (armature). It is used for shunt and compound motors.

 L (+) o----[OLR coil]----o Handle pivot
                           \
           studs: 1  2  3  4  5  ... ON
           o--R--o--R--o--R--o--R--o
           |     (starting resistance)
           |                       |
       to A (armature)        [NVC]--- to F (field)
                                 (holding magnet)

 Supply -ve ------ motor armature and field return

Parts

• Starting resistance divided into sections between studs.
• Handle with a soft-iron piece, held in OFF position by a spring.
• No-volt coil (NVC) / hold-on coil: in series with the shunt field.
• Overload release (OLR) coil: in series with the line, with an armature that shorts the NVC when it is lifted.

Working

1. The handle is moved from OFF to stud 1. The field gets full supply voltage through the NVC, so full flux is produced. All starting resistance is in series with the armature, so starting current is limited.
2. As the motor speeds up, $E_b$ grows and current falls; the handle is moved stud by stud, cutting out resistance.
3. At the last (ON) stud all resistance is cut out of the armature circuit. The NVC, energised by field current, holds the handle against the spring.

Protections

• No-voltage protection: if supply fails or the field circuit opens, the NVC loses its magnetism and the spring returns the handle to OFF. So the motor cannot restart at full voltage without the starter when supply returns. It also prevents dangerous over-speeding if the field opens.
• Overload protection: if current exceeds a set value, the OLR lifts its armature and short-circuits the NVC. The handle is released to OFF and the motor is disconnected.

Drawback: for speed control by field resistance, a large field resistance weakens the NVC current, and the handle may release unwantedly. This is overcome by the four-point starter, where the NVC is connected directly across the supply.

A DC motor converts DC electrical energy into mechanical energy. It works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force, $F = B I l$ newtons, whose direction is given by Fleming's left-hand rule.

Working principle

1. The field winding produces a steady flux between N and S poles.
2. DC supply is given to the armature through brushes and commutator, so current flows in the armature conductors.
3. Each conductor carries current in the field and experiences a force $F = BIl$. Conductors under the N pole and under the S pole carry opposite currents (because of the commutator), so their forces act in the same rotational sense and produce a torque.
4. When a conductor moves from one pole to the next, the commutator reverses its current, so the torque stays in one direction and the rotation is continuous.
5. As the armature rotates, a back emf $E_b$ is induced which opposes the supply: $V = E_b + I_a R_a$.

        N pole
   ---------------
    (x) (x)  -> F      x = current into page
       armature        . = current out of page
    (.) (.)  <- F
   ---------------
        S pole
  Forces form a couple -> rotation

Torque equation

Method 1: from force on conductors

Let $r$ = armature radius (m), $l$ = conductor length (m), $I_a/A$ = current per conductor. Average flux density:

$$B = \frac{P\phi}{2\pi r l}$$

Force per conductor $= B \dfrac{I_a}{A} l$. Torque per conductor $= B \dfrac{I_a}{A} l\, r$.

Total torque for $Z$ conductors:

$$\begin{aligned} T_a &= Z \cdot \frac{P\phi}{2\pi r l} \cdot \frac{I_a}{A} \cdot l \cdot r \\ &= \frac{P \phi Z I_a}{2\pi A}\ \text{N·m} \end{aligned}$$

Method 2: from power balance

Mechanical power developed $= E_b I_a$, and also $= T_a \omega = T_a \dfrac{2\pi N}{60}$. So

$$T_a \cdot \frac{2\pi N}{60} = E_b I_a = \frac{P\phi Z N}{60 A} I_a$$ $$T_a = \frac{P \phi Z I_a}{2\pi A} = 0.159\, \frac{P \phi Z I_a}{A}\ \text{N·m}$$

Conclusions

• For a given machine, $T_a = k \phi I_a$, i.e. $T_a \propto \phi I_a$.
• Shunt motor ($\phi$ nearly constant): $T_a \propto I_a$.
• Series motor ($\phi \propto I_a$ before saturation): $T_a \propto I_a^2$.
• In terms of speed: $T_a = 9.55 \dfrac{E_b I_a}{N}$ N·m. The useful shaft torque is $T_{sh} = 9.55 \dfrac{P_{out}}{N}$, which is less than $T_a$ by the torque lost to iron and friction losses.

