Voltage Equation of DC Motor MCQ Quiz - Objective Question with Answer for Voltage Equation of DC Motor - Download Free PDF

Explanation:

The ratio of back EMF to voltage of a DC motor is an indication of its efficiency.

To understand why efficiency is the correct answer, let's delve into the working of a DC motor and how back EMF and applied voltage affect its performance.

Working Principle of a DC Motor:

A DC motor operates on the principle that when a current-carrying conductor is placed in a magnetic field, it experiences a force. In a DC motor, the armature winding (the current-carrying conductor) is placed within the magnetic field created by the stator (field winding). When a DC voltage is applied to the armature winding, current flows through it, generating a magnetic field that interacts with the stator’s magnetic field, producing a torque that causes the armature to rotate.

Back EMF (Electromotive Force):

As the armature rotates, it cuts through the magnetic field lines, which induces a voltage in the armature winding known as back EMF (E_b). According to Lenz's Law, this induced voltage opposes the applied voltage (V) and is given by:

E_b = k * φ * ω

Where:

• k is a constant that depends on the construction of the motor.
• φ is the magnetic flux.
• ω is the angular velocity of the armature.

The back EMF increases with the speed of the motor. When the motor is running at a steady speed, the back EMF almost balances the applied voltage, and the current drawn by the motor is reduced to a level just sufficient to overcome friction, windage losses, and provide the necessary torque for the load.

Efficiency of a DC Motor:

Efficiency (η) of a DC motor is defined as the ratio of the mechanical power output to the electrical power input. It can be expressed as:

η = (P_out / P_in) * 100%

Where:

• P_out is the mechanical power output.
• P_in is the electrical power input.

The electrical power input (P_in) is given by:

P_in = V * I

Where:

• V is the applied voltage.
• I is the current drawn by the motor.

The mechanical power output (P_out) is given by:

P_out = (E_b * I) - (I² * R_a)

Where:

• E_b is the back EMF.
• I is the current drawn by the motor.
• R_a is the armature resistance.

From the above equations, we can see that the back EMF (E_b) plays a crucial role in determining the efficiency of the motor. When the motor is running at a higher efficiency, the back EMF will be closer to the applied voltage (V), meaning that most of the input electrical power is being converted into useful mechanical power, with minimal losses.

Therefore, the ratio of back EMF to the applied voltage is a direct indication of the efficiency of the motor. A higher ratio suggests higher efficiency, as more of the electrical input is being effectively converted into mechanical output with fewer losses.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Starting Torque

The starting torque of a DC motor is the torque it produces when it starts from a standstill. At the start, the back EMF is zero because the motor is not yet moving. Therefore, the starting torque is primarily determined by the applied voltage and the current flowing through the armature windings. The ratio of back EMF to voltage does not provide information about the starting torque.

Option 2: Speed Regulation

Speed regulation refers to the ability of the motor to maintain a constant speed under varying load conditions. While back EMF does play a role in speed regulation, the ratio of back EMF to voltage alone does not directly indicate speed regulation. Speed regulation depends on various factors, including the design of the motor and the control methods used.

Option 3: Running Torque

Running torque is the torque produced by the motor when it is running at a steady speed. While back EMF affects the current and hence the torque, the ratio of back EMF to voltage is not a direct measure of running torque. Running torque is influenced by the load on the motor and the current flowing through the armature.

Conclusion:

In conclusion, the ratio of back EMF to the applied voltage of a DC motor is an indication of its efficiency. A higher ratio implies that the motor is converting a greater portion of the electrical input into useful mechanical output with fewer losses. Understanding this relationship is crucial for analyzing the performance and efficiency of DC motors in various applications.

Concept:

The EMF in a DC shunt motor is given by:

\(E_b=V-I_aR_a\)

where, E_b = Back EMF

V = Terminal voltage

I_a = Armature current

R_a = Armature rasistance 

Calculation:

Given, V = 250 V

I_L = 20 A

R_sh = 200 Ω 

R_a = 0.3 Ω 

\(I_{sh}={250\over 200}=1.25 A\)

I_a = 20 - 1.25 = 18.75  A

I) The speed-torque relationship of a DC series motor is non-linear: TRUE

The Speed (N) vs Torque (Ta) Characteristic is similar to the Speed (N) vs Current (Ia) Characteristic because of torque is proportional to the square of current up to magnetic saturation and proportional to the current after magnetic saturation.

• It is clear that series motor develops high torque at low speed and vice-versa.
• It is because an increase in torque requires an increase in armature current, which is also the field current.
   

