# DC Motor Problems and Solution – Step by Step

- Link: https://studyelectrical.com/2024/05/dc-motor-problems-and-solution.html
- Published: 2024-05-21T18:40:42+00:00
- Author: Tesla

Here are a collection of DC motor problems and solution. You can find problems on
various topics related to dc motor such as [types of dc motors](https://studyelectrical.com/2014/12/types-of-dc-motors.html),
[speed control of dc motor](https://studyelectrical.com/2013/11/speed-control-of-dc-motors.html),
torque of dc motor, characteristics of dc motor etc.

### DC Motor Problem 1

**A 220-V d.c. machine has an armature resistance of 0.5 Ω. If the full-load armature
current is 20 A, find the induced e.m.f. when the machine acts as (i) generator (
ii) motor.**

**Solution:**

To find the induced EMF in a 220-V DC machine with an armature resistance of 0.5
Ω and a full-load armature current of 20 A, we need to consider the machine operating
both as a generator and as a motor.

Given Data:

 * Voltage, $ V = 220 \, \text{V} $
 * Armature resistance, $ R_a = 0.5 \, \Omega $
 * Full-load armature current, $ I_a = 20 \, \text{A} $

**(i) When the machine acts as a generator:**

In a generator, the terminal voltage $ V $ is given by:
$$ V = E_g – I_a R_a $$Where
$ E_g $ is the induced EMF in the generator.

![DC motor problems and solutions](https://studyelectrical.com/wp-content/uploads/
2024/05/image.png)

Generator

Rearrange the equation to solve for $ E_g $:
$$ E_g = V + I_a R_a $$

Substitute the known values:
$ E_g = 220 \, \text{V} + 20 \, \text{A} \times 0.5\,\
Omega $$ E_g = 220 \, \text{V} + 10 \, \text{V} $$ E_g = 230 \, \text{V} $

Therefore, the induced EMF when the machine acts as a generator is 230 V.

**(ii) When the machine acts as a motor:**

In a motor, the terminal voltage $ V $ is given by:
$$ V = E_b + I_a R_a $$, Where
$ E_b $ is the [back EMF in the dc motor](https://studyelectrical.com/2015/02/back-emf-and-significance-in-dc-motor.html).

![DC motor problem solution](https://studyelectrical.com/wp-content/uploads/2024/
05/image-1.png)

Rearrange the equation to solve for $ E_b $:
$$ E_b = V – I_a R_a $$

Substitute the known values:
$ E_b = 220 \, \text{V} – 20 \, \text{A} \times 0.5\,\
Omega $$ E_b = 220 \, \text{V} – 10 \, \text{V} $$ E_b = 210 \, \text{V} $

Therefore, the induced EMF when the machine acts as a motor is 210 V.

### DC Motor Problem 2

**A DC motor takes an armature current of 110 A at 480 V. The armature circuit resistance
is 0.2 Ω. The machine has 6-poles and the armature is lap-connected with 864 conductors.
The flux per pole is 0.05 Wb. Calculate (i), the speed and (ii) the gross torque
developed by the armature.**

**Solution:**

To calculate the speed and the gross torque developed by the armature of the DC 
motor, we will use the given parameters:

 * Armature current, $ I_a = 110 \text{ A} $
 * Armature voltage, $ V = 480 \text{ V} $
 * Armature resistance, $ R_a = 0.2 \, \Omega $
 * Number of poles, $ P = 6 $
 * Number of conductors, $ Z = 864 $
 * Flux per pole, $ \Phi = 0.05 \text{ Wb} $
 * Type of winding: lap-connected

#### (i) Speed of the Motor

**Calculate the Back EMF $ E_b $:**
The back EMF $ E_b $ can be calculated using
the voltage equation of the motor:$$E_b = V – I_a R_a$$

Substituting the given values:
$E_b = 480 \text{ V} – (110 \text{ A} \times 0.2 \,\
Omega)$$E_b = 480 \text{ V} – 22 \text{ V} = 458 \text{ V}$

**Calculate the Speed $ n $ in RPM:**
The back EMF for a lap-wound motor is given
by:$$E_b = \frac{P \Phi Z n}{60 A}$$

Where:

 * $ P $ = Number of poles
 * $ \Phi $ = Flux per pole in Weber
 * $ Z $ = Total number of conductors
 * $ n $ = Speed in RPM
 * $ A $ = Number of parallel paths (For lap winding, $ A = P $)

Rearrange to solve for $ n $:
$$n = \frac{E_b \times 60 \times A}{P \times \Phi \
times Z}$$

Substitute the values:
$n = \frac{458 \text{ V} \times 60 \times 6}{6 \times 0.05\
text{ Wb} \times 864}$

Simplify the equation:
$n = \frac{458 \times 60}{0.05 \times 864}$$n \approx 636.11\
text{ RPM}$

Therefore, the speed of the motor is approximately 636.11 RPM.