Given: $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 32$ A, $N_1 = 1500$ rpm. Extra 75 Ω in field circuit, load torque raised by 20%.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Initial condition

$$\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A} \\ I_{a1} &= 32 - 2 = 30\ \text{A} \\ E_{b1} &= 250 - 30 \times 0.2 = 244\ \text{V} \end{aligned}$$

New field current and flux

$$I_{sh2} = \frac{250}{125 + 75} = 1.25\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.25}{2} = 0.625$$

New armature current

$T_2 = 1.2\, T_1$ and $T \propto \phi I_a$:

$$\begin{aligned} \phi_2 I_{a2} &= 1.2\, \phi_1 I_{a1} \\ I_{a2} &= \frac{1.2 \times 30}{0.625} = 57.6\ \text{A} \end{aligned}$$

New speed

$$\begin{aligned} E_{b2} &= 250 - 57.6 \times 0.2 = 238.48\ \text{V} \\ \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ N_2 &= 1500 \times \frac{238.48}{244} \times \frac{1}{0.625} = 2345.7\ \text{rpm} \end{aligned}$$

Answer: New speed $N_2 \approx 2346$ rpm (armature current rises to 57.6 A).

Given: $V = 200$ V, $R_a = 0.12\ \Omega$, $R_{sh} = 250\ \Omega$, rotational loss = 250 W, $I_L = 10$ A, $N = 1500$ rpm.

Currents and back emf

$$\begin{aligned} I_{sh} &= \frac{200}{250} = 0.8\ \text{A} \\ I_a &= I_L - I_{sh} = 10 - 0.8 = 9.2\ \text{A} \\ E_b &= V - I_a R_a = 200 - 9.2 \times 0.12 = 198.896\ \text{V} \end{aligned}$$

(i) Mechanical power developed

$$P_m = E_b I_a = 198.896 \times 9.2 = 1829.84\ \text{W}$$

(ii) Power output

$$P_{out} = P_m - P_{rot} = 1829.84 - 250 = 1579.84\ \text{W}$$

(iii) Load (shaft) torque

$$\begin{aligned} \omega &= \frac{2\pi N}{60} = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s} \\ T_L &= \frac{P_{out}}{\omega} = \frac{1579.84}{157.08} = 10.06\ \text{N·m} \end{aligned}$$

(iv) Full-load efficiency

$$\begin{aligned} P_{in} &= V I_L = 200 \times 10 = 2000\ \text{W} \\ \eta &= \frac{1579.84}{2000} \times 100 = 78.99\% \end{aligned}$$

Check of losses: armature copper $= 9.2^2 \times 0.12 = 10.16$ W, field copper $= 200 \times 0.8 = 160$ W, rotational 250 W; total 420.16 W $= 2000 - 1579.84$ W.

Answer: (i) $P_m \approx 1829.8$ W, (ii) $P_{out} \approx 1579.8$ W, (iii) $T_L \approx 10.06$ N·m, (iv) $\eta \approx 79.0\%$.

Need for a starter

In a DC motor $I_a = \dfrac{V - E_b}{R_a}$. At start the speed is zero, so $E_b = 0$ and $I_a = V/R_a$. Since $R_a$ is very small (e.g. 0.5 Ω), a 220 V motor would draw about 440 A, many times rated current. This would burn the winding, cause heavy sparking at the commutator, give a damaging torque jerk and a dip in supply voltage. A starter puts a resistance in series with the armature at start and cuts it out step by step as $E_b$ builds up. It also gives no-voltage and overload protection.

Four-point starter

It has four terminals: L (line), A (armature), F (field) and N (no-volt coil return to supply). The key change from a three-point starter is that the no-volt coil (NVC) is connected directly across the supply through a protective resistance, not in series with the field.

 L o--[OLR]--o pivot/handle
              \
   studs 1 2 3 4 5 ... ON
   o-R-o-R-o-R-o-R-o
   |                      F o---- shunt field
   A o---- armature         (brass arc from stud 1)
                          
   [NVC]--[R protective]--o N ---- supply (-)

Working

1. The handle is moved from OFF to stud 1. Current divides into three parallel paths: (a) armature through the full starting resistance, (b) shunt field through the brass arc, getting full voltage, and (c) the NVC with its protective resistance, across the supply.
2. As speed rises, $E_b$ increases; the handle is moved stud by stud, cutting out the starting resistance.
3. At the ON position the armature is connected straight to the supply and the NVC holds the handle against its spring.

Protections

• No-voltage release: if supply fails, the NVC is de-energised and the spring returns the handle to OFF.
• Overload release (OLR): on excess current the OLR attracts its armature and shorts the NVC, releasing the handle.

Why four-point

In a three-point starter the NVC carries field current. When a large field regulator resistance is used to raise speed, field current falls, NVC becomes weak and the handle may release while the motor is running. In a four-point starter the NVC current does not depend on field current, so field control over a wide range is possible without false tripping.

