DC Motor

Why the starting current is large

Reason: the armature current of a DC motor is

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

At the instant of starting the armature is at rest, so the back emf $E_b = \dfrac{\Phi ZNP}{60A} = 0$. The current is then limited only by the very small armature resistance:

$$I_{a(start)} = \frac{V}{R_a}$$

For example, a 220 V motor with $R_a = 0.5\ \Omega$ would draw $220/0.5 = 440$ A, about 15–20 times its full-load current (say 25 A).

Starter

Necessity of a starter: this large current would

• burn the armature winding and damage the commutator and brushes (heavy sparking),
• cause a large voltage dip on the supply and affect other consumers,
• produce a very high starting torque that can shock and damage the shaft, gears and load.

A starter inserts an external resistance in series with the armature at starting and removes it step by step as the motor gains speed and $E_b$ builds up. It limits the starting current to about 1.25–2 times full-load current:

$$R_{start} = \frac{V}{I_{start}} - R_a$$

It also gives protection (no-volt and overload release).

3-point starter

Parts of a 3-point starter: three terminals L (line), A (armature) and F (field); a starting resistance divided into sections with studs; a spring-loaded handle (starting arm) with a brass arc; a no-volt release coil (NVC), an overload release coil (OLR) and a soft-iron piece on the handle.

   L (+)
    |      OLR (overload coil)
    +------[~~~]---+
                   |   handle (spring loaded)
   OFF  1   2   3  4  ON   o----- soft iron
        |   |   |  |   |     piece, held by NVC
        +[R]+[R]+[R]+---+---------> A (armature)
                        |
             +--- brass arc ---[NVC]----[ Field ]--> F
   L (-) ---------------------------- supply -

Working:

1. With the handle at OFF, the circuit is open. The handle is moved slowly to stud 1. The full starting resistance is now in series with the armature, and the shunt field gets full supply voltage through the brass arc and NVC. This gives full flux, so the motor starts with high torque and low current.
2. As the motor speeds up, $E_b$ rises. The handle is moved stud by stud, cutting out resistance sections so the armature current stays within limits.
3. At the last stud (ON), the whole resistance is cut out and the motor is directly across the supply. The soft-iron piece on the handle touches the NVC electromagnet, which holds the handle there against the spring.
4. No-volt protection: if the supply fails, the NVC loses its magnetism, the spring returns the handle to OFF, and the motor does not restart with the resistance out when the supply returns.
5. Overload protection: if the armature current exceeds a set value, the OLR electromagnet lifts its armature and short-circuits the NVC. The NVC releases the handle, and the spring returns it to OFF.

Drawback: the NVC is in series with the field. If the field current is reduced for speed control, the NVC may be too weak to hold the handle. A 4-point starter places the NVC in a separate branch across the supply to avoid this.

Working principle

A DC motor converts electrical energy into mechanical energy. It works on the principle that a current-carrying conductor placed in a magnetic field experiences a mechanical force:

$$F = BIl\ \text{newton}$$

The direction of the force is given by Fleming's left-hand rule (forefinger = field, middle finger = current, thumb = force).

       N                       S
   +--------+             +--------+
   |   ->   |   (x) F up  |        |
   |  field |  (.) F down |        |
   +--------+   armature  +--------+
              conductors

1. The field winding produces flux $\Phi$ in the air gap.
2. Current from the supply enters the armature conductors through the brushes and commutator.
3. Conductors under the N pole carry current in one direction and those under the S pole in the opposite direction, so forces on both sides act in the same sense and produce a torque that rotates the armature.
4. The commutator reverses the current in a conductor as it passes from one pole to the next, so the torque stays unidirectional.
5. The rotating conductors cut flux and induce a back emf $E_b$ opposing the supply, so $V = E_b + I_aR_a$.

Torque equation

Let $\Phi$ = flux per pole, $Z$ = armature conductors, $P$ = poles, $A$ = parallel paths, $I_a$ = armature current, $r$ = armature radius, $l$ = effective length.

Current per conductor $= I_a/A$. Force on one conductor $F = B\,l\,\dfrac{I_a}{A}$. Torque of one conductor $= F\times r$.

Total torque of $Z$ conductors, with average flux density $B = \dfrac{P\Phi}{2\pi r l}$:

$$T_a = Z\cdot B\,l\,\frac{I_a}{A}\cdot r = Z\cdot\frac{P\Phi}{2\pi r l}\cdot l\cdot\frac{I_a}{A}\cdot r$$ $$T_a = \frac{ZP}{2\pi A}\,\Phi I_a = 0.159\,\frac{ZP}{A}\,\Phi I_a\ \text{N·m}$$

Writing $K_a = \dfrac{ZP}{2\pi A}$:

$$T_a = K_a\Phi I_a$$

Equivalent form from power: $T_a\omega = E_bI_a$ with $E_b = \dfrac{\Phi ZNP}{60A}$ gives the same result. For a shunt motor ($\Phi$ constant) $T \propto I_a$; for an unsaturated series motor ($\Phi\propto I_a$) $T \propto I_a^2$.