II) Both the armature current and the field current of a DC series motor decrease with increasing load torque.: FALSE

Torque (Ta) [Load] vs Current (Ia) Characteristic:

We know that:

Ta ∝ ϕIa

Up to magnetic saturation, ϕ ∝ Ia

So that, Ta ∝ \(I_a^2\)

After magnetic saturation, ϕ is constant

So that Ta ∝ Ia

Statement III) & IV): Comparison of Efficiency:

The actual operating efficiency of a series motor is typically less than that of a shunt motor due to design differences, load variations, speed regulation issues, higher field losses, and the need for additional control and protection mechanisms.

The correct answer is Both I and II.

Key Points

• DC Shunt Motors: A DC shunt motor is a type of direct current (DC) motor where the field windings are connected in parallel (shunt) with the armature windings.
• Statement I: At low speeds, DC shunt motors are comparable with synchronous motors.
  • DC shunt motors offer a nearly constant speed under varying load conditions due to their shunt field winding configuration.
  • Synchronous motors, on the other hand, operate at a constant speed regardless of load variations.
  • At low speeds, the performance of DC shunt motors can be similar to synchronous motors in terms of speed regulation.
  • Hence, statement I is correct.
• Statement II: The outstanding feature of a DC shunt motor however is its superb wide range flexible speed control above and below the base speed using solid-state controlled rectifiers.
  • DC shunt motors are known for their excellent speed control capabilities.
  • Using solid-state controlled rectifiers, the speed of DC shunt motors can be varied over a wide range, both above and below the base speed.
  • This flexibility in speed control is one of the major advantages of DC shunt motors in various applications.
  • Hence, statement II is correct.

Additional Information

• DC Shunt Motor Applications: Due to their consistent speed and excellent speed control, DC shunt motors are commonly used in industrial applications such as lathes, machine tools, and elevators.
• Synchronous Motors: These motors are typically used in applications that require constant speed regardless of load variations, such as in clocks, record players, and conveyor systems.
• Solid-State Controlled Rectifiers: These devices are used to control the voltage and current supplied to the motor, allowing precise speed control. They are a key component in modern motor control systems.
• Speed Control Methods: In addition to solid-state rectifiers, other methods for controlling the speed of DC shunt motors include armature voltage control and field flux control.

Armature resistance:

• The armature resistance in a DC machine refers to the electrical resistance in the winding of the armature. It's denoted by ��" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">����" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">���� �� ��
• This resistance is essential to consider because it affects the voltage drop across the armature and hence influences the machine's performance.

Field resistance:

• The field resistance in a DC machine refers to the resistance present in the field winding of the machine. It's denoted by ��" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">����" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">���� �� �� _f.
• This winding is responsible for establishing the magnetic field within the machine, which interacts with the armature to produce motion or induce a voltage when the machine is rotated.

Given that, 

\(V_T = 230V; r_a = 0.5\Omega\)

\(R_r = 115|Omega; I_L = 50A\)

\(I_r = \frac{V_r}{R_r}\)

\(\frac{230}{115} = 2A\)

For generator,

Armature current, \(I_a = I_f + I_L\)

\(I_a = 50A\)

\(E = V_T + I_aR_a\)

=\(230 + 52\times0.5\)

= 256V.

For moto, 

\(I_s = I_L-I_f\)

= 48 A.

\(E = V_T -I_aR_S\)

= \(230-48\times0.5\)

= 206V.

Here, option 1 is correct.

I) The speed-torque relationship of a DC series motor is non-linear: TRUE

The Speed (N) vs Torque (Ta) Characteristic is similar to the Speed (N) vs Current (Ia) Characteristic because of torque is proportional to the square of current up to magnetic saturation and proportional to the current after magnetic saturation.

• It is clear that series motor develops high torque at low speed and vice-versa.
• It is because an increase in torque requires an increase in armature current, which is also the field current.
   

II) Both the armature current and the field current of a DC series motor decrease with increasing load torque.: FALSE

Torque (Ta) [Load] vs Current (Ia) Characteristic:

We know that:

Ta ∝ ϕIa

Up to magnetic saturation, ϕ ∝ Ia

So that, Ta ∝ \(I_a^2\)

After magnetic saturation, ϕ is constant

So that Ta ∝ Ia

Statement III) & IV): Comparison of Efficiency:

The actual operating efficiency of a series motor is typically less than that of a shunt motor due to design differences, load variations, speed regulation issues, higher field losses, and the need for additional control and protection mechanisms.

