Back EMF MCQ Quiz in தமிழ் - Objective Question with Answer for Back EMF - இலவச PDF ஐப் பதிவிறக்கவும்

The correct answer is 180V.

Concept:

For DC motor, generating emf is given by:

Vg = Vt + IaRa

For DC motor, back emf is given by:

Vb = Vt - IaRa 

Vg = Generating emf

Vb = Back emf

Vt = Terminal Voltage

Ia = Armature Current

Ra = Armature Resistance

Calculation:

Vt = Terminal Voltage = 200 V

Ra = 

Ia = 25 A

So, back emf:

Vb = Vt - IaRa

Vb = 200 - 0.8 x 25

Vb = 180 V

In a DC motor, Back emf always opposes the supply voltage. 

back emf (E_b) ∝ N ϕ 

so the back emf (Eb) of the DC motor is directly proportional to speed.

if speed increases, emf will also increase.

The correct option is 1.

N₁ = 1000 rpm

V = 200 V

R_a = 1 Ω

No load Back emf (E_b1) = 200 V

N₂ = 500 rpm

T₃ = 0.5 T_­2

N₃ = 520 rpm

⇒ E_b2 = 100 V

E_b2 = V – I_a2

100 = V – I_a2

⇒ E_b3 = 104

E_b3 = V – I_a3

104 = V – I_a3

Concept:

Circuit diagram for DC shunt motor is given below.

Back emf (E_b) = V – I_aR_a

Vt = terminal voltage DC

IL = line current

IF = field current

Calculation:

Given that, voltage (V) = 250 V

Armature resistance (R_a) = 0.1 Ω

Shunt field resistance (R_sh) = 250 Ω

Full load current (I_L) = 41 A

Shunt field current (I_sh) = 250/250 = 1 A

Armature current (I_a) = I_L – I_sh = 41 – 1 = 40 A

Back emf (E_b) = V – I_aR_a = 250 – 40 × 0.1 = 246 V

CONCEPT:

• When a current-carrying conductor is placed inside the magnetic field, it experiences a force.

• A rectangular slab which has area A and N number of turns is rotating in a magnetic field B with an angular velocity ω, then the EMF induced on it is given by

⇒ e = NBAω Sinωt

Where N = Number of turns, ω = Angular velocity, and B = Magnetic field

• When the coil is placed perpendicular to the field, then the EMF indued can be written as 

⇒ e = NBAω              [∵ θ = ωt = 900]

• Due to the rotation of armateur in the magnetic field back EMF is induced and is given by

⇒ e = E - iR

Where E = applied voltage, I = Current and R = Resistance

EXPLANATION:

• The equation of back emf can be written as

⇒ NBAω  = E - iR

• The above equation implies the idea that back EMF depends on ω, which in turn means more ω means more the induced EMF
• Hence back EMF will be maximum induced at the maximum speed of the motor
• Hence, option 3  is the answer

Armature current (I_a1) = 50 A

Voltage (V) = 250 V

Armature resistance (R_a) = 0.2Ω

Back emf E₁ = V – I_a1 R_a

= 250 – 50(0.2) = 240 V

Given that, N₂ = 1.4 N₁, T₂ = 1.4 T₁

We know that, E_b ∝ Nϕ, T ∝ ϕ I_a

E_b2 = V – I_a2 R_a

Let I_a increases by a factor of x.

⇒ I_a2 = x I_a1

⇒ E_b2 = 250 – 50x(0.2) = 250 – 10x.

Let 

⇒ (1.4) 240 (y) = 250 – 10x

⇒ 10x + 336y = 250     → (1)

     → (2)

From equations 1) and 2)

⇒ 336y² – 250y + 14 = 0

⇒ y = 0.683

Change in flux 

= -31.7%

Explanation:

Back EMF in a DC Motor

Definition: Back EMF (Electromotive Force), also known as counter EMF, is the voltage that is induced in the armature windings of a DC motor when it rotates. This induced voltage opposes the applied voltage (supply voltage) and is a result of the motor's operation as a generator while it is running. The back EMF is a fundamental characteristic of DC motors and plays a critical role in their operation.