Back emf

When the armature of a DC motor rotates, its conductors cut the magnetic flux, so an emf is induced in them (generator action). By Lenz's law this emf opposes the supply voltage which is the cause of the current. It is called the back emf or counter emf:

$$E_b = \frac{\Phi ZNP}{60A}\ \text{V}$$

The voltage equation of the motor is

$$V = E_b + I_aR_a \quad\Rightarrow\quad I_a = \frac{V - E_b}{R_a}$$

How back emf adjusts the torque to the load

The torque is $T = K_a\Phi I_a$, so the motor must draw the armature current that its load requires. Back emf adjusts this current automatically:

 Load up -> speed N falls -> Eb falls -> (V - Eb) rises
        -> Ia rises -> torque rises -> speed stops falling
 Load down -> N rises -> Eb rises -> Ia falls -> torque falls

1. When the load on the shaft increases, the load torque exceeds the motor torque, so the speed falls.
2. $E_b\propto N$ falls, so $(V - E_b)$ increases and a larger armature current $I_a = (V - E_b)/R_a$ flows.
3. The torque $K_a\Phi I_a$ rises until it equals the new load torque, and the motor settles at a slightly lower speed.
4. When the load decreases, the reverse happens: speed rises, $E_b$ rises, $I_a$ falls and torque falls to match the lighter load.

So back emf acts as a self-regulating governor that makes the motor draw only the current needed by the load. At no load $E_b\approx V$ and the current is just enough to supply the losses. The power converted is $E_bI_a = T\omega$.

Back emf is also why the starting current is large: at start $E_b = 0$, so $I_a = V/R_a$ and a starter is required.

The speed of a shunt motor is controlled either by changing the flux, or the armature circuit resistance, or the armature voltage. From the motor equations:

$$N = \frac{V - I_aR_a}{K\Phi} \propto \frac{E_b}{\Phi}$$

So the speed of a shunt motor can be changed by varying the flux $\Phi$, the armature circuit resistance, or the applied voltage.

1. Armature (rheostatic) control

A variable resistance $R_x$ is connected in series with the armature (field kept constant at full value):

$$N = \frac{V - I_a(R_a + R_x)}{K\Phi}$$

• The speed is reduced below the normal (base) speed; the greater $R_x$, the lower the speed.
• Torque for a given $I_a$ is unchanged ($T\propto\Phi I_a$), so it suits constant-torque loads.
• Disadvantages: large $I_a^2R_x$ loss, poor efficiency, and poor speed regulation (the speed changes with load); it is used for small and short-time duty.

2. Field (flux) control

A rheostat $R_f$ is connected in series with the shunt field:

$$I_f = \frac{V}{R_{sh} + R_f}, \qquad N \propto \frac{1}{\Phi}$$

• Increasing $R_f$ lowers $\Phi$, so the speed rises above the base speed (typically up to 2:1 or 3:1).
• It is simple, cheap and efficient (field loss is small, since $I_f$ is small).
• Limits: commutation worsens and the machine can become unstable (armature reaction) at very weak fields; the torque for a given $I_a$ falls as $\Phi$ falls, so it suits constant-power loads.

   Speed
     ^       field control (N up)
     |      /
  base N ---+--- 
     |      \  armature control (N down)
     +--------------------> Rheostat setting

3. Armature voltage control (Ward-Leonard)

Varying the voltage applied to the armature (with a separate source) changes the speed smoothly from zero up to the base speed.

Comparison

A 3-point starter is a device that limits the starting current of a DC shunt (or compound) motor by connecting a resistance in series with the armature at starting, and cutting it out in steps as the motor speeds up. It also gives no-volt and overload protection.

Reason it is needed: at standstill $E_b = 0$, so $I_a = V/R_a$ is very large.

Parts of a 3-point starter: three terminals L (line), A (armature) and F (field); a starting resistance divided into sections with studs; a spring-loaded handle (starting arm) with a brass arc; a no-volt release coil (NVC), an overload release coil (OLR) and a soft-iron piece on the handle.