Drawback: if the field circuit opens, the NVC still holds the handle, so there is no protection against over-speeding due to field failure.

In a series motor flux is proportional to the series field (line) current, below saturation. So

$$N \propto \frac{E_b}{\phi} \propto \frac{E_b}{I}, \qquad E_b = V - I(R_a + R_{se}) - V_{brush}$$

Given: $V = 220$ V, $I_1 = 38$ A, $N_1 = 600$ rpm, $R_a + R_{se} = 0.4 + 0.2 = 0.6\ \Omega$, brush drop 3 V.

$$E_{b1} = 220 - 38 \times 0.6 - 3 = 194.2\ \text{V}$$

a) Speed at 19 A

$$\begin{aligned} E_{b2} &= 220 - 19 \times 0.6 - 3 = 205.6\ \text{V} \\ N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\ &= 600 \times \frac{205.6}{194.2} \times \frac{38}{19} = 1270.4\ \text{rpm} \end{aligned}$$

b) Speed at 1 A (load removed)

$$\begin{aligned} E_{b3} &= 220 - 1 \times 0.6 - 3 = 216.4\ \text{V} \\ N_3 &= 600 \times \frac{216.4}{194.2} \times \frac{38}{1} = 25\,406\ \text{rpm} \end{aligned}$$

Answer: a) $N \approx 1270$ rpm; b) $N \approx 25\,400$ rpm.

This dangerously high value (about 42 times rated speed) shows why a series motor must never be started or run without load: as current falls, flux falls and speed rises without limit. (In practice residual flux and windage keep it a little lower, but it would still be destructive.)

Necessity of testing DC machines

Testing is done to find the performance of a DC machine without (or before) putting it in service:

• To find losses (constant and variable) and hence efficiency at any load.
• To check temperature rise and heating under load, which decides the rating.
• To check commutation and sparking at the brushes.
• To find characteristics (OCC, load and speed-torque curves) and confirm the design and name-plate values.
• For acceptance and quality control by manufacturer and buyer.

Direct loading of large machines wastes much energy and needs a large load, so indirect tests such as Swinburne's (no-load) test and the retardation test are used.

Swinburne's test

The machine runs as a shunt motor on no load at rated voltage and speed. Input $V I_0$ minus the small no-load armature copper loss gives the constant losses (iron + mechanical) directly. Efficiency at any load is then calculated.

Retardation (running-down) test

The machine is run above rated speed and the supply to the armature is cut off (field kept excited for iron loss). The armature slows down as its stored kinetic energy supplies the rotational losses. The time to fall through a known speed range (say 1.1 $N$ to 0.9 $N$) is measured:

$$P_{rot} = \omega J \frac{d\omega}{dt}$$

$J$ is found by repeating the test with a known extra load (or a flywheel). Running down with field off gives mechanical loss alone, so iron loss is separated.

Difference

In a separately excited motor $T = k\phi I_a$ and $E_b = k\phi\omega$. The developed power $E_b I_a = T\omega$.

Given: $T = 300$ N·m (constant), $V = 600$ V, $N_1 = 1500$ rpm, $R_a = 0.5\ \Omega$.

Part 1: Armature current

$$\omega_1 = \frac{2\pi \times 1500}{60} = 157.08\ \text{rad/s}$$ $$E_{b1} I_{a1} = T \omega_1 = 300 \times 157.08 = 47\,124\ \text{W}$$

With $E_{b1} = 600 - 0.5 I_{a1}$:

$$\begin{aligned} (600 - 0.5 I_{a1}) I_{a1} &= 47\,124 \\ 0.5 I_{a1}^2 - 600 I_{a1} + 47\,124 &= 0 \\ I_{a1} &= \frac{600 - \sqrt{600^2 - 4(0.5)(47\,124)}}{2(0.5)} = 84.49\ \text{A} \end{aligned}$$

(The other root, about 1115 A, is not a practical operating point.)

$$E_{b1} = 600 - 0.5 \times 84.49 = 557.76\ \text{V}$$

Part 2: Flux reduced by 10% ($\phi_2 = 0.9\phi_1$)

Torque is constant, so $\phi_1 I_{a1} = \phi_2 I_{a2}$:

$$I_{a2} = \frac{84.49}{0.9} = 93.88\ \text{A}$$ $$E_{b2} = 600 - 0.5 \times 93.88 = 553.06\ \text{V}$$

Speed:

$$\begin{aligned} N_2 &= N_1 \times \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} \\ &= 1500 \times \frac{553.06}{557.76} \times \frac{1}{0.9} = 1652.6\ \text{rpm} \end{aligned}$$

Answer: Initially $I_a \approx 84.5$ A. With 10% less flux, $I_a \approx 93.9$ A and speed $\approx 1653$ rpm.

