1

Ui U

t R

f

Ra

M

If

Ia

Equivalent wiring diagram of the motor:

I

DC motors

1. Parallel (shunt) excited DC motor

A shunt excited DC motor’s terminal voltage is 500 V. The armature resistance is 0,5 Ω, field

resistance is 250 Ω. On a certain load it takes 20 A current from the DC supply and it runs at 409

1/min. Calculate the induced voltage, the output power, the efficiency and the torque of the

motor if the mechanical and the magnetic losses are 900 W together.

Input power:

𝑃𝑖𝑛 = 𝑈𝑡 ∙ 𝐼 = 500 ∙ 20 = 10000 𝑊

The field current:

𝐼𝑓 =𝑈𝑡

𝑅𝑓=

500

250= 2 𝐴

The armature current (according to KCL):

𝐼𝑎 = 𝐼 − 𝐼𝑓 = 20 − 2 = 18 𝐴

The losses of the motor:

The induced voltage (according to KVL):

𝑈𝑡 − 𝑈𝑖 − 𝐼𝑎 ∙ 𝑅𝑎 = 0

𝑈𝑖 = 𝑈𝑡 − 𝐼𝑎 ∙ 𝑅𝑎 = 500 − 18 ∙ 0,5 = 491 𝑉

The internal power:

𝑃𝑏 = 𝑈𝑖 ∙ 𝐼𝑎 = 491 ∙ 18 = 8838 𝑊

The difference between the input and the internal power is due to the electrical losses:

𝑃1 − 𝑃𝑖𝑛𝑡 = 10000 − 8838 = 1132 𝑊 = 𝑃𝑎𝑣 + 𝑃𝑓

𝑃𝑓 = 𝑈𝑡 ∙ 𝐼𝑓 = 500 ∙ 2 = 1000 𝑊 the field power,

𝑃𝑎𝑣 = 𝐼𝑎2 ∙ 𝑅𝑎 = 182 ∙ 0,5 = 162 𝑊 the armature losses.

2

We can get the output power if we subtract the mechanical and the iron losses from the internal

power:

𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛𝑡 − 𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ = 8838 − 900 = 7938 𝑊

The efficiency: 𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛=

7938

10000= 0,7938

To calculate the torque of the motor we have to calculate the angular speed first:

Ω =2𝜋 ∙ 𝑛

60=

2𝜋 ∙ 409

60= 42,83 1/𝑠

𝑀𝑠 =𝑃𝑜𝑢𝑡

Ω=

7938

42,83= 185,33 𝑁𝑚

2. Parallel (shunt) excited DC generator

Let’s see the previous machine operating as generator. Terminal voltage of a shunt excited DC

generator is 500 V, the armature resistance is 0,5 Ω, field resistance is 250 Ω. A 25 Ω load is

connected to its terminals. Calculate the induced voltage, the output power and the efficiency of

the generator if the mechanical and the magnetic losses are 900 W together.

What speed should the shaft rotate if the driving torque is 100 Nm?

The current consumption of the load:

𝐼𝑙 =𝑈𝑘

𝑅𝑡=

500

25= 20 𝐴

The field current:

𝐼𝑓 =𝑈𝑡

𝑅𝑓=

500

250= 2 𝐴

The armature current (according to KCL):

𝐼𝑎 = 𝐼𝑙 + 𝐼𝑓 = 20 + 2 = 22𝐴

The output power of the generator:

𝑃2 = 𝑈𝑡 ∙ 𝐼𝑙 = 500 ∙ 20 = 10000 𝑊

The induced voltage (according to KVL):

Ui U

t R

f

Ra

G

If I

a

Rl

Il

3

𝑈𝑡 − 𝑈𝑖 + 𝐼𝑎 ∙ 𝑅𝑎 = 0

𝑈𝑖 = 𝑈𝑡 + 𝐼𝑎 ∙ 𝑅𝑎 = 500 + 22 ∙ 0,5 = 511 𝑉

The internal power:

