# DC Motor Back EMF MCQ || Back EMF in DC Motor Questions and Answers

1. The back EMF of the DC motor is ________ to speed.

1. Square
2. Directly proportional
3. Inversely proportional
4. Not equal

Answer:2. Directly proportional

Explanation:

• In dc motor also generator action takes place.
• Because of this generator action, the rotating conductors cut the flux and emf induced in the conductors.
• This induced emf is called back emf. It always opposes the supply voltage.

The back emf induced in the dc motor can be given by,

Eb = (ϕZNP) / (60 A)

Where

ϕ = flux/pole

Z = total number of conductors

A = number of parallel paths

N = speed in RPM

P = number of poles

In the back emf expression Z, P, and A are constant once machine design is completed

⇒ Eb ∝ Nϕ

⇒ Eb ∝ N

∴ The back emf in the dc motor is directly proportional to the speed.

2. At certain loading conditions, back e.m.f. in DC motor was found half of the supply voltage. Then power delivered by DC motor is

1. half of the rated power
2. maximum
3. minimum
4. double of the rated power

Answer: 2. Maximum

Explanation:

Relation between Mechanical power (Pm), Supply Voltage (Vt), and Back emf (Eb):

The back emf in the dc motor is expressed as:

Eb = V – IaRa …….(1)

Eb = back emf

Ia = armature current

Vt = terminal voltage

Ra = resistance of armateur

The Power developed on the motor is expressed by

Pm = EbIa = VIa – Ia2Ra ……(2)

On differentiating of the given equation

$\frac{{{\rm{d}}{{\rm{P}}_{\rm{m}}}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}} = \frac{{\rm{d}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}}\left( {{\rm{V}}{{\rm{I}}_{\rm{a}}} – {\rm{I}}_{\rm{a}}^2{{\rm{R}}_{\rm{a}}}} \right)$

For maximum power develop

$\frac{{{\rm{d}}{{\rm{P}}_{\rm{m}}}}}{{{\rm{d}}{{\rm{I}}_{\rm{a}}}}} = {\rm{V}} – 2{{\rm{I}}_{\rm{a}}}{{\rm{R}}_{\rm{a}}} = 0$

V = 2 IaRa ⇒ IaRa = V / 2

From the back emf equation

Eb = V – IaRa = V – V / 2

${{\rm{E}}_{\rm{b}}} = \frac{{\rm{V}}}{2}$ …….(3)

The maximum power is developed in the motor when the back emf is equal to half of the supply voltage.

and the maximum power is

(Pm)max = VIa – I­a2Ra = Ia (V – I­aRa)

${\left( {{{\rm{P}}_{\rm{m}}}} \right)_{{\rm{max}}}} = \frac{{{\rm{V}}{{\rm{I}}_{\rm{a}}}}}{2}$

(Pm)max = EbIa

3. With the increase in speed of a D.C. motor

1. Both back emf as well as line current increases
2. Both back emf as well as line current fall
3. Back emf increases but line current fall
4. Back emf fall but line current increases

Answer:4. Back emf fall but line current increases

Explanation:

Significance of Back EMF & Speed:

In the DC motor also generator action (mechanical energy converted into electrical energy) takes place. Due to generator action, the rotating conductor cuts the flux and emf induced, this induced emf is called back EMF.

It always opposes the supply voltage.

The induced back emf is A.C in the armature winding. But with respect to brushes, the back emf is D.C.

In a DC shunt motor, Back EMF is given by

Eb = V – IaRa

Where Ia is armature current

Ra is armature resistance

Also,

The back emf of a dc motor is directly proportional to speed.

Eb ∝ Nϕ

If the speed of a DC motor increases, there will be an increase in back emf also.

Current drawn in the DC motor is given by,

Ia = (V − Eb)/Ra

When the speed of a DC motor increases, emf will get an increase, and hence the current drawn will get decrease.

4. The direction of induced EMF in the DC motor is determined By the _____

1. Fleming’s Right-Hand Rule
2. Fleming’s Left-hand Rule
3. Maxwell Thumb Rule
4. Any of the above

Answer:1. Fleming’s Right-Hand Rule

Explanation:

The Fleming right-hand rule determines the direction of the induced EMF. According to Fleming’s Right-Hand Rule, if we hold our thumb, middle finger, and index finger of the right hand by an angle of 90°, then the index finger represents the direction of the magnetic field.

• When the current-carrying conductor is placed in a magnetic field, the torque induces on the conductor, the torque rotates the conductor which cuts the flux of the magnetic field.
• According to the Electromagnetic Induction Phenomenon “when the conductor cuts the magnetic field, EMF induces in the conductor”.
• It is seen that the direction(Right-hand rule) of the induced emf is opposite to the applied voltage. Thereby the emf is known as the counter emf or back emf.