For a series motor (unsaturated), $\phi \propto I$, so $T \propto I^2$ and $N \propto E_b / I$.

Given: $V = 250$ V, $N_1 = 600$ rpm, $I_1 = 40$ A, $R_a + R_{se} = 0.2\ \Omega$, $N_2 = 300$ rpm, $T_2 = 0.5\,T_1$.

Step 1: New current

$$\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = 0.5 \;\Rightarrow\; I_2 = \frac{40}{\sqrt 2} = 28.28\ \text{A}$$

Step 2: Back emfs

$$E_{b1} = 250 - 40 \times 0.2 = 242\ \text{V}$$ $$\begin{aligned} \frac{N_2}{N_1} &= \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2} \\ E_{b2} &= E_{b1} \times \frac{N_2}{N_1} \times \frac{I_2}{I_1} \\ &= 242 \times \frac{300}{600} \times \frac{28.28}{40} = 85.56\ \text{V} \end{aligned}$$

Step 3: Series resistance

$$\begin{aligned} E_{b2} &= V - I_2 (0.2 + R) \\ 0.2 + R &= \frac{250 - 85.56}{28.28} = 5.814\ \Omega \\ R &= 5.814 - 0.2 = 5.61\ \Omega \end{aligned}$$

Answer: Extra series resistance $R \approx 5.61\ \Omega$ (current falls to 28.28 A).

For a DC series motor below saturation, flux is proportional to the current: $\phi \propto I$. Hence

$$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{\phi_1}{\phi_2} = \frac{E_{b2}}{E_{b1}} \times \frac{I_1}{I_2}$$

Given: $V = 230$ V, $I_1 = 100$ A, $N_1 = 1000$ rpm, $R_a = 0.15\ \Omega$, $R_{se} = 0.1\ \Omega$, $I_2 = 25$ A. Brush drop neglected.

  + o---Rse---Ra---(M)---o -
        I = Ia = Ise

Step 1: Back emf at 100 A

$$\begin{aligned} E_{b1} &= V - I_1 (R_a + R_{se}) \\ &= 230 - 100 \times 0.25 = 205\ \text{V} \end{aligned}$$

Step 2: Back emf at 25 A

$$\begin{aligned} E_{b2} &= 230 - 25 \times 0.25 \\ &= 230 - 6.25 = 223.75\ \text{V} \end{aligned}$$

Step 3: New speed

$$\begin{aligned} N_2 &= 1000 \times \frac{223.75}{205} \times \frac{100}{25} \\ &= 1000 \times 1.0915 \times 4 = 4365.9\ \text{rpm} \end{aligned}$$

Answer: Speed at 25 A $\approx 4366$ rpm.

The speed rises more than four times when current falls to one quarter. This is the typical series characteristic: at light load the flux is weak and the speed becomes very high, so a series motor is always coupled to its load directly or through gears, never through a belt.

If the field circuit of a running DC shunt motor opens, the flux falls suddenly to the small residual flux. Since $N \propto \dfrac{E_b}{\phi}$ and $T \propto \phi I_a$, both speed and current change drastically.

What happens

1. Flux collapses: $\phi$ drops to residual value, maybe 2-5% of normal.
2. Back emf drops: at that instant speed cannot change (inertia), so $E_b = k\phi N$ falls almost to zero.
3. Armature current rises sharply: $I_a = \dfrac{V - E_b}{R_a}$ approaches $V/R_a$, many times the rated value, like a start without a starter.
4. Torque and speed:
   • With a light load, the large $I_a$ times the small $\phi$ still gives enough torque, so the motor accelerates rapidly. To build enough back emf with tiny flux, speed must rise to a very high value. The motor runs away (over-speeds).
   • With a heavy load, the torque $k\phi_{res} I_a$ may be less than load torque. The motor slows or stalls while drawing a very large current.

Implications

• Mechanical damage: centrifugal forces at over-speed can throw out armature windings, damage the commutator and bearings, and burst the rotor.
• Electrical damage: very large armature current overheats the winding and insulation.
• Heavy sparking and flashover at the commutator due to large current and weak field (poor commutation).
• Supply disturbance: fuses/breakers trip; voltage dip affects other loads.
• Safety hazard to persons nearby and to the driven machine.
• Inductive kick: opening a highly inductive field can produce a high voltage arc at the break, damaging the field insulation.