𝑃𝑖𝑛𝑡 = 𝑈𝑖 ∙ 𝐼𝑎 = 511 ∙ 22 = 11242 𝑊

The difference between the internal and the output power is due to the electrical losses:

𝑃1 − 𝑃𝑖𝑛𝑡 = 11242 − 10000 = 1242 𝑊 = 𝑃𝑎 + 𝑃𝑓

𝑃𝑓 = 𝑈𝑡 ∙ 𝐼𝑓 = 500 ∙ 2 = 1000 𝑊 is the field power and 𝑃𝑎 = 𝐼𝑎2 ∙ 𝑅𝑎 = 222 ∙ 0,5 = 242 𝑊 is the

armature losses.

To get the input power we have to add the iron and the mechanical losses to the internal power:

𝑃1 = 𝑃𝑏 + 𝑃𝑣𝑎𝑠+𝑠ú𝑟𝑙 = 11242 + 900 = 12142 𝑊

The efficiency: 𝜂 =𝑃2

𝑃1=

10000

12142= 0,8235

To calculate the speed of the motor we have to calculate the angular speed first:

Ω =𝑃1

𝑀ℎ=

12142

100= 121,42 1/𝑠

𝑛 =Ω

2𝜋∙ 60 =

121,42

2𝜋∙ 60 = 1159,47 1/𝑚𝑖𝑛

3. Idling

Calculate the idling speed of the previous motor (1st exercise):

The basic equations of the DC machines:

𝑼𝒊 = 𝒌 ∙ 𝚽 ∙ 𝛀, we can write this equation as:

𝑈𝑖 = 𝑘 ∙ Φ ∙2π

60∙ 𝑛

⇒ Instead of k we can use 𝑐 = 𝑘 ∙

2π

60 as “machine factor”, so we can write:

𝑈𝑖 = 𝑐 ∙ Φ ∙ n (sometimes using this equation is more convenient than transform Ω to n or n to

Ω)

𝑴𝒃 = 𝒌 ∙ 𝚽 ∙ 𝐈𝒂

(k is the machine factor, Φ is the flux, Ω is the angular speed and Ia is the armature current)

Idling means that the machine rotates unloaded, i.e. the torque of the motor is zero:

𝑀𝑠 = 0 ⇒ 𝑃𝑜𝑢𝑡 = 0

a) First we don’t care the losses and calculate the so called “ideal” idling speed in lossless case.

If there are no losses, there are neither iron nor mechanical losses (Piron=0, Pmech=0), the internal

power of the motor is zero too. (Pint=0)

𝑃𝑖𝑛𝑡 = 𝑀𝑖𝑛𝑡 ∙ Ω ⇒ 𝑀𝑖𝑛𝑡 = 0 ⇒ 𝐼𝑎 =𝑀𝑏

𝑘∙Φ= 0, since the flux wasn’t change. (The field circuit

remains to be connected to the power supply.)

4

If there no armature current, there is no voltage drop due to the armature resistance, so the

induced voltage is equal to the terminal voltage. So the ideal idling speed can be calculated as:

𝑛0𝑖𝑑 =𝑈𝑡

𝑐 ∙ Φ

we can get the product of the machine constant and the flux (cΦ) from the results of the 1st

exercise:

When the speed was 409 1/min (the angular speed is Ω=42,83 1/s), the induced voltage was

Ui=491V:

𝑐 ∙ Φ =𝑈𝑖

𝑛=

491

409= 1,2 𝑉𝑚𝑖𝑛, or

𝑘 ∙ Φ =𝑈𝑖

Ω=

491

42,82= 11,463 𝑉𝑠, and

𝑛0𝑖𝑑 =𝑈𝑡

𝑐 ∙ Φ=

500

1,2= 416,67 1/𝑚𝑖𝑛

b. Let’s take into account the losses:

The torque is zero: Mt=0.

