300+ DC Motor MCQ [Free PDF] – Objective Question Answer for DC Motor Quiz

Answer 2:- Decreases

Explanation:-

D.C. machines are generally much more adaptable to adjustable speed service. The ready availability of D.C. motors to adjustment of their operating speed over wide ranges and by a variety of methods is one of the important reasons for the strong competitive position of D. C. machinery in modern industrial applications.

Voltage Equation of DC Motor

The voltage equation of the DC Motor is given as

V = E_b + I_aR_a

The voltage V applied across the motor armature has to
(i) overcome the back e.m.f. E_b and
(ii) supply the armature ohmic drop I_aR_a

The above equation can also be written as

E_b = V − I_aR_a

Back EMF of DC Motor

When the armature of a d.c. the motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator. The induced e.m.f. acts in opposite direction to the applied voltage  V (Lenz’s law) and is known as back or counter e.m.f. Eb. 

${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$

Where

P – Number of poles of the machine

ϕ – Flux per pole in Weber.

Z – Total number of armature conductors.

N – Speed of armature in revolution per minute (r.p.m).

A – Number of parallel paths in the armature winding.

From the above equation, it is clear that the EMF of DC Motor is Directly proportional to the Number of poles of the machine (P), Flux per pole in Weber (ϕ), Total number of armature conductors(Z), and  Speed of armature (N)

From the voltage equation and Back EMF equation, we can conclude that

$\begin{array}{l}\dfrac{{P\Phi ZN}}{{60A}} = V – {\rm{ }}{I_a}{R_a}\\\\N = \dfrac{{V – {\rm{ }}{I_a}{R_a}}}{\Phi } \times \left( {\dfrac{{60A}}{{PZ}}} \right)r.p.m\\\\{E_b} = V – {\rm{ }}{I_a}{R_a}\\\\\therefore N = K\dfrac{{{E_b}}}{\Phi }\end{array}$

From the above equation it is clear that by increasing the speed of DC motor the back EMF increased but the armature current decrease.

Answer. 2. When the field is Weakend

Explanation:-

The speed of DC shunt can be controlled by various Methods such as

1. Flux Control Method
2. Armature voltage control method

Flux Control Method

The speed of DC Shunt Motor is inversely proportional to the flux. The flux is dependent on the current through the shunt field winding. Thus flux can be controlled by adding a rheostat (variable resistance) in series with the shunt field winding. At Starting the rheostat R is kept at the minimum. The supply voltage is at its rated value. So the current through shunt field winding is also at its rated value. Hence the speed is also rated speed i.e at normal speed. When the resistance R is increased due to which shunt field current Ish decreases, hence decreasing the flux produced. As N =(I/Φ), the speed of the motor increases beyond its rated value. Now as we know that the speed of DC motor is directly proportional to the flux hence when the speed is increased the back EMF also increased. Thus by this method, the speed control above rated value is possible.

Answer 1. 18.75 A, 245 V

Explanation:-

Given Data:-

Supply Voltage V = 250 V
Line Current I_L = 20A
Shunt Resistance R_sh = 200Ω
Armature Resistance R_a = 0.3 Ω
Armature Current I_a =?
Back EMF E_b =?

The Diagram of the given question is shown in figure

(i) Armature current

The Line current of DC shunt motor is the sum of Armature current and Shunt field current

I_L = I_a + I_sh

And Shunt Field Resistance is given as

I_sh = V/R_sh = 250/200 = 1.25 A

∴ I_a = I_L − I_sh

I_a = 20 − 1.25 = 18.75 A

(ii) Back EMF

The Back EMF of DC Motor is given by

E_b = V − I_aR_a

= 250 − 18.75 × 0.3
= 250 − 5.625

= 244.375 V ≅ 245 V

Answer.4 1374 RPM

Explanation:

Given Data

Supply Voltage V = 250 V

Number of Poles P = 4

Total number of armature conductors Z = 200

Number of parallel paths in the armature winding A = 2 (for wave winding number of parallel path is always 2)

Line current I_L = 60 A

Armature current Ia = 60 A ( In DC series motor the line current is equal to the armature current i.e I_L = I_a)

Flux Φ = 25 x 10^-3

Armature Resistance R_a = 0.15Ω

Series Resistance R_se = 0.2Ω

The Back EMF of DC series Motor is given by

E_b = V − I_a(R_a + R_se)

= 250 − 60 (0.15 + 0.2)
= 250 − 21

E_b = 229

Back EMF in DC Motor E_b. 