Protection

• A three-point starter has its no-volt coil in series with the field, so a field break releases the handle and disconnects the motor.
• Field-failure relays, over-speed trips, and overload/over-current relays or circuit breakers.
• The field circuit should never contain a fuse or switch that can open it alone.

 N, Ia
  |          ___ speed (runaway)
  |        _/
  |      _/   ___ armature current
  |    _/   _/
  |___/___/__________________ time
      ^ field opens

A DC motor converts DC electrical energy into mechanical energy. Its operating principle is that a current-carrying conductor placed in a magnetic field experiences a force $F = B I l$, whose direction is given by Fleming's left-hand rule.

Operating principle

1. The field winding on the poles, fed with DC, sets up a steady magnetic flux.
2. DC supply is fed to the armature conductors through brushes and commutator.
3. Each armature conductor carrying current in the field experiences a force $F = BIl$. Conductors under the N pole carry current in one direction and those under the S pole in the opposite direction, so all forces act in the same rotational sense and produce a torque $T = \dfrac{P\phi Z I_a}{2\pi A}$.
4. As the armature turns, a conductor moving from N pole to S pole has its current reversed by the commutator, so the torque remains unidirectional and the armature keeps rotating.
5. The rotating conductors cut the flux and an emf is induced which opposes the supply: the back emf $E_b$, with $V = E_b + I_a R_a$. Back emf limits the current and makes the motor self-regulating with load.

          N
   +---------------+
   |  x  x  x  x   |  <- force (left)
   |   ( armature ) |     rotation
   |  .  .  .  .   |  -> force (right)
   +---------------+
          S
   x: current in, .: current out

Functions of commutator segments

The commutator is a cylinder of copper segments insulated by mica, each joined to coil ends.

• Mechanical inverter: the supply is DC, but the current in each conductor must reverse as it passes from one pole to the next to keep torque in one direction. The commutator reverses the current in each coil at the correct moment.
• Unidirectional torque: by this reversal, conductors under a given pole always carry current in the same direction.
• Connection of coils: it joins the coils in series to form the parallel paths (lap or wave).
• Sliding contact surface for brushes, transferring current to the rotating armature.

Functions of carbon brushes

Brushes are blocks of carbon or graphite held in brush holders pressed on the commutator by springs.

• Lead current from the stationary supply terminals to the rotating commutator.
• Short-circuit the coil under commutation briefly, helping current reversal.
• Good commutation: carbon has fairly high contact resistance, which helps current reversal (resistance commutation) and reduces sparking.
• Self-lubricating and soft: graphite lubricates the commutator, wears itself rather than the costly commutator, and is easily replaced.

Given: $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 110\ \Omega$, $N_1 = 1400$ rpm at half load, extra 30 Ω in field, load torque +10%.

Note: the current at half load is not given. Since $R_a$ is very small, the $I_a R_a$ drop (about 1-2 V) hardly affects the speed, so the main answer neglects it. A check with an assumed current follows.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Flux ratio

$$\begin{aligned} I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{sh2} = \frac{220}{110 + 30} = 1.5714\ \text{A} \\ \frac{\phi_2}{\phi_1} &= \frac{110}{140} = 0.7857 \end{aligned}$$

New speed (neglecting $I_a R_a$, so $E_{b1} \approx E_{b2}$)

$$N_2 = N_1 \times \frac{\phi_1}{\phi_2} = 1400 \times \frac{140}{110} = 1781.8\ \text{rpm}$$

Check with an assumed half-load line current of 23 A

$$\begin{aligned} I_{a1} &= 23 - 2 = 21\ \text{A}, \quad E_{b1} = 220 - 21 \times 0.08 = 218.32\ \text{V} \\ I_{a2} &= \frac{1.1 \times 21}{0.7857} = 29.4\ \text{A}, \quad E_{b2} = 220 - 29.4 \times 0.08 = 217.65\ \text{V} \\ N_2 &= 1400 \times \frac{217.65}{218.32} \times \frac{1}{0.7857} = 1776.3\ \text{rpm} \end{aligned}$$

The 10% torque increase only raises $I_a$; its effect on speed is under 0.5%.

Answer: New speed $\approx 1782$ rpm (about 1776 rpm if a half-load current of 23 A is assumed).

At starting, speed is zero so back emf $E_b = 0$; only $R_a$ limits the armature current.