The consumed power of the motor is equal to sum of the filed power and the losses.

The internal power:

𝑃𝑖𝑛𝑡 = 𝑃𝑖𝑟𝑖𝑛+𝑚𝑒𝑐ℎ = 900 𝑊, since the iron and the mechanical losses are independent to the

load.

The internal torque: 𝑀𝑖𝑛𝑡 =𝑃𝑖𝑛𝑡

Ω0=

𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ

Ω0≈

𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ

Ω=

900

42,83= 21 𝑁𝑚

To maintain the 21 Nm torque, the motor needs armature current: 𝐼𝑎0 = 𝑀𝑖𝑛𝑡

𝑘∙Φ=

21

11,46= 1,83 𝐴

So the idling induced voltage:

𝑈𝑖0 = 𝑈𝑡 − 𝐼𝑎0 ∙ 𝑅𝑎 = 500 − 1,83 ∙ 0,5 = 499,083 𝑉

The idling speed:

𝑛0 =𝑈𝑖0

𝑐 ∙ Φ=

499,083

1,2= 415,9 1/𝑚𝑖𝑛

(It is almost the same than the result of the lossless case. It means that it is more than enough to

apply the estimation of the lossless case.)

4. Starting the motor

Calculate the required resistance of the starter resistor if the maximum starting armature current

is one and a half times greater that the nominal value of the armature current.

At the moment of starting there is no flux inside the machine so the induced voltage is zero. It

means that the all terminal voltage drops on the low resistance armature:

𝑈𝑡 − 𝑈𝑖 − 𝐼𝑎 ∙ 𝑅𝑎 = 0

5

𝑈𝑡 = 𝐼𝑎 ∙ 𝑅𝑎 ⇒ 𝐼𝑎 =𝑈𝑡

𝑅𝑎=

500

0,5= 1000 𝐴 It would be an impermissible high current.

The simplest way to reduce this high current is increasing the resistance of the armature by a

serial connected resistor.

The maximum armature current:

𝐼𝑎𝑖 = 1,5 ∙ 𝐼𝑎 = 1,5 ∙ 18 = 27 𝐴

The required starter resistance:

𝐼𝑎𝑖 =𝑈𝑡

𝑅𝑎 + 𝑅𝑠𝑡

𝑅𝑖 =𝑈𝑘

𝐼𝑎𝑖− 𝑅𝑎 =

500

27− 0,5 = 18,01 𝐴

5. Speed control

𝑛 =𝑈𝑖

𝑐 ∙ Φ=

𝑈𝑘 − 𝐼𝑎 ∙ 𝑅𝑎

𝑐 ∙ Φ

The speed depends on:

a. Ut terminal voltage

b. the resistance of the armature circuit

c. the flux, which is the dependent upon the field current

So the speed can be controlled with one of the three upper parameters

a. Let’s examine how the terminal voltage influences the speed in cases of the 3 different

exciting mode

a1. Separately excited DC motor

500 V terminal voltage separately excited DC motor’s armature resistance is 0,5 Ω. It takes 20 A

current from the DC supply and its speed is 1000 1/min. Calculate the speed if the terminal voltage is

decreased to 400 V while the load is unchanged.

Since the field current is independent of

the armature, the changing of the terminal

voltage doesn’t influence the flux.

𝑐Φ =𝑈𝑖1

𝑛1=

𝑈𝑖2

𝑛2= á𝑙𝑙.