#### (ii) Gross Torque Developed by the Armature

The torque $ T $ developed by the armature can be calculated using the power equation:

$$P_{mech} = T \cdot \omega$$

Where $ P_{mech} $ is the mechanical power in watts and $ \omega $ is the angular
speed in radians per second.

First, convert the speed from RPM to rad/s:
$$\omega = \frac{2 \pi n}{60}$$Substituting
$ n = 636.11 $:$\omega = \frac{2 \pi \times 636.11}{60} \approx 66.66 \text{ rad/
s}$

The electrical power converted to mechanical power by the armature (neglecting losses
other than armature resistance) is given by:
$$P_{mech} = E_b \times I_a$$Substituting
the known values:$$P_{mech} = 458 \text{ V} \times 110 \text{ A} = 50380 \text{ 
W}$$

Now, solve for the torque:
$T = \frac{P_{mech}}{\omega} = \frac{50380 \text{ W}}{
66.66 \text{ rad/s}}$$ T = \approx 755.85 \text{ Nm}$

Therefore, the gross torque developed by the armature is approximately 755.85 Nm.

### DC Motor Problem 3

A 250-V, 4-pole, wave-wound DC series motor has 782 conductors on its armature. 
It has armature and series field resistance of 0.75 ohm. The motor takes a current
of 40 A. Estimate its speed and gross torque developed if it has a flux per pole
of 25 mWb.

**Solution:**

**Calculate Back EMF ((E_b)):**
$E_b = V – I_a(R_a + R_{series})$$E_b = 250 \, \
text{V} – 40 \, \text{A} \times (0.75 \, \Omega + 0.75 \, \Omega)$ $E_b = 250 \,\
text{V} – 60 \, \text{V} = 190 \, \text{V}$

**Calculate Speed ((n)):**
$$n = \frac{E_b \times 60 \times 2}{P \times \Phi \times
Z}$$$n = \frac{190 \, \text{V} \times 60 \times 2}{4 \times 25 \, \text{mWb} \times
782}$$n = \frac{22800}{782} \approx 29.17 \, \text{RPM}$

**Convert RPM to rad/s:**
$$\omega = \frac{2 \pi \times 29.17}{60} \approx 3.05 \,\
text{rad/s}$$

**Calculate Gross Torque ((T)):**
$$P_{mech} = E_b \times I_a$$$P_{mech} = 190 \,\
text{V} \times 40 \, \text{A} = 7600 \, \text{W}$$T = \frac{P_{mech}}{\omega} = \
frac{7600 \, \text{W}}{3.05 \, \text{rad/s}} $$T \approx 2491.8 \, \text{Nm}$

Therefore, the speed of the motor is approximately 29.17 RPM, and the gross torque
developed by the armature is approximately 2491.8 Nm.

### DC Motor Problem 4

A 400 V DC motor takes an armature current of 100 A when its speed is 1000 rpm If
the armature
resistance is 0·25 ohm, calculate the torque produced in Nm.

**Solution**:

Sure, here are the equations in (\LaTeX) format:

 1. **Calculate the back EMF (($E_b$)):**

$$
E_b = V – I_a R_a$$

 2. **Calculate the mechanical power output (($P_{\text{mech}}$)):**

$$
P_{\text{mech}} = E_b \times I_a$$

 3. **Calculate the angular speed (($\omega$)):**

$$
\omega = \frac{2 \pi N}{60}$$

 4. **Calculate the torque (($T$)):**

$$
T = \frac{P_{\text{mech}}}{\omega}$$

Putting all the calculations together:

$$
E_b = 400\,\text{V} – (100\,\text{A} \times 0.25\,\Omega) = 400\,\text{V} – 25\,\
text{V} = 375\,\text{V}$$

$$
P_{\text{mech}} = 375\,\text{V} \times 100\,\text{A} = 37500\,\text{W}$$

$$
\omega = \frac{2 \pi \times 1000}{60} \approx 104.72\,\text{rad/s}$$

$$
T = \frac{37500\,\text{W}}{104.72\,\text{rad/s}} \approx 358.13\,\text{Nm}$$

Thus, the torque produced by the motor is approximately 358.13 Nm.

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## Comments

### By: KUMORDZIE K WILSON

- Link: https://studyelectrical.com/2024/05/dc-motor-problems-and-solution.html#comment-33005
- Published: 2024-06-22T20:39:37+00:00

THANK YOU FOR HELPING ME. I REALY APPRECIATE YOUR WORK