Given: $V = 220$ V, $R_a = 0.08\ \Omega$, $R_{sh} = 100\ \Omega$.

Part 1: Starting current at normal voltage

$$\begin{aligned} I_{a,st} &= \frac{V}{R_a} = \frac{220}{0.08} = 2750\ \text{A} \\ I_{sh} &= \frac{220}{100} = 2.2\ \text{A} \\ I_{L,st} &= 2750 + 2.2 = 2752.2\ \text{A} \end{aligned}$$

Part 2: Starting resistance

"Reduced by 80%" is read as: new starting armature current = 20% of 2750 A. Supply voltage is 10% lower, $V' = 0.9 \times 220 = 198$ V.

$$I'_{a,st} = 0.2 \times 2750 = 550\ \text{A}$$ $$\begin{aligned} I'_{a,st} &= \frac{V'}{R_a + R_{st}} \\ R_a + R_{st} &= \frac{198}{550} = 0.36\ \Omega \\ R_{st} &= 0.36 - 0.08 = 0.28\ \Omega \end{aligned}$$

(If line current is used instead: $I'_L = 0.2 \times 2752.2 = 550.44$ A, $I_{sh} = 1.98$ A, $I'_a = 548.46$ A, giving $R_{st} = 0.281\ \Omega$, practically the same.)

Answer: Starting current $\approx 2750$ A in the armature (2752.2 A from the line); starting resistance required $R_{st} \approx 0.28\ \Omega$.

Given: $V = 220$ V, $I_{L1} = 25$ A, $N_1 = 1200$ rpm, $R_a = 0.1\ \Omega$, $R_{sh} = 110\ \Omega$; 50 Ω added in field; load torque reduced by 10%.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Initial condition

$$\begin{aligned} I_{sh1} &= \frac{220}{110} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 220 - 23 \times 0.1 = 217.7\ \text{V} \end{aligned}$$

New field

$$I_{sh2} = \frac{220}{110 + 50} = 1.375\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{1.375}{2} = 0.6875$$

New armature current ($T_2 = 0.9\,T_1$)

$$I_{a2} = \frac{0.9 \times 23}{0.6875} = 30.11\ \text{A}$$

New speed

$$\begin{aligned} E_{b2} &= 220 - 30.11 \times 0.1 = 216.99\ \text{V} \\ N_2 &= 1200 \times \frac{216.99}{217.7} \times \frac{1}{0.6875} = 1739.8\ \text{rpm} \end{aligned}$$

Answer: New speed $\approx 1740$ rpm.

Why a series motor is used for high torque

In a series motor the field carries the full armature current, so (below saturation) $\phi \propto I_a$ and

$$T \propto \phi I_a \propto I_a^2$$

• Doubling the current gives about four times the torque, so a large starting torque is obtained with moderate current. A shunt motor ($T \propto I_a$) would need far more current for the same torque.
• At heavy load the speed falls automatically, so power $T\omega$ stays nearly constant and the supply is not overloaded.
• Hence series motors suit traction (electric trains, trams), cranes, hoists, elevators and conveyors, where high starting torque is needed. They must never run without load, because speed becomes dangerously high.

Compound motor characteristics

A compound motor has both shunt and series fields. In a cumulative compound motor the series flux aids the shunt flux; in a differential one it opposes it.

1. Electrical characteristic ($T$ vs $I_a$)

• Cumulative: $\phi = \phi_{sh} + \phi_{se}$ rises with load, so torque rises faster than in a shunt motor (between shunt and series curves). Good starting torque.
• Differential: $\phi$ falls with load, so torque rises slowly; starting torque is poor.

2. Speed characteristic ($N$ vs $I_a$)

• Cumulative: since flux rises with load, speed falls more than a shunt motor but has a definite no-load speed (shunt field), so no runaway.
• Differential: flux falls with load, so speed stays nearly constant or rises; can be unstable.

3. Mechanical characteristic ($N$ vs $T$): obtained from the two above; cumulative gives a drooping curve, differential a flat/rising one.

 T                           N
 |      series                |\_ differential
 |     /  cumul.              |  \_______ shunt
 |    /  /  shunt             |\_________
 |   /  / /                   |  \____ cumulative
 |  / / / differential        |       \__
 | ///_-----                  |          \ series
 |/____________ Ia            |____________ Ia

Adding resistance in the armature circuit does not change the field current, so flux is constant. Then $T \propto I_a$ and $N \propto E_b$.

Given: $V = 220$ V, $I_{L1} = 20$ A, $N_1 = 1400$ rpm, $R_a = 0.2\ \Omega$, $R_{sh} = 220\ \Omega$, extra 0.1 Ω in armature, torque +10%.