Since is the load (i.e. the torque of the

motor) doesn’t change, neither the

armature current changes. (𝑀𝑏 = 𝑘 ∙ Φ ∙

I𝑎)

𝑈𝑖1 = 𝑈𝑘1 − 𝐼𝑎 ∙ 𝑅𝑎 = 500 − 20 ∙ 0,5 = 490 𝑉

Ui=0 Uf R

f

Ra

M

If

Iai I

Rst

Ui U

k R

g

Ra

M

Ig

Ia

Ug

6

𝑈𝑖2 = 𝑈𝑘2 − 𝐼𝑎 ∙ 𝑅𝑎 = 400 − 20 ∙ 0,5 = 390 𝑉

so the speed is:

𝑛2 =𝑈𝑖2

𝑈𝑖1∙ 𝑛1 =

390

490∙ 1000 = 795,91 1/𝑚𝑖𝑛

a2. Serial excited DC motor

500 V terminal voltage serial excited DC motor’s armature and field resistance is 5 Ω together. On a

certain load it takes 20 A current from the DC supply and its speed is 1000 1/min. Calculate the speed

if the terminal voltage is decreased to 400 V while the load is unchanged.

The flux of this motor is generated by the armature

current. The armature current depends on the load. If

the load is constant, the armature current, so the flux

stays constant too. (𝑀𝑖𝑛𝑡 = 𝑘 ∙ Φ ∙ I𝑎 = 𝑘′ ∙ 𝐼𝑎2)

𝑐Φ =𝑈𝑖1

𝑛1=

𝑈𝑖2

𝑛2= 𝑐𝑜𝑛𝑠𝑡.

𝑈𝑖1 = 𝑈𝑡1 − 𝐼𝑎 ∙ (𝑅𝑎 + 𝑅𝑓) = 500 − 20 ∙ 5 = 400 𝑉

𝑈𝑖2 = 𝑈𝑡2 − 𝐼𝑎 ∙ (𝑅𝑎 + 𝑅𝑓) = 400 − 20 ∙ 5 = 300 𝑉

𝑛2 =𝑈𝑖2

𝑈𝑖1∙ 𝑛1 =

300

400∙ 1000 = 750 1/𝑚𝑖𝑛.

a3. Parallel excited DC motor

Let’s take the motor of the 1st example. Calculate the speed if the terminal voltage is decreased to

400V while the load is unchanged.

We can observe that changing the terminal voltage

changes the field current:

𝐼𝑓2 =𝑈𝑡2

𝑅𝑓=

400

250= 1,6 𝐴

The flux depends on the field current we can say that the

flux changes approximately proportional to the field

current.

The data of the 1st example: Ut1=500 V, Ui1=491 V, If1=2A, n1=409 1/min, cΦ1 =1,2 Vmin.

when the field current is If2=1,6A ,

𝑐𝜙2 =𝐼𝑔2

𝐼𝑔1∙ 𝑐𝜙1 =

1,6

2∙ 1,2 = 0,96 𝑉𝑚𝑖𝑛

Ui U

k

Rg Ra

M

Ia= I

g

Ui U

k R

g

Ra

M

Ig

Ia

7

Because of the decreased flux the motor needs higher armature current to maintain the same

torque:

𝑀𝑖𝑛𝑡 = 𝑘 ∙ 𝜙1 ∙ 𝐼𝑎1 = 𝑘 ∙ 𝜙2 ∙ 𝐼𝑎2 = 𝑐𝑜𝑛𝑠𝑡.

𝐼𝑎2 =𝑘 ∙ 𝜙1

𝑘 ∙ 𝜙2∙ 𝐼𝑎1 =

𝑐 ∙ 𝜙1

𝑐 ∙ 𝜙2∙ 𝐼𝑎1 =

1,2

0,96∙ 18 = 22,5 𝐴

The induced voltage:

𝑈𝑖2 = 𝑈𝑘2 − 𝐼𝑎2 ∙ 𝑅𝑎 = 400 − 22,5 ∙ 0,5 = 388,75 𝑉

The speed:

𝑛2 =𝑈𝑖2

𝑐 ∙ Φ2=

388,75

0,96= 404,94 1/𝑚𝑖𝑛

We can observe that this method has no effect to the shunt excited motor’s speed.

b. Speed control by increasing the armature resistance

Calculate the resistance connected in series with the armature if we want to decrease the speed of

the motor of the 1st example to 300 1/min while the load is unchanged.