   L (+)
    |      OLR (overload coil)
    +------[~~~]---+
                   |   handle (spring loaded)
   OFF  1   2   3  4  ON   o----- soft iron
        |   |   |  |   |     piece, held by NVC
        +[R]+[R]+[R]+---+---------> A (armature)
                        |
             +--- brass arc ---[NVC]----[ Field ]--> F
   L (-) ---------------------------- supply -

Working:

1. With the handle at OFF, the circuit is open. The handle is moved slowly to stud 1. The full starting resistance is now in series with the armature, and the shunt field gets full supply voltage through the brass arc and NVC. This gives full flux, so the motor starts with high torque and low current.
2. As the motor speeds up, $E_b$ rises. The handle is moved stud by stud, cutting out resistance sections so the armature current stays within limits.
3. At the last stud (ON), the whole resistance is cut out and the motor is directly across the supply. The soft-iron piece on the handle touches the NVC electromagnet, which holds the handle there against the spring.
4. No-volt protection: if the supply fails, the NVC loses its magnetism, the spring returns the handle to OFF, and the motor does not restart with the resistance out when the supply returns.
5. Overload protection: if the armature current exceeds a set value, the OLR electromagnet lifts its armature and short-circuits the NVC. The NVC releases the handle, and the spring returns it to OFF.

Drawback: the NVC is in series with the field. If the field current is reduced for speed control, the NVC may be too weak to hold the handle. A 4-point starter places the NVC in a separate branch across the supply to avoid this.

In a series motor the field current is the armature current, so (unsaturated) the flux is proportional to the current: $\Phi\propto I$. The back emf then follows $E_b \propto \Phi N \propto I N$.

Total resistance $R = R_a + R_{se} = 0.06 + 0.04 = 0.1\ \Omega$.

Condition 1: 1200 rpm, 25 A

$$E_{b1} = V - I_1R = 220 - 25(0.1) = 217.5\ \text{V}$$

Condition 2: 800 rpm, current $I_2$

$$E_{b2} = 220 - 0.1I_2$$

Using $\dfrac{E_{b2}}{E_{b1}} = \dfrac{\Phi_2N_2}{\Phi_1N_1} = \dfrac{I_2N_2}{I_1N_1}$:

$$\frac{220 - 0.1I_2}{217.5} = \frac{I_2}{25}\times\frac{800}{1200}$$ $$220 - 0.1I_2 = 5.800\,I_2 \;\Rightarrow\; I_2 = \frac{220}{0.1 + 5.800} = 37.29\ \text{A}$$

Check: $E_{b2} = 220 - 0.1(37.29) = 216.27\ \text{V}$.

Answer: the motor draws $I_2 = 37.29\ \text{A}$ at 800 rpm (assuming an unsaturated magnetic circuit).

The field is across the supply, so the flux is constant. Constant torque means constant armature current. The speed is lowered by an external armature resistance $R_x$, so $N\propto E_b$.

Initial condition:

$$I_f = \frac{200}{100} = 2\ \text{A}, \qquad I_a = 50 - 2 = 48\ \text{A}$$ $$E_{b1} = 200 - 48(0.1) = 195.20\ \text{V}$$

At 800 rpm:

$$E_{b2} = E_{b1}\times\frac{800}{1000} = 195.20\times 0.8 = 156.16\ \text{V}$$

The torque is constant, so $I_a = 48$ A:

$$V = E_{b2} + I_a(R_a + R_x) \;\Rightarrow\; R_a + R_x = \frac{200 - 156.16}{48} = 0.9133\ \Omega$$ $$R_x = 0.9133 - 0.1 = 0.8133\ \Omega$$

Answer: external armature resistance $R_x = 0.8133\ \Omega$.

For a fan or centrifugal pump, torque is proportional to the square of speed: $T\propto N^2$. The flux is constant, so $T\propto I_a$ and therefore $I_a\propto N^2$. The speed is lowered by an external armature resistance $R_x$, and $N\propto E_b$.

Initial condition (1000 rpm):

$$I_f = \frac{200}{100} = 2\ \text{A}, \qquad I_{a1} = 50 - 2 = 48\ \text{A}$$ $$E_{b1} = 200 - 48(0.1) = 195.20\ \text{V}$$

At 800 rpm:

$$I_{a2} = I_{a1}\left(\frac{800}{1000}\right)^2 = 48\times 0.64 = 30.72\ \text{A}$$ $$E_{b2} = E_{b1}\times\frac{800}{1000} = 156.16\ \text{V}$$ $$R_a + R_x = \frac{V - E_{b2}}{I_{a2}} = \frac{200 - 156.16}{30.72} = 1.4271\ \Omega$$ $$R_x = 1.4271 - 0.1 = 1.3271\ \Omega$$

Answer: external armature resistance $R_x = 1.3271\ \Omega$ (compare with $0.8133\ \Omega$ for a constant-torque load).

Production of unidirectional torque

When both the armature (through the brushes) and the field winding are supplied with direct current, the armature conductors carry current in a magnetic field, so each conductor experiences a force $F = BIl$ (Fleming's left-hand rule).

• Conductors under the N pole carry current in one direction (into the page); those under the S pole carry it in the opposite direction (out of the page).
• Because the field direction under the two poles is also opposite, the forces on the two sets of conductors act in the same rotational sense, so a net torque turns the armature.