${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$

Where

P – Number of poles of the machine

ϕ – Flux per pole in Weber.

Z – Total number of armature conductors.

N – Speed of armature in revolution per minute (r.p.m).

A – Number of parallel paths in the armature winding.

$\begin{array}{l}N = \dfrac{{{E_b} \times 60A}}{{P\Phi Z}}\\\\ = \dfrac{{229 \times 60 \times 2}}{{4 \times 25 \times {{10}^{ – 3}} \times 200}}\\\\N = 1374{\text{ R}}{\rm{.P}}{\rm{.M}}\end{array}$

Answer C. 320 RPM

Explanation:-

Given that

Supply voltage Vs = 500 V
Speed N₁ = 250 RPM
Armature current I_a1= 200 A
Armature Resistance R₁ = 0.12Ω

Back EMF of DC Motor is given as

E_b = V − I_aR_a

E_b = 500 − 200 × 0.12 = 478V

Now when the resistance is inserted in the field, Flux reduce to 80%

Flux Φ₂ = 0.8Φ₁

Armature current I_a2 = 100A

Back EMF 

E_b = V − I_aR_a

E_b = 500 − 100 × 0.12 = 488V

As we Know that In DC motor the back EMF is directly proportional to the speed and flux.

E ∝ Nφ

$\begin{array}{l}\dfrac{{{E_{b2}}}}{{{E_{b1}}}} = \dfrac{{{N_2}}}{{{N_1}}} \times \dfrac{{{\Phi _2}}}{{{\Phi _1}}}\\\\{N_2} = {N_1} \times \dfrac{{{E_{b2}}}}{{{E_{b1}}}} \times \dfrac{{{\Phi _2}}}{{{\Phi _1}}}\\\\{N_2} = 250 \times \dfrac{{488}}{{476}} \times \left( {\dfrac{{{\Phi _1}}}{{0.8{\Phi _1}}}} \right)\\\\{N_2} = 320{\rm{ R}}{\text{.P}}{\rm{.M}}\end{array}$

Answer 2. Inversely proportional to flux per pole

Explanation:-

Voltage Equation of DC Motor

The voltage equation of the DC Motor is given as

V = E_b + I_aR_a

The voltage V applied across the motor armature has to
(i) overcome the back e.m.f. E_b and
(ii) supply the armature ohmic drop I_aR_a

The above equation can also be written as

E_b = V − I_aR_a

Back EMF of DC Motor

When the armature of a d.c. the motor rotates under the influence of the driving torque, the armature conductors move through the magnetic field and hence e.m.f. is induced in them as in a generator. The induced e.m.f. acts in opposite direction to the applied voltage  V (Lenz’s law) and is known as back or counter e.m.f. Eb. 

${E_b} = \dfrac{{P\Phi ZN}}{{60A}}$

Where

P – Number of poles of the machine

ϕ – Flux per pole in Weber.

Z – Total number of armature conductors.

N – Speed of armature in revolution per minute (r.p.m).

A – Number of parallel paths in the armature winding.

From the above equation, it is clear that the EMF of DC Motor is Directly proportional to the Number of poles of the machine (P), Flux per pole in Weber (ϕ), Total number of armature conductors(Z), and  Speed of armature (N)

From the voltage equation and Back EMF equation, we can conclude that

$\begin{array}{l}\dfrac{{P\Phi ZN}}{{60A}} = V – {\rm{ }}{I_a}{R_a}\\\\N = \dfrac{{V – {\rm{ }}{I_a}{R_a}}}{\Phi } \times \left( {\dfrac{{60A}}{{PZ}}} \right)r.p.m\\\\{E_b} = V – {\rm{ }}{I_a}{R_a}\\\\\therefore N = K\dfrac{{{E_b}}}{\Phi }\end{array}$

The above equation shows that speed is directly proportional to back e.m.f. E b and inversely to the flux Φ 

N ∝ E_b /Φ.

$N = \dfrac{{{E_b}}}{\Phi } \propto \dfrac{{V – {\rm{ }}{I_a}{R_a}}}{\Phi }$

But as the value of armature resistance R_a and series field resistance R_se is very small, the drop I_aR_a and Ia (R_a + R_se) is very small compared to applied voltage V. Hence, neglecting these voltage drops the speed equation can be modified as,

$N \propto \dfrac{V}{\Phi }$

Answer 3. Iron and Friction losses

Explanation:-

The torque is developed in the armature and hence gross torque produced is denoted asT_a.