𝑐Φ =𝑈𝑖1

𝑛1=

𝑈𝑖2

𝑛2= 𝑐𝑜𝑛𝑠𝑡.

(since 𝐼𝑓 =𝑈𝑓

𝑅𝑓= 𝑐𝑜𝑛𝑠𝑡. )

𝑈𝑖2 =𝑛2

𝑛1∙ 𝑈𝑖1 =

300

409∙ 491 = 360,15 𝑉

𝑈𝑖1 = 𝑈𝑡1 − 𝐼𝑎 ∙ 𝑅𝑎 = 500 − 18 ∙ 0,5 = 491 𝑉

𝑈𝑖2 = 𝑈𝑡2 − 𝐼𝑎 ∙ (𝑅𝑎 + 𝑅𝑐) ⇒ 𝑅𝑘 =𝑈𝑘2 − 𝑈𝑖2

𝐼𝑎− 𝑅𝑎 =

500 − 360,15

18− 0,5 = 7,27Ω

Calculate the losses of the control resistor:

𝑃𝑅𝑐= 𝐼𝑎

2 ∙ 𝑅𝑐 = 182 ∙ 7,27 = 2355,48 𝑊

c. Speed control with changing the flux

We can control the flux with the field current. The simplest way the control the field current is

changing the resistance of the field circuit applying a resistor connected in series with the field

winding.

Connect a 150 Ω resistor in series to the field

circuit and calculate the speed of the motor of the

1st example. (The load is constant)

Ut R

f

Ra

M

If

Ia I

Rc

Ui

Ui U

t

Rf

Ra

M

If

Ia I

Rc

8

The field current:

𝐼𝑓2 =𝑈𝑡

𝑅𝑔 + 𝑅𝑐=

500

250 + 150= 1,25 𝐴

We suppose that the flux is proportional to the field current:

𝐼𝑔1 = 2 𝐴 ⇒ 𝑐𝛷1 = 1,2 𝑉𝑚𝑖𝑛

and when

𝐼𝑔2 = 1,25 𝐴 ⇒ 𝑐𝛷2 =𝐼𝑔2

𝐼𝑔1∙ 𝑐𝛷1 =

1,25

2∙ 1,2 = 0,75 𝑉𝑚𝑖𝑛

Since the flux decreases, the motor need to take more current to maintain the same torque:

𝑀𝑏 = 𝑘 ∙ 𝜙1 ∙ 𝐼𝑎1 = 𝑘 ∙ 𝜙2 ∙ 𝐼𝑎2 = 𝑐𝑜𝑛𝑠𝑡.

𝐼𝑎2 =𝑘 ∙ 𝜙1

𝑘 ∙ 𝜙2∙ 𝐼𝑎1 =

𝑐 ∙ 𝜙1

𝑐 ∙ 𝜙2∙ 𝐼𝑎1 =

1,2

0,75∙ 18 = 28,8 𝐴

The induced voltage:

𝑈𝑖2 = 𝑈𝑡2 − 𝐼𝑎2 ∙ 𝑅𝑎 = 500 − 28,8 ∙ 0,5 = 485,6 𝑉

The speed:

𝑛2 =𝑈𝑖2

𝑐 ∙ Φ2=

485,6

0,75= 647,47 1/𝑚𝑖𝑛

We can see that this method causes significant change of the speed.

Calculate the losses of the control resistance:

𝑃𝑅𝑐= 𝐼𝑔

2 ∙ 𝑅𝑠𝑧 = 1,252 ∙ 150 = 234,375 𝑊

this is ten times smaller than the losses of the previous method.

6. Nominal power of a 550 V serial excited DC motor is 50 kW. The armature resistance is 0,2 Ω, the

field resistance is 0,1Ω. It takes 100 A current from the DC supply and its speed is 650 1/min.

Calculate the efficiency, the losses, the induced voltage and the torque of the motor!

The nominal power always means the maximum

output power given in the motor catalogue.