When the armature turns, a conductor moves from under the N pole to under the S pole. If its current stayed the same, the force on it would reverse. The commutator prevents this: its segments reverse the connection of the conductor to the brushes at the moment it crosses the neutral axis, so the current in the conductor reverses at the same time as the field. The product $B\times I$ keeps the same sign, so the torque stays in the same direction (unidirectional).

   Under N pole        Under S pole
   current  (x)  --->  current  (.)  after commutation
   force  -> up        force  -> up   (same sense of rotation)

Factors affecting the torque

From $T = \dfrac{ZP}{2\pi A}\Phi I_a$, the torque depends on:

1. Flux per pole $\Phi$ (field current and magnetic circuit).
2. Armature current $I_a$ (load and supply voltage).
3. Number of armature conductors $Z$.
4. Number of poles $P$.
5. Number of parallel paths $A$ (type of winding: lap $A = P$, wave $A = 2$).

For a given machine $Z$, $P$, $A$ are fixed, so $T\propto\Phi I_a$. For a shunt motor $T\propto I_a$; for an unsaturated series motor $T\propto I_a^2$. Armature reaction (which weakens the flux) and the brush position also change the torque.

In a series motor the field winding carries the armature current, so the flux depends on the load: $\Phi\propto I_a$ (below saturation). The speed is

$$N \propto \frac{E_b}{\Phi} \approx \frac{V - I_a(R_a + R_{se})}{\Phi}$$

• At no load $I_a$ becomes very small, so the flux $\Phi$ becomes very small (only the residual value).
• The numerator $E_b\approx V$ stays nearly constant, so the speed $N\propto 1/\Phi$ rises to a dangerously high value (several times rated speed).
• This is called racing. The centrifugal forces can burst the armature, throw the winding out of the slots and damage the commutator and bearings.
• The torque $T\propto\Phi I_a\propto I_a^2$ is small at no load, so the motor has no load to hold the speed down.

Hence a series motor must always be started and run with a load connected (preferably directly coupled or geared, never by a belt that may slip or break).

False. A DC series motor must never be started (or run) at no load.

Justification:

• The series field carries the armature current, so $\Phi\propto I_a$. At no load $I_a$ is very small, so the flux is very small.
• Speed $N\propto E_b/\Phi$, and $E_b\approx V$, so as $\Phi\to$ residual value the speed rises to a very high, dangerous value (racing).
• The excessive centrifugal force can damage the armature winding, commutator and bearings.
• Torque $T\propto I_a^2$ is small at no load, so there is no load to restrain the speed.

So the series motor should always be started with a load coupled directly or through gears, never with a belt (the belt may slip or break).

Why a series motor cannot be started without load

In a series motor the field winding carries the armature current, so the flux depends on the load: $\Phi\propto I_a$ (below saturation). The speed is

$$N \propto \frac{E_b}{\Phi} \approx \frac{V - I_a(R_a + R_{se})}{\Phi}$$

• At no load $I_a$ becomes very small, so the flux $\Phi$ becomes very small (only the residual value).
• The numerator $E_b\approx V$ stays nearly constant, so the speed $N\propto 1/\Phi$ rises to a dangerously high value (several times rated speed).
• This is called racing. The centrifugal forces can burst the armature, throw the winding out of the slots and damage the commutator and bearings.
• The torque $T\propto\Phi I_a\propto I_a^2$ is small at no load, so the motor has no load to hold the speed down.

Hence a series motor must always be started and run with a load connected (preferably directly coupled or geared, never by a belt that may slip or break).

Speed control of a DC shunt motor

From the motor equations:

$$N = \frac{V - I_aR_a}{K\Phi} \propto \frac{E_b}{\Phi}$$

So the speed of a shunt motor can be changed by varying the flux $\Phi$, the armature circuit resistance, or the applied voltage.

1. Armature (rheostatic) control

A variable resistance $R_x$ is connected in series with the armature (field kept constant at full value):

$$N = \frac{V - I_a(R_a + R_x)}{K\Phi}$$

• The speed is reduced below the normal (base) speed; the greater $R_x$, the lower the speed.
• Torque for a given $I_a$ is unchanged ($T\propto\Phi I_a$), so it suits constant-torque loads.
• Disadvantages: large $I_a^2R_x$ loss, poor efficiency, and poor speed regulation (the speed changes with load); it is used for small and short-time duty.

2. Field (flux) control

A rheostat $R_f$ is connected in series with the shunt field:

$$I_f = \frac{V}{R_{sh} + R_f}, \qquad N \propto \frac{1}{\Phi}$$

• Increasing $R_f$ lowers $\Phi$, so the speed rises above the base speed (typically up to 2:1 or 3:1).
• It is simple, cheap and efficient (field loss is small, since $I_f$ is small).
• Limits: commutation worsens and the machine can become unstable (armature reaction) at very weak fields; the torque for a given $I_a$ falls as $\Phi$ falls, so it suits constant-power loads.