Working Principle:

When a DC motor operates, electrical energy is supplied to the armature windings, creating a magnetic field. This magnetic field interacts with the field produced by the permanent magnets or field windings, resulting in a torque that causes the armature to rotate. As the armature rotates, the conductors within it cut through the magnetic field, inducing an electromotive force (EMF) in accordance with Faraday's Law of Electromagnetic Induction. This induced EMF is known as back EMF because it opposes the current flow that is driving the motor.

The back EMF is given by the formula:

E_b = (P × Φ × Z × N) / (60 × A)

Where:

• E_b = Back EMF (volts)
• P = Number of poles
• Φ = Flux per pole (webers)
• Z = Total number of armature conductors
• N = Speed of the armature (RPM)
• A = Number of parallel paths in the armature winding

Significance of Back EMF:

• Opposition to Current: Back EMF opposes the applied voltage and regulates the current flowing through the armature. This self-regulating mechanism ensures that the motor draws only the necessary current, preventing excessive current flow that could damage the motor.
• Energy Conversion: Back EMF is a direct consequence of energy conversion in the motor. As electrical energy is converted into mechanical energy, the motor also acts as a generator, producing the back EMF.
• Speed Control: The magnitude of the back EMF is proportional to the speed of the armature. As the motor speed increases, the back EMF increases, reducing the net voltage and current in the armature. This relationship helps maintain a stable operating speed under varying load conditions.
• Efficiency: The presence of back EMF minimizes energy losses by limiting the current flow in the armature, improving the efficiency of the motor.

Importance in DC Motors:

Back EMF is an essential feature of DC motors. Without back EMF, the motor would draw an excessive amount of current from the power supply, leading to overheating and potential damage. It provides a natural feedback mechanism that ensures the motor operates safely and efficiently. Additionally, back EMF is used in speed control and monitoring systems to assess the motor's operating conditions.

Correct Option Analysis:

The correct option is:

Option 2: Back EMF

This option correctly identifies the induced EMF in a DC motor that opposes the current flow in the armature conductors as back EMF. The term "back EMF" precisely describes this phenomenon, which is a fundamental characteristic of DC motors and is responsible for regulating current, maintaining efficiency, and ensuring safe operation.

Additional Information

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

Option 1: Field EMF

This option is incorrect. Field EMF refers to the electromotive force associated with the field windings of a motor or generator. It is not the induced EMF in the armature conductors that opposes the current flow. Field EMF is related to the creation of the magnetic field, not the regulation of armature current.

Option 3: Induced Voltage

While it is true that back EMF is an induced voltage, this term is too generic and does not specifically describe the opposing EMF in a DC motor. Induced voltage can refer to any voltage generated by electromagnetic induction, including those in transformers, generators, or other electrical devices. Back EMF is a more precise term for the phenomenon in question.

Option 4: Supply EMF

This option is incorrect. Supply EMF refers to the voltage provided by the power source to the motor. It is the applied voltage that drives the current through the armature windings. Back EMF, on the other hand, is the voltage induced within the motor that opposes the supply EMF.

Conclusion:

Understanding the concept of back EMF is crucial for comprehending the operation of DC motors. Back EMF is the induced voltage in the armature windings that opposes the applied voltage, regulating the current and ensuring efficient operation. It is distinct from other types of EMF, such as field EMF or supply EMF, and plays a vital role in the performance and safety of DC motors. The correct answer, option 2, accurately identifies this phenomenon as back EMF.

Concept

The circuit diagram of DC shunt generator is shown below:

The current across the field resistance is given by:

The load current is given by:

I = 100 A

The armature current is:

I_a = 102.5 A

Concept:

• For speed control of DC shunt motor, flux can be varied by connecting a variable resistance in series with the shunt field winding
• By increasing the resistance of the field rheostat, the shunt field current I_sh can be reduced and hence the field flux. Thus, by the flux control method, the speed of a DC shunt can only be increased above the normal speed.

Number of poles (p) = 8

Flux (ϕ) = 0.03 wb

Number of parallel paths (A) = 8

(∵ since it is lap wound machine)

Number of conductors (Z) = 660

⇒ I_a = 101.5 A

E_b = V – I_a(R_a + R_se)

= 240 – (101.5) (0.2 + 0.02)

= 217.66 V

⇒ N = 659.57 rpm