The nominal torque of the motor:

𝑀𝑠 =𝑃𝑜𝑢𝑡

Ω=

𝑃2

2𝜋 ∙ 𝑛∙ 60 =

50000

2𝜋 ∙ 650∙ 60

= 734,56𝑁𝑚

The input power:

𝑃𝑖𝑛 = 𝑈𝑡 ∙ 𝐼𝑎 = 550 ∙ 100 = 55000 𝑊

The efficiency:

Ui U

k

Rg Ra

M

Ia= I

g

9

𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛=

50000

55000= 0,909

The induced voltage:

𝑈𝑖 = 𝑈𝑡 − 𝐼𝑎 ∙ (𝑅𝑎 + 𝑅𝑓) = 550 − 100 ∙ (0,1 + 0,2) = 520 𝑉

The internal power:

𝑃𝑏 = 𝑈𝑖 ∙ 𝐼𝑎 = 520 ∙ 100 = 52000 𝑊

The losses:

Armature losses: 𝑃𝑎 = 𝐼𝑎2 ∙ 𝑅𝑎 = 1002 ∙ 0,2 = 2000 𝑊

The field power: 𝑃𝑓 = 𝐼𝑓2 ∙ 𝑅𝑓 = 1002 ∙ 0,1 = 1000 𝑊

The sum of the iron and the mechanical power:

𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ = 𝑃𝑏 − 𝑃2 = 52000 − 50000 = 2000 𝑊

How does the speed change if the load decreases in half?

The load means the torque of the motor. The relationship between the torque and the current:

𝑀𝑏 = 𝑘𝛷 ∙ 𝐼𝑎

The serial excited motor’s flux depends on the armature current. We suppose that the flux is

proportional to the armature current, so the torque is proportional to the square of the current:

𝑀𝑏 = 𝑘′ ∙ 𝐼𝑎2

Calculating the current consumption:

the torque decreases to Mt2= M𝑡1

2=

734,56

2= 367,28 𝑁𝑚, so

𝑀𝑏1

𝐼𝑎12 =

𝑀𝑏2

𝐼𝑎22 ⇒ 𝐼𝑎2 = 𝐼𝑎1 ∙ √

𝑀𝑏2

𝑀𝑏1= 100 ∙ √0,5 = 70.71 𝐴

when the speed is n1=650 1/min , the induced voltage is Ui1=520 V volt, so:

𝑐Φ1 =𝑈𝑖1

𝑛1=

520

650= 0,8 𝑉𝑚𝑖𝑛

The flux is generated by the armature current so when the current consumption decreases to

Ia2=70,71A, the flux decreases too:

𝑐𝜙2 =𝐼𝑎2

𝐼𝑎1∙ 𝑐𝜙1 =

70,71

100∙ 0,8 = 0,5657 𝑉𝑚𝑖𝑛

the induced voltage:

𝑈𝑖2 = 𝑈𝑡 − 𝐼𝑎2 ∙ (𝑅𝑎 + 𝑅𝑔) = 550 − 70,71 ∙ (0,1 + 0,2) = 528,787 𝑉

the speed:

10

𝑛2 =𝑈𝑖2

𝑐𝛷2=

528,787

0,5657= 934,75 1/𝑚𝑖𝑛

We can observe that the load influences significantly the speed of the motor !

7. Serial excited generator

Terminal voltage of a serial excited generator at a 25 Ω load is 500 V. The resistance of the armature

and the field circuit is 1 Ω together. Calculate the efficiency of the generator if sum of the iron and the

mechanical losses is 1,5 kW. Calculate the speed of the generator if the driving torque is 240 Nm.

The current consumed by the load, the

armature current and the field current is the

same.