   Speed
     ^       field control (N up)
     |      /
  base N ---+--- 
     |      \  armature control (N down)
     +--------------------> Rheostat setting

3. Armature voltage control (Ward-Leonard)

Varying the voltage applied to the armature (with a separate source) changes the speed smoothly from zero up to the base speed.

Comparison

A DC series motor can run on AC because the torque direction does not depend on the direction of the supply current.

• In a series motor the same current flows through the field winding and the armature, so flux $\Phi\propto I$ and the torque is

$$T\propto\Phi I_a\propto I^2$$

• When the AC supply reverses, the current in the armature and the field both reverse together. The product $\Phi I_a$ keeps the same sign, so the torque is always in one direction (pulsating at twice the supply frequency, smoothed by the rotor inertia).

Such a motor is called a universal motor (works on both DC and AC). For good performance on AC the motor needs these changes:

1. The field core and yoke are laminated to reduce eddy-current and hysteresis loss.
2. The field winding has fewer turns and the armature has more, to reduce the reactance voltage drop and improve power factor.
3. A compensating winding is used to neutralise armature reaction and reduce commutation sparking.

It is used in vacuum cleaners, mixers, drills and sewing machines.

Two main characteristics are drawn at rated (constant) supply voltage: torque vs armature current ($T$–$I_a$) and speed vs torque ($N$–$T$ mechanical characteristic).

Basic relations: $T\propto\Phi I_a$, $\;N\propto\dfrac{E_b}{\Phi} = \dfrac{V - I_aR_a}{\Phi}$.

DC shunt motor

The field is across the supply, so $\Phi$ is practically constant.

• $T$–$I_a$: $T\propto I_a$, a straight line through the origin (it curves down slightly at heavy load because of armature reaction). Starting torque is moderate because starting current is limited.
• $N$–$T$: $N = \dfrac{V - I_aR_a}{K\Phi}$ falls only slightly (about 3–5%) from no load to full load. The speed is nearly constant.

DC series motor

The field carries the armature current, so $\Phi\propto I_a$ (before saturation).

• $T$–$I_a$: $T\propto I_a^2$ for low current (a parabola); after saturation $T\propto I_a$ (a straight line). It gives a very high starting torque.
• $N$–$T$: $N\propto\dfrac{V - I_a(R_a + R_{se})}{I_a}$, so speed falls rapidly as load torque increases. At light load the speed becomes dangerously high, so the motor must never run unloaded.

  T                         N
  |         series          |\  series
  |       _.-'              | \
  |     ,'    shunt         |  \____
  |   ,'    _.-'            |-----------__ shunt
  | ,'  _.-'                |
  +------------- Ia         +------------- T

Comparison

DC series motor

Characteristics: very high starting torque ($T\propto I_a^2$), speed falls as load rises (automatic power adjustment), dangerous at no load.

Suitable for: loads that need high starting torque and can be directly coupled: traction, cranes, hoists, trolleys, vacuum cleaners and similar.

DC shunt motor

Characteristics: almost constant speed from no load to full load, moderate starting torque, easy speed control by field rheostat.

Suitable for: constant-speed loads that start with light load: lathes, drilling machines, fans, blowers, pumps and paper machines.

DC compound motor

Characteristics (cumulative): good starting torque (better than shunt) with a safe no-load speed, and a drooping speed with load.

Suitable for: loads with sudden changes and heavy starting torque: shears, presses, punches, rolling mills, elevators and conveyors (flywheel loads).

Application selection

Why the starting current is high

Reason: the armature current of a DC motor is

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

At the instant of starting the armature is at rest, so the back emf $E_b = \dfrac{\Phi ZNP}{60A} = 0$. The current is then limited only by the very small armature resistance:

$$I_{a(start)} = \frac{V}{R_a}$$

For example, a 220 V motor with $R_a = 0.5\ \Omega$ would draw $220/0.5 = 440$ A, about 15–20 times its full-load current (say 25 A).

Speed at 51 A

The field is across the supply, so $I_{sh} = 250/250 = 1$ A and the flux is constant. Then $N\propto E_b$.

Condition 1 (1000 rpm, 2 A):

$$I_{a1} = 2 - 1 = 1\ \text{A}, \qquad E_{b1} = V - I_{a1}R_a = 250 - 1(0.2) = 249.8\ \text{V}$$

Condition 2 (51 A):

$$I_{a2} = 51 - 1 = 50\ \text{A}, \qquad E_{b2} = 250 - 50(0.2) = 240.0\ \text{V}$$ $$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \;\Rightarrow\; N_2 = 1000\times\frac{240.0}{249.8} = 960.8\ \text{rpm}$$

Percentage speed drop

$$\frac{N_1 - N_2}{N_1}\times 100 = \frac{1000 - 960.8}{1000}\times 100 = 3.92\%$$

Answer: speed at 51 A $= 960.8\ \text{rpm}$; speed drop $= 3.92\%$.