The mechanical power developed in the armature is transmitted to the load through the shaft of the motor. It is impossible to transmit the entire power developed by the armature to the load. This is because while transmitting the power through the shaft, there is a power loss due the friction, windage, and iron loss. The torque required to overcome these losses is called lost torque, denoted as T_f. These losses are also called stray losses. The torque which is available at the shaft for doing the useful work is known as load torque or shaft torque denoted as T_sh.

∴ T_a = T_sh + T_f

The shaft torque magnitude is always less than the armature torque, (T_sh <  T_a).

Answer 1. Flux per pole × Armature Resistance

Explanation:-

By the term, torque is meant the turning or twisting moment of a force about an axis. It is measured by the product of the force and the radius at which this force acts. Consider a pulley of radius r meter acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m.

The angular speed of the pulley is

ω = 2πN/60 rad/sec

Work done by this force in one revolution

= Force × distance = F × 2πR Joule

The power developed = Work Done/Time

= (F × 2πR)/60/N

= (F × R) × (2πN)/60

The power developed = T × ω watt or P = T ω Watt

Armature Torque

Let Ta be the torque developed by the armature of a motor running at N r.p.s. If Ta is in N/M, then
power developed

P =  Ta × 2π N watt

The electrical power converted into mechanical power in the armature = E_bI_a watt

Power in armature = Armature torque × ω

E_bI_a = Ta × 2πN/60
or
Ta = 9.55E_bI_a /N (N-m)

The back EMF (E_b) of the motor is given by

E_B = PΦZN/60A

Φ = flux per pole in Weber

P = number of poles

Z = total number of armature conductor

N = rotation speed of the armature in revolution per minute (r.p.m)

A = number of parallel paths

$\begin{array}{l}\dfrac{{P\Phi ZN}}{{60A}} \times {I_a} = {T_a} \times \dfrac{{2\pi N}}{{60}}\\\\{T_a} = \dfrac{1}{{2\pi }}\Phi {I_a} \times \dfrac{{PZ}}{A}\\\\{T_a} = 0.159\Phi {I_a}.\dfrac{{PZ}}{A}N – M\end{array}$

Hence From the above equation it is clear that the torque of the DC series motor is directly proportional to the armature current (I_a) and Flux per pole(Φ).

T ∝ I_a × Φ

Answer 1. 76.32 N-m

Explanation:-

Given Data

Number of Poles P = 4

The number of parallel Path A = 4  (In Lap wound the number of brushes required by this winding is always equal to the number of poles.)

Total number of armature conductor Z = 480

Flux per Pole Φ = 20 mWb = 20 x 10^-3 Wb,

Armature current I_a = 50 A

$\begin{array}{l}{T_a} = 0.159\Phi {I_a}\dfrac{{PZ}}{A}\\\\ = 0.159 \times 20 \times {10^{ – 3}} \times 50\dfrac{{4 \times 480}}{4}\\\\{T_a} = 76.32{\text{ N – m}}\end{array}$

Answer. 2. When the armature current increases

Explanation:-

When the load is connected to the DC motor, the armature winding of the DC motor carries a current. Every current-carrying conductor produces its own flux so the armature of the DC motor also produces its own flux when carrying a current. So there are two fluxes present in the air gap, one due to armature current while the second is produced by the field winding called main flux. The flux produced by the armature is called armature flux

So the effect of the armature flux on the main flux affects its value and the distribution is called armature reaction.

Armature reaction occurs in DC motors and is caused by the stator magnetic field being distorted, or altered, in reaction to the armature magnetic field. The armature reaction is actually a bending of the motor magnetic field so that the brushes are no longer aligned with the neutral magnetic plane of the motor. If the brushes are not in alignment with this magnetic plane, the current conducted to the armature does not split equally in the armature conductors and therefore causes a voltage difference at the brushes. This causes sparking where the brush meets the commutator. In a motor with a constant load.

We know that reducing the flux leads to an increase in speed, so we can now see that in a machine with a pronounced armature reaction, when the load on the shaft is increased and the armature current increases to produce more torque, the field is simultaneously reduced and the motor speeds up. Though this behavior is not a true case of instability, it is not generally regarded as desirable!

Large motors often carry additional windings fitted into slots in the pole-faces and connected in series with the armature. These ‘compensating’ windings produce an m.m.f. in opposition to the armature m.m.f., thereby reducing or eliminating the armature reaction effect.