𝐼𝑎 = 𝐼𝑓 = 𝐼𝑙 =𝑈𝑡

𝑅𝑙=

500

10= 50 𝐴

The output power:

𝑃𝑜𝑢𝑡 = 𝑈𝑡 ∙ 𝐼𝑙 = 500 ∙ 50 = 25000 𝑊

The induced voltage:

𝑈𝑖 = 𝑈𝑡 + 𝐼𝑎 ∙ (𝑅𝑎 + 𝑅𝑓) = 500 + 50 ∙ 1 = 550 𝑉

The internal power:

𝑃𝑖𝑛𝑡 = 𝑈𝑖 ∙ 𝐼𝑎 = 550 ∙ 50 = 57500 𝑊

The input power:

𝑃𝑖𝑛 = 𝑃𝑖𝑛𝑡 + 𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ = 57500 + 1500 = 29000 𝑊

The efficiency:

𝜂 =𝑃2

𝑃1=

25000

29000= 0,862

The speed:

𝑛 =60

2𝜋∙ Ω =

60

2𝜋∙

𝑃1

𝑀ℎ=

60

2𝜋∙

29000

240= 1153,87 1/𝑚𝑖𝑛

8. Terminal voltage of a 400 V separately excited DC motor is 400 V. It takes 22 A, its speed is

955 1/min. The resistance of the armature is 2 Ω, the field resistance is 200Ω. Calculate the

induced voltage, the losses, the efficiency and the torque of the motor if the sum of

mechanical and the iron losses is the 4 % of the

output power.

The input power:

Ui U

k

Rg Ra

G

Ia=Ig=It

Rt

Ui U

t R

f

Ra

M

If I

a

Uf

11

𝑃𝑖𝑛 = 𝑈𝑡 ∙ 𝐼𝑎 = 400 ∙ 22 = 8800 𝑊

The field power:

𝑃𝑓 = 𝑈𝑓 ∙ 𝐼𝑓 =𝑈𝑓

2

𝑅𝑓=

4002

200= 800 𝑊

The induced voltage:

𝑈𝑖 = 𝑈𝑡 − 𝐼𝑎 ∙ 𝑅𝑎 = 400 − 22 ∙ 2 = 356 𝑉

The internal power:

𝑃𝑖𝑛𝑡 = 𝑈𝑖 ∙ 𝐼𝑎 = 356 ∙ 22 = 7832 𝑊

The difference between the input and the internal power is due to the armature losses:

𝑃1 − 𝑃𝑏 = 𝑃𝑎 = 𝐼𝑎2 ∙ 𝑅𝑎 = 222 ∙ 2 = 968 𝑊

We get the output power if we subtract the mechanical and the iron losses from the internal

power:

𝑃𝑜𝑢𝑡 = 𝑃𝑖𝑛𝑡 − 𝑃𝑖𝑟𝑜𝑛+𝑚𝑒𝑐ℎ = 𝑃𝑖𝑛𝑡 − 0,04 ∙ 𝑃𝑜𝑢𝑡

𝑃2 =𝑃𝑏

1,04=

7832

1,04= 7530,77 𝑊

𝑃𝑣𝑎𝑠+𝑠ú𝑟𝑙 = 7832 − 7530,77 = 301,23 𝑊

When we calculate the efficiency, we have to take into account the power of the field circuit

which is independent to the armature circuit. The power of the field circuit have to add to the

input power:

𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛 + 𝑃𝑓=

7530,77

8800 + 800= 0,784

The torque:

𝑀𝑠 =𝑃𝑜𝑢𝑡

Ω=

𝑃𝑜𝑢𝑡

2𝜋 ∙ 𝑛∙ 60 =

7530

2𝜋 ∙ 955∙ 60 = 75,29𝑁𝑚

10. Other questions:

a) Calculate the speed of the motor if the terminal voltage of the armature is decreased to 300

V?

b) Calculate the speed of the motor if the filed voltage is decreased to 300 V.

c) Calculate the speed of the motor if a 45 Ω resistor is connected in series to the armature.

d) Calculate the resistance of the starter resistor if the maximum allowed armature current is 1,5

times greater than the nominal armature current.