For a dc motor from the power equation, it is known that,

Pm = Gross mechanical power developed = EbIa = VIa - Ia2Ra

Where,

V = Applied voltage

E_b = Back emf

I_a = Armature current

R_a = Armature resistance

For maximum Pm,

\(\begin{array}{l} \frac{{d{P_m}}}{{d{I_a}}} = 0\\ \Rightarrow 0 = V - 2{I_a}{R_a}\\ \Rightarrow {I_a}{R_a} = \frac{V}{2} \end{array}\)

Substituting in voltage equation (V = Eb + IaRa), we get

\(\begin{array}{l} V = {E_b} + {I_a}{R_a} = {E_b} + \left( {\frac{V}{2}} \right)\\ \Rightarrow {E_b} = \frac{V}{2} \end{array}\)

So that when a DC motor generates maximum power, ratio of applied voltage to back emf is 2 : 1

Concept:

Equivalent circuit of separately excited D.C motor,

V = E_b + I_aR_a

Explanation:

Let,

Torque T₁ at N₁ = 1000 rpm

Torque T₂ = at N₂ = 500 rpm

∴ T ∝ N² {Given}

⇒ \(\frac{T_2}{T_1}=\left(\frac{N_2}{N_1}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{500}{1000}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\frac{1}{4}\)

Load torque can be expressed as,

T = k ϕ. I_a

T ∝ ϕ I_a

T ∝ I_a {∴ ϕ = constant}

⇒ \(\frac{T_2}{T_1}=\frac{Ia_2}{Ia_1}\)

⇒ \(\frac{Ia_2}{Ia_1}=\frac{1}{4}\)

⇒ \(Ia_2=\frac{10}{4}=2.5 \) A

Therefore, 

\(E_{b_1}=V_1-I{a_1}R_a=100-10=90\) V

∴ \(E_b=\frac{NP\phi Z}{60\ A}\)

⇒ \(\frac{E_{b_1}}{E_{b_2}}=\frac{N_1}{N_2}\)

⇒ \(E_{b_2}=E_{b_1}\frac{N_2}{N_1}\)

⇒ \(E_{b_2}=\frac{90×500}{1000}\)

⇒ \(E_{b_2}=45\) V

Therefore,

\(V_2=E_{b_2}+I_{a_2}R_a\)

⇒ V₂ = 45 + 2.5 × 1

⇒ V₂ = 47.5 V

Hence, The armature voltage is 47.5 V

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{N_1}}}{{{N_2}}}\)          E₁ is 200V itself since I_a is zero at no load

So E₂ can be obtained as 186.67V

Power P = \(\frac{{2\pi 1400 \times 5}}{{60}} = 733\ W\)

E₂I₂ = 733         I₂ = 3.92 A

\(\begin{array}{l} 186.67 = 200 - 3.92{R_a}\\ {R_a} = 3.4\ {\rm{\Omega }} \end{array}\)

Armature resistance:

• The armature resistance in a DC machine refers to the electrical resistance in the winding of the armature. It's denoted by ��" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">����" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">���� �� ��
• This resistance is essential to consider because it affects the voltage drop across the armature and hence influences the machine's performance.

Field resistance:

• The field resistance in a DC machine refers to the resistance present in the field winding of the machine. It's denoted by ��" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">����" id="MathJax-Element-4-Frame" role="presentation" style="position: relative;" tabindex="0">���� �� �� _f.
• This winding is responsible for establishing the magnetic field within the machine, which interacts with the armature to produce motion or induce a voltage when the machine is rotated.

Given that, 

\(V_T = 230V; r_a = 0.5\Omega\)

\(R_r = 115|Omega; I_L = 50A\)

\(I_r = \frac{V_r}{R_r}\)

\(\frac{230}{115} = 2A\)

For generator,

Armature current, \(I_a = I_f + I_L\)

\(I_a = 50A\)

\(E = V_T + I_aR_a\)

=\(230 + 52\times0.5\)

= 256V.

For moto, 

\(I_s = I_L-I_f\)

= 48 A.

\(E = V_T -I_aR_S\)

= \(230-48\times0.5\)

= 206V.

Here, option 1 is correct.

I) The speed-torque relationship of a DC series motor is non-linear: TRUE

The Speed (N) vs Torque (Ta) Characteristic is similar to the Speed (N) vs Current (Ia) Characteristic because of torque is proportional to the square of current up to magnetic saturation and proportional to the current after magnetic saturation.

• It is clear that series motor develops high torque at low speed and vice-versa.
• It is because an increase in torque requires an increase in armature current, which is also the field current.
   

II) Both the armature current and the field current of a DC series motor decrease with increasing load torque.: FALSE

Torque (Ta) [Load] vs Current (Ia) Characteristic:

We know that:

Ta ∝ ϕIa

Up to magnetic saturation, ϕ ∝ Ia

So that, Ta ∝ \(I_a^2\)

After magnetic saturation, ϕ is constant

So that Ta ∝ Ia

Statement III) & IV): Comparison of Efficiency:

The actual operating efficiency of a series motor is typically less than that of a shunt motor due to design differences, load variations, speed regulation issues, higher field losses, and the need for additional control and protection mechanisms.