Initial condition

$$\begin{aligned} I_{sh} &= \frac{220}{220} = 1\ \text{A}, \quad I_{a1} = 20 - 1 = 19\ \text{A} \\ E_{b1} &= 220 - 19 \times 0.2 = 216.2\ \text{V} \end{aligned}$$

New condition

Flux constant, torque up 10%:

$$\begin{aligned} I_{a2} &= 1.1 \times 19 = 20.9\ \text{A} \\ E_{b2} &= 220 - 20.9 \times (0.2 + 0.1) = 220 - 6.27 = 213.73\ \text{V} \end{aligned}$$

New speed

$$N_2 = N_1 \times \frac{E_{b2}}{E_{b1}} = 1400 \times \frac{213.73}{216.2} = 1384.0\ \text{rpm}$$

Answer: New speed $\approx 1384$ rpm.

For a series motor, the same current means the same flux, so speed is directly proportional to back emf: $\dfrac{N_2}{N_1} = \dfrac{E_{b2}}{E_{b1}}$.

Given: $V = 200$ V, $I = 15$ A, $N_1 = 800$ rpm, $R_a = 0.6\ \Omega$, $R_{se} = 0.4\ \Omega$, extra series resistance 5 Ω.

Back emfs

$$\begin{aligned} E_{b1} &= V - I (R_a + R_{se}) = 200 - 15 \times 1.0 = 185\ \text{V} \\ E_{b2} &= 200 - 15 \times (1.0 + 5) = 200 - 90 = 110\ \text{V} \end{aligned}$$

New speed

$$N_2 = 800 \times \frac{110}{185} = 475.7\ \text{rpm}$$

Answer: Speed with 5 Ω in series $\approx 476$ rpm.

A diverter is a resistance $R_d$ in parallel with the series field. It takes part of the armature current, so the field current and flux fall and the speed rises.

Given: $V = 500$ V, $N_1 = 500$ rpm, $I_1 = 60$ A, $R_{se} = 0.3\ \Omega$, $R_a = 0.2\ \Omega$, $N_2 = 600$ rpm, torque constant, no saturation ($\phi \propto I_{se}$).

          +--Rse--+
  + o-----|       |----Ra---(M)----o -
          +--Rd---+
   Ia2 splits: Ise2 in Rse, Ia2-Ise2 in Rd

Step 1: Initial back emf

$$E_{b1} = 500 - 60 \times (0.3 + 0.2) = 470\ \text{V}$$

Step 2: Torque condition

$T \propto \phi I_a \propto I_{se} I_a$:

$$I_{se2} I_{a2} = 60 \times 60 = 3600$$

Step 3: Speed condition

$$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \times \frac{I_{se1}}{I_{se2}} = 1.2$$

with $E_{b2} = 500 - 0.2 I_{a2} - 0.3 I_{se2}$ (drop across field and diverter $= 0.3 I_{se2}$).

Substituting $I_{se2} = 3600/I_{a2}$:

$$\begin{aligned} E_{b2} \times \frac{60\, I_{a2}}{3600} &= 1.2 \times 470 = 564 \\ E_{b2} I_{a2} &= 33\,840 \\ 500 I_{a2} - 0.2 I_{a2}^2 - 1080 &= 33\,840 \\ 0.2 I_{a2}^2 - 500 I_{a2} + 34\,920 &= 0 \\ I_{a2} &= 71.91\ \text{A} \end{aligned}$$

(The other root, about 2428 A, is not practical.)

$$I_{se2} = \frac{3600}{71.91} = 50.06\ \text{A}, \quad E_{b2} = 470.60\ \text{V}$$

Check: $N_2 = 500 \times \dfrac{470.60}{470} \times \dfrac{60}{50.06} = 600$ rpm.

Step 4: Diverter resistance

$$\begin{aligned} I_d &= I_{a2} - I_{se2} = 71.91 - 50.06 = 21.84\ \text{A} \\ R_d &= \frac{I_{se2} R_{se}}{I_d} = \frac{50.06 \times 0.3}{21.84} = 0.688\ \Omega \end{aligned}$$

Answer: Diverter resistance $R_d \approx 0.69\ \Omega$ (armature current becomes 71.9 A, field current 50.1 A).

With constant supply voltage, the shunt field current and flux stay constant, so $N \propto E_b$.