Importance of back emf

• The back emf $E_b = \dfrac{\Phi ZNP}{60A}$ opposes the supply voltage, so $I_a = (V - E_b)/R_a$. It limits the armature current to a safe value once the motor runs.
• It is self-regulating: when the load increases the speed and $E_b$ fall, more current flows and the torque rises to meet the load; the reverse happens for a lighter load.
• It is the means of energy conversion: the electrical power converted to mechanical power is $E_bI_a$.
• At starting $E_b = 0$, so a starter is needed to limit the current.
• Its value shows the speed of the motor, since $N\propto E_b/\Phi$.

Numerical

For a series motor (unsaturated) $\Phi\propto I$, so $E_b\propto N I$.

Condition 1: 1800 rpm, 40 A

$$E_{b1} = V - IR = 240 - 40(0.2) = 232.0\ \text{V}$$

Condition 2: 2400 rpm, 10 A, with an extra series resistance $R_x$

$$\frac{E_{b2}}{E_{b1}} = \frac{N_2I_2}{N_1I_1} \;\Rightarrow\; E_{b2} = 232.0\times\frac{2400}{1800}\times\frac{10}{40} = 77.33\ \text{V}$$

The total resistance in the circuit is found from the voltage equation:

$$V = E_{b2} + I_2(R + R_x) \;\Rightarrow\; 240 = 77.33 + 10(0.2 + R_x)$$ $$0.2 + R_x = \frac{240 - 77.33}{10} = 16.267 \;\Rightarrow\; R_x = 16.07\ \Omega$$

Answer: an external resistance of $16.07\ \Omega$ must be connected in series with the armature.

The field current is constant ($V$ and $R_f$ are fixed), so the flux is constant and $N\propto E_b$.

Field current: $I_f = \dfrac{230}{115} = 2$ A.

Case 1 (600 rpm, line current 60 A):

$$I_{a1} = 60 - 2 = 58\ \text{A}, \qquad E_{b1} = 230 - 58(0.2) = 218.4\ \text{V}$$

Case 2 (armature current 30 A):

$$E_{b2} = 230 - 30(0.2) = 224.0\ \text{V}$$ $$N_2 = N_1\frac{E_{b2}}{E_{b1}} = 600\times\frac{224.0}{218.4} = 615.4\ \text{rpm}$$

Answer: speed at armature current 30 A $= 615.4\ \text{rpm}$.

The current is the same in both cases, so the flux is the same ($\Phi\propto I$). Hence $N\propto E_b$.

Case 1: 1000 rpm, 250 V, 20 A, $R = 1\ \Omega$ (armature plus series field).

$$E_{b1} = V - IR = 250 - 20(1) = 230\ \text{V}$$

Case 2: extra series resistance 6 Ω, so total $R = 1 + 6 = 7\ \Omega$ and the same $I = 20$ A.

$$E_{b2} = 250 - 20(7) = 110\ \text{V}$$ $$N_2 = N_1\frac{E_{b2}}{E_{b1}} = 1000\times\frac{110}{230} = 478.3\ \text{rpm}$$

Answer: the motor runs at $478.3\ \text{rpm}$.

The current is unchanged, so the flux is unchanged ($\Phi\propto I$) and $N\propto E_b$.

Total motor resistance: $R = R_a + R_{se} = 0.6 + 0.4 = 1.0\ \Omega$.

Case 1: 800 rpm, $I = 15$ A.

$$E_{b1} = 200 - 15(1.0) = 185\ \text{V}$$

Case 2: 5 Ω added in series, so the total resistance is $1.0 + 5 = 6.0\ \Omega$ and $I = 15$ A.

$$E_{b2} = 200 - 15(6.0) = 110\ \text{V}$$ $$N_2 = 800\times\frac{110}{185} = 475.7\ \text{rpm}$$

Answer: the motor runs at $475.7\ \text{rpm}$.

Use $\Phi\propto I_f$ (saturation neglected), $T\propto\Phi I_a$ and $E_b\propto\Phi N$. With the diverter in parallel with the field only, the armature current is the line current $I_a$, and the field current is a fraction $k = \dfrac{R_d}{R_d + R_f}$ of it.

        Ia         +----[ Rd ]----+
   ---->----[Ra]---+              +----> 
                   +---[ Rf ]-----+     
                      If = k.Ia

Initial condition (no diverter)

$$I_{a1} = I_{f1} = 60\ \text{A}, \qquad E_{b1} = 500 - 60(0.2 + 0.3) = 470\ \text{V}$$

New condition (600 rpm, same load torque)

Let the new armature current be $I_{a2} = x$ and $I_{f2} = kx$.