5. With the increase in speed of a D.C. motor back emf _______ but line current _______

1. Increase, Increase
2. Decrease, Increase
3. Decrease, Decrease
4. Increase, Decrease

Answer:4. Increase, Decrease

Explanation:

In the DC motor also generator action (mechanical energy converted into electrical energy) takes place. Due to generator action, the rotating conductor cuts the flux and emf induced, this induced emf is called back EMF.

It always opposes the supply voltage.

The induced back emf is A.C in the armature winding. But with respect to brushes, the back emf is D.C.

In a DC shunt motor, Back EMF is given by

Eb = V – IaRa

Where Ia is armature current

Ra is armature resistance

Also,

The back emf of a dc motor is directly proportional to speed.

Eb ∝ Nϕ

If the speed of a DC motor increases, there will be an increase in back emf also.

Current drawn in the DC motor is given by,

Ia = (V − Eb)/Ra

When the speed of a DC motor increases, emf will get an increase, and hence the current drawn will get decrease.

6. A 230 V dc motor has an armature circuit resistance of 0.6 Ω. If the full load armature current is 30 A and the no-load armature current is 4 A, the change in back emf from no load to full load will be

1. 32.2 V
2. 64 V
3. 1.5 V
4. 15.6 V

Answer:4. 15.6

Explanation:

In a DC shunt motor, Back EMF is given by

Eb = V – IaRa

Where, Ia is armature current

Ra is armature resistance

Application:

Given,

V = 230 volts

Ra = 0.6 Ω

Case 1: When the motor is under no load,

Ia = 4 A

∴ EbNL = 230 – (4 × 0.6) = 227.6 volts

Case 2: When the motor is under full load,

Ia = 30 A

∴ EbFL = 230 – (30 × 0.6) = 212 volts

Hence, change in back emf (ΔEb) from no load to full load will be,

ΔEb = 227.6 – 212 = 15.6 volts

7. A separately excited DC motor runs at 1000 rpm on no-load when its armature terminals are connected to a 200 V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1 Ω. The no-load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rational losses, the full load armature current (in Ampere) is _________.

1. 4 A
2. 8 A
3. 2 A
4. 16 A

Answer:2. 8 A

Explanation:

N1 = 1000 rpm

V = 200 V

Ra = 1 Ω

No load Back emf (Eb1) = 200 V

N2 = 500 rpm

T3 = 0.5 T­2

N3 = 520 rpm

Eb1/Eb2 = 1000/500

200/Eb2 = 1000/500

Eb2 = 100 V

Eb1/Eb2 = 1000/500

⇒ Eb3 = 104

Eb3 = V – Ia3

104 = V – Ia3

T2/T3 ∝ Ia2/Ia3

T2/0.5T2 = Ia2/Ia3

Ia3 = Ia2/2 = 8A

8. A 10 hp, 240 V dc shunt motor, having armature-circuit resistance of 0.5 Ω and a full-load current of 40 A, is started by a starter, such that sections of required resistances in series with the armature-circuit should limit the starting current to 150% of the full-load current. The steady-state emf developed by the machine at full-load when the arm of the starter is moved to the next step is

1. 120 V
2. 100 V
3. 80 V
4. 60 V

Answer:3. 80 V

Explanation:

Given that, V = 240 V

Full load current I = 40 A

150% of full load current = 40 × 1.5 = 60 A

Starting resistance to limit current up to 150%

= 240/60 = 4 Ω

Back emf developed at full load when the starter is moved to next step = 240 – 40 × 4 = 80 V

9. The back EMF of DC motor is _______ of the applied voltage.

1. Equal
2. Half
3. Twice
4. Zero

Answer:2. Half

Explanation:

The back emf of the DC motor is equal to half of the applied voltage for maximum gross mechanical power. It is impossible to achieve this condition practically. Since half of the power input is wasted as heat in the armature, efficiency will be less than 50 percent taking account of other mechanical losses also.

10. If the field winding of a DC shunt motor gets disconnected while in normal operation then _______

1. Flux is reduced to Zero
2. Motor will run at high Speed
3. Motor gets damaged
4. All  of the above

Answer:4. All  of the above

Explanation:

If the field winding of a DC shunt motor gets disconnected while in normal operation then the flux will drop to almost zero value (residual value) and, therefore, the speed will increase to a tremendously high value and the motor may get damaged due to heavy centrifugal forces.

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