For a dc motor from the power equation, it is known that,

Pm = Gross mechanical power developed = EbIa = VIa - Ia2Ra

Where,

V = Applied voltage

E_b = Back emf

I_a = Armature current

R_a = Armature resistance

For maximum Pm,

\(\begin{array}{l} \frac{{d{P_m}}}{{d{I_a}}} = 0\\ \Rightarrow 0 = V - 2{I_a}{R_a}\\ \Rightarrow {I_a}{R_a} = \frac{V}{2} \end{array}\)

Substituting in voltage equation (V = Eb + IaRa), we get

\(\begin{array}{l} V = {E_b} + {I_a}{R_a} = {E_b} + \left( {\frac{V}{2}} \right)\\ \Rightarrow {E_b} = \frac{V}{2} \end{array}\)

So that when a DC motor generates maximum power, ratio of applied voltage to back emf is 2 : 1

Concept:

Equivalent circuit of separately excited D.C motor,

V = E_b + I_aR_a

Explanation:

Let,

Torque T₁ at N₁ = 1000 rpm

Torque T₂ = at N₂ = 500 rpm

∴ T ∝ N² {Given}

⇒ \(\frac{T_2}{T_1}=\left(\frac{N_2}{N_1}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{500}{1000}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\frac{1}{4}\)

Load torque can be expressed as,

T = k ϕ. I_a

T ∝ ϕ I_a

T ∝ I_a {∴ ϕ = constant}

⇒ \(\frac{T_2}{T_1}=\frac{Ia_2}{Ia_1}\)

⇒ \(\frac{Ia_2}{Ia_1}=\frac{1}{4}\)

⇒ \(Ia_2=\frac{10}{4}=2.5 \) A

Therefore, 

\(E_{b_1}=V_1-I{a_1}R_a=100-10=90\) V

∴ \(E_b=\frac{NP\phi Z}{60\ A}\)

⇒ \(\frac{E_{b_1}}{E_{b_2}}=\frac{N_1}{N_2}\)

⇒ \(E_{b_2}=E_{b_1}\frac{N_2}{N_1}\)

⇒ \(E_{b_2}=\frac{90×500}{1000}\)

⇒ \(E_{b_2}=45\) V

Therefore,

\(V_2=E_{b_2}+I_{a_2}R_a\)

⇒ V₂ = 45 + 2.5 × 1

⇒ V₂ = 47.5 V

Hence, The armature voltage is 47.5 V

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{N_1}}}{{{N_2}}}\)          E₁ is 200V itself since I_a is zero at no load

So E₂ can be obtained as 186.67V

Power P = \(\frac{{2\pi 1400 \times 5}}{{60}} = 733\ W\)

E₂I₂ = 733         I₂ = 3.92 A

\(\begin{array}{l} 186.67 = 200 - 3.92{R_a}\\ {R_a} = 3.4\ {\rm{\Omega }} \end{array}\)

The correct answer is Both I and II.

Key Points

• DC Shunt Motors: A DC shunt motor is a type of direct current (DC) motor where the field windings are connected in parallel (shunt) with the armature windings.
• Statement I: At low speeds, DC shunt motors are comparable with synchronous motors.
  • DC shunt motors offer a nearly constant speed under varying load conditions due to their shunt field winding configuration.
  • Synchronous motors, on the other hand, operate at a constant speed regardless of load variations.
  • At low speeds, the performance of DC shunt motors can be similar to synchronous motors in terms of speed regulation.
  • Hence, statement I is correct.
• Statement II: The outstanding feature of a DC shunt motor however is its superb wide range flexible speed control above and below the base speed using solid-state controlled rectifiers.
  • DC shunt motors are known for their excellent speed control capabilities.
  • Using solid-state controlled rectifiers, the speed of DC shunt motors can be varied over a wide range, both above and below the base speed.
  • This flexibility in speed control is one of the major advantages of DC shunt motors in various applications.
  • Hence, statement II is correct.

Additional Information

• DC Shunt Motor Applications: Due to their consistent speed and excellent speed control, DC shunt motors are commonly used in industrial applications such as lathes, machine tools, and elevators.
• Synchronous Motors: These motors are typically used in applications that require constant speed regardless of load variations, such as in clocks, record players, and conveyor systems.
• Solid-State Controlled Rectifiers: These devices are used to control the voltage and current supplied to the motor, allowing precise speed control. They are a key component in modern motor control systems.
• Speed Control Methods: In addition to solid-state rectifiers, other methods for controlling the speed of DC shunt motors include armature voltage control and field flux control.