Torque constant: $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $60\times 60 = kx\cdot x$:

$$kx^2 = 3600 \;\Rightarrow\; I_{f2} = kx = \frac{3600}{x}$$

Back emf:

$$E_{b2} = 500 - 0.2x - 0.3(kx) = 500 - 0.2x - \frac{1080}{x}$$

Speed ratio:

$$\frac{E_{b2}}{E_{b1}} = \frac{\Phi_2N_2}{\Phi_1N_1} = \frac{I_{f2}}{60}\times\frac{600}{500} \;\Rightarrow\; E_{b2} = 470\times 1.2\times\frac{3600/x}{60} = \frac{33840}{x}$$

Equating the two expressions for $E_{b2}$:

$$500 - 0.2x - \frac{1080}{x} = \frac{33840}{x} \;\Rightarrow\; 0.2x^2 - 500x + 34920 = 0$$ $$x = \frac{500 - \sqrt{500^2 - 4(0.2)(34920)}}{2(0.2)} = 71.91\ \text{A}$$

(The smaller root is taken because the motor current stays near the original value.)

Diverter resistance

$$I_{f2} = \frac{3600}{x} = 50.06\ \text{A}, \qquad k = \frac{I_{f2}}{I_{a2}} = 0.6962$$ $$\frac{R_d}{R_d + R_f} = k \;\Rightarrow\; R_d = \frac{kR_f}{1 - k} = \frac{0.6962\times 0.3}{1 - 0.6962} = 0.688\ \Omega$$

Check: $E_{b2} = 500 - 0.2(71.91) - 0.3(50.06) = 470.60\ \text{V}$, and $470\times 1.2\times(50.06/60) = 470.60\ \text{V}$.

Answer: the diverter resistance is $R_d \approx 0.688\ \Omega$ (the motor then draws about $71.91$ A).

For an unsaturated series motor $\Phi\propto I$, so the torque $T\propto\Phi I\propto I^2$, and the speed $N\propto E_b/\Phi\propto E_b/I$.

Case 1: 800 rpm, $I_1 = 100$ A, $R = 0.1\ \Omega$.

$$E_{b1} = 220 - 100(0.1) = 210\ \text{V}$$

Case 2: torque $T_2 = T_1/2$.

$$\frac{T_2}{T_1} = \left(\frac{I_2}{I_1}\right)^2 = \frac12 \;\Rightarrow\; I_2 = \frac{100}{\sqrt{2}} = 70.71\ \text{A}$$ $$E_{b2} = 220 - I_2R = 220 - (70.71)(0.1) = 212.93\ \text{V}$$ $$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}}\times\frac{I_1}{I_2} \;\Rightarrow\; N_2 = 800\times\frac{212.93}{210}\times\frac{100}{70.71} = 1147\ \text{rpm}$$

Answer: the motor runs at about $1147\ \text{rpm}$ when developing half the torque.

Assume the flux is proportional to the field current (unsaturated). The load torque is constant, so $T\propto\Phi I_a$ is constant. Let $R_x$ be the added field resistance.

Initial condition

$$I_{f1} = \frac{250}{125} = 2\ \text{A}, \qquad I_{a1} = 25 - 2 = 23\ \text{A}, \qquad E_{b1} = 250 - 23(0.5) = 238.5\ \text{V}$$

New condition (1100 rpm)

Let $I_{f2} = s$ (A).

Constant torque: $\Phi_1I_{a1} = \Phi_2I_{a2}$, so $2\times 23 = s\,I_{a2}$:

$$I_{a2} = \frac{46}{s}$$

Back emf:

$$E_{b2} = 250 - 0.5I_{a2} = 250 - \frac{23}{s}$$

Speed ratio: $E_b\propto\Phi N$:

$$E_{b2} = E_{b1}\times\frac{\Phi_2}{\Phi_1}\times\frac{N_2}{N_1} = 238.5\times\frac{s}{2}\times\frac{1100}{900} = 145.750\,s$$

Equating:

$$250 - \frac{23}{s} = 145.750\,s \;\Rightarrow\; 145.750\,s^2 - 250s + 23 = 0$$ $$s = \frac{250 + \sqrt{250^2 - 4(145.750)(23)}}{2(145.750)} = 1.6177\ \text{A}$$

(The larger root is taken; the smaller one gives an unrealistically weak field.)

Added resistance

$$R_f + R_x = \frac{250}{s} = \frac{250}{1.6177} = 154.54\ \Omega$$ $$R_x = 154.54 - 125 = 29.54\ \Omega$$

Check: $I_{a2} = 46/1.6177 = 28.44$ A, $E_{b2} = 250 - 0.5(28.44) = 235.78$ V, and $145.750(1.6177) = 235.78$ V.

Answer: additional field resistance $\approx 29.54\ \Omega$.

The flux is constant for a shunt motor, so $N\propto E_b$.

No load: the no-load armature current is small, so the armature drop is neglected and $E_{b0}\approx V = 250$ V at 1000 rpm.