Given: $V = 250$ V, $R_a = 1\ \Omega$, $R_{sh} = 125\ \Omega$; no load: $I_{L0} = 5$ A, $N_0 = 1500$ rpm; load: $I_L = 25$ A.

Field current

$$I_{sh} = \frac{250}{125} = 2\ \text{A}$$

No-load (running light)

$$\begin{aligned} I_{a0} &= 5 - 2 = 3\ \text{A} \\ E_{b0} &= 250 - 3 \times 1 = 247\ \text{V} \end{aligned}$$

On load

$$\begin{aligned} I_a &= 25 - 2 = 23\ \text{A} \\ E_b &= 250 - 23 \times 1 = 227\ \text{V} \end{aligned}$$

Speed on load

$$N = N_0 \times \frac{E_b}{E_{b0}} = 1500 \times \frac{227}{247} = 1378.5\ \text{rpm}$$

(Armature reaction is neglected, so flux is taken as unchanged.)

Answer: Speed at 25 A $\approx 1378.5$ rpm.

Adding resistance in series with the armature leaves the field current unchanged, so flux is constant; then $T \propto I_a$ and $N \propto E_b$.

Given: $V = 240$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 120\ \Omega$, $N_1 = 1500$ rpm, $I_{L1} = 5$ A; extra 0.1 Ω in armature; load torque reduced by 20%.

Initial condition

$$\begin{aligned} I_{sh} &= \frac{240}{120} = 2\ \text{A}, \quad I_{a1} = 5 - 2 = 3\ \text{A} \\ E_{b1} &= 240 - 3 \times 0.4 = 238.8\ \text{V} \end{aligned}$$

New condition

$$\begin{aligned} I_{a2} &= 0.8 \times 3 = 2.4\ \text{A} \\ E_{b2} &= 240 - 2.4 \times (0.4 + 0.1) = 240 - 1.2 = 238.8\ \text{V} \end{aligned}$$

New speed

$$N_2 = 1500 \times \frac{238.8}{238.8} = 1500\ \text{rpm}$$

The extra drop in the added resistance exactly balances the reduced drop from the smaller current ($3 \times 0.4 = 2.4 \times 0.5 = 1.2$ V), so the speed is unchanged.

Answer: New speed $= 1500$ rpm (no change).

Given: $V = 250$ V, $R_a = 0.4\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 25$ A at half load, $N_1 = 1200$ rpm; 50 Ω added in field; load torque +20%.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Initial condition

$$\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 250 - 23 \times 0.4 = 240.8\ \text{V} \end{aligned}$$

New field

$$I_{sh2} = \frac{250}{125 + 50} = 1.4286\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{125}{175} = 0.7143$$

New armature current ($T_2 = 1.2\,T_1$)

$$I_{a2} = \frac{1.2 \times 23}{0.7143} = 38.64\ \text{A}$$

New speed

$$\begin{aligned} E_{b2} &= 250 - 38.64 \times 0.4 = 234.54\ \text{V} \\ N_2 &= 1200 \times \frac{234.54}{240.8} \times \frac{1}{0.7143} = 1636.4\ \text{rpm} \end{aligned}$$

Answer: New speed $\approx 1636$ rpm.

Given: $V = 250$ V, $R_a = 0.2\ \Omega$, $R_{sh} = 125\ \Omega$, $I_{L1} = 25$ A at half load, $N_1 = 1400$ rpm; 50 Ω added in field; load torque +20%.

For a shunt motor, take flux proportional to field current (no saturation). Then

$$T \propto \phi I_a, \qquad N \propto \frac{E_b}{\phi}, \qquad E_b = V - I_a R_a$$

Initial condition

$$\begin{aligned} I_{sh1} &= \frac{250}{125} = 2\ \text{A}, \quad I_{a1} = 25 - 2 = 23\ \text{A} \\ E_{b1} &= 250 - 23 \times 0.2 = 245.4\ \text{V} \end{aligned}$$

New field

$$I_{sh2} = \frac{250}{175} = 1.4286\ \text{A}, \qquad \frac{\phi_2}{\phi_1} = \frac{125}{175} = 0.7143$$

New armature current ($T_2 = 1.2\,T_1$)

$$I_{a2} = \frac{1.2 \times 23}{0.7143} = 38.64\ \text{A}$$

New speed

$$\begin{aligned} E_{b2} &= 250 - 38.64 \times 0.2 = 242.27\ \text{V} \\ N_2 &= 1400 \times \frac{242.27}{245.4} \times \frac{1}{0.7143} = 1935.0\ \text{rpm} \end{aligned}$$

Answer: New speed $\approx 1935$ rpm.