Loaded (line current 50 A):

$$I_{sh} = \frac{250}{250} = 1\ \text{A}, \qquad I_a = 50 - 1 = 49\ \text{A}$$ $$E_b = V - I_aR_a = 250 - 49(0.2) = 240.2\ \text{V}$$ $$N = 1000\times\frac{E_b}{E_{b0}} = 1000\times\frac{240.2}{250} = 960.8\ \text{rpm}$$

Answer: loaded speed $\approx 960.8\ \text{rpm}$.

Assume the flux is proportional to field current. The load torque is constant, so $\Phi I_a$ is constant. Field weakening therefore increases the armature current.

Initial:

$$I_{f1} = \frac{250}{250} = 1\ \text{A}, \qquad E_{b1} = 250 - 20(0.25) = 245\ \text{V}$$

After inserting 250 Ω in the field:

$$I_{f2} = \frac{250}{250 + 250} = 0.5\ \text{A}, \qquad \frac{\Phi_2}{\Phi_1} = \frac{0.5}{1} = 0.5$$

Constant torque ($T\propto\Phi I_a$):

$$I_{a2} = I_{a1}\times\frac{\Phi_1}{\Phi_2} = 20\times 2 = 40\ \text{A}$$ $$E_{b2} = 250 - 40(0.25) = 240\ \text{V}$$

Speed (from $E_b\propto\Phi N$):

$$N_2 = N_1\times\frac{E_{b2}}{E_{b1}}\times\frac{\Phi_1}{\Phi_2} = 1500\times\frac{240}{245}\times 2 = 2939\ \text{rpm}$$

Answer: new speed $\approx 2939\ \text{rpm}$.

The field is across the supply, so the flux is constant. Then torque $\propto I_a$ and speed $\propto E_b$.

Initial condition:

$$I_f = \frac{240}{120} = 2\ \text{A}, \qquad I_{a1} = 5 - 2 = 3\ \text{A}$$ $$E_{b1} = 240 - 3(0.4) = 238.8\ \text{V}$$

New condition: load torque falls by 20%, so $T_2 = 0.8T_1$ and

$$I_{a2} = 0.8\times 3 = 2.4\ \text{A}$$

With the extra 0.1 Ω in the armature circuit, the armature circuit resistance is $0.4 + 0.1 = 0.5\ \Omega$:

$$E_{b2} = 240 - 2.4(0.5) = 238.8\ \text{V}$$ $$N_2 = N_1\times\frac{E_{b2}}{E_{b1}} = 1500\times\frac{238.8}{238.8} = 1500\ \text{rpm}$$

Answer: the new speed is $1500\ \text{rpm}$ (the series resistance drop exactly offsets the reduced armature drop, so the speed is unchanged).

The flux is constant (shunt field across the supply), so torque $\propto I_a$ and speed $\propto E_b$.

Field current: $I_f = 240/120 = 2$ A.

Half-load condition:

$$I_{a1} = 27 - 2 = 25\ \text{A}, \qquad E_{b1} = 240 - 25(0.4) = 230\ \text{V}, \qquad N_1 = 600\ \text{rpm}$$

With the extra 1 Ω, the armature circuit resistance is $0.4 + 1 = 1.4\ \Omega$.

(i) Load torque constant at half-load value

The torque is unchanged, so $I_{a2} = I_{a1} = 25$ A.

$$E_{b2} = 240 - 25(1.4) = 205\ \text{V}$$ $$N_2 = 600\times\frac{205}{230} = 534.8\ \text{rpm}$$

(ii) Load torque increased to full-load torque

Full-load torque is twice the half-load torque, so

$$I_{a2} = 2\times 25 = 50\ \text{A}$$ $$E_{b2} = 240 - 50(1.4) = 170\ \text{V}$$ $$N_2 = 600\times\frac{170}{230} = 443.5\ \text{rpm}$$

Answer: (i) $534.8\ \text{rpm}$; (ii) $443.5\ \text{rpm}$.

The field is across the supply, so the flux is constant and $N\propto E_b$. Constant load torque means constant armature current.

Initial condition:

$$I_f = \frac{220}{100} = 2.2\ \text{A}, \qquad I_a = 40 - 2.2 = 37.8\ \text{A}$$ $$E_{b1} = 220 - 37.8(0.02) = 219.244\ \text{V}$$

At 1200 rpm:

$$E_{b2} = E_{b1}\times\frac{1200}{1400} = 187.923\ \text{V}$$ $$R_a + R_x = \frac{V - E_{b2}}{I_a} = \frac{220 - 187.923}{37.8} = 0.8486\ \Omega$$ $$R_x = 0.8486 - 0.02 = 0.8286\ \Omega$$

Answer: resistance to be inserted in the armature circuit $R_x = 0.8286\ \Omega$.