# DC Motor Speed Regulation MCQ || Speed Regulation of DC Motor Questions and Answers

1. The PWM control of DC motor varies _______

1. Linearly with speed
2. Inversely with speed
3. Parabolically with speed
4. Exponentially with speed

Explanation:

• PWM (Pulse Width Modulation)  is often used for DC motor control using a microcontroller is Pulse Width Modulation (PWM) method. The speed of the electric motor depends on the modulator voltage. The greater the voltage, the faster the rotation of an electric motor.
• The method is to apply a pulse train to the power terminals of the motor. The average voltage obtained at the terminals is then proportional to the duty cycle of the pulse train, which is proportional to the speed of rotation (rpm) of the motor.
• Thus, as the duty cycle is increased, the motor rpm increases and vice versa. When the power supply is constant, it runs at 100% of its power rating (at no load).
• As the duty cycle reduces, the speed and the power reduce. In the PWM technique duty ratio is a linear function with respect to speed.

2. The speed of DC shunt motor can be increased above its rated speed by

1. Increasing the supply voltage within its rated value
2. Increasing the flux per pole
3. Decreasing the flux per pole
4. Connecting additional resistance in series of the armature

Answer:3. Decreasing the flux per pole

Explanation:

Field control method:

By adding external resistance in the field circuit, we can control the speed of the DC shunt motor above the normal speed.

• In this method, speed variation is accomplished by means of a variable resistance inserted in series with the shunt field.
• An increase in controlling resistances reduces the field current with a reduction in flux and an increase in speed.
• This method of speed control is independent of the load on the motor. Power wasted in controlling resistance is very less as field current is a small value.

We know that,

N ∝ Eb

N ∝ (V − Ia R/φ

ϕ ↓  → N ↑

By varying flux, we can increase the speed more than its base speed. This method is a constant power and variable torque drive.

3. A separately excited dc motor has an armature resistance Ra = 0.05 Ω. The field excitation is kept constant. At an armature voltage of 100 V, the motor produces a torque of 500 Nm at zero speed. Neglecting all mechanical losses, the no-load speed of the motor (in radian/s) for an armature voltage of 150 V is

1. 600 RPM
2. 1200 RPM
3. 300 RPM
4. 900 RPM

Explanation:

At zero speed (N = 0)

induced voltage (E) = 0

T =  KφIa1

⇒ T = KIa1 {∴ ϕ = Constant}

E = Vt − Ia1R1

Ia1 = 100 − 0/0.05 = 2000A

∴ K = T/Ia1 = 500/2000 = 1/4

E2Ia2 = Tω

⇒ Ia2 = 0

So, induced voltage

E2 = Kφω2 = Kω2 =  150

ω2 = E2/K = 150 × 4 /1

ω2 = 600 RPM

4. A 200 V d.c. shunt motor with armature resistance of 0.2 Ω and carrying a current of 50 A is running at 960 r.p.m. If the flux is reduced by 10% at constant torque and with negligible iron and friction losses, the speed will be near.

1. 1280 r.p.m.
2. 1170 r.p.m.
3. 1100 r.p.m.
4. 1060 r.p.m.

Explanation:

Given V = 200 volt

Ia = 50 A

ra = 0.2 A

N = 960 RPM

Eb = 200 – 50 × 0.2 = 190

Back EMF Eb = KnϕN

Knϕ = Eb/N = 190/960

Torque is constant

T = constant

KaϕIa =constant

When flux decreases by 10%

I’a = 50/0.9 = 55.55

Eb = 200 – 0.2 × 55.55 = 188.88 V

$N’ = \dfrac{{188.88}}{{0.9 \times 190}} \times 960$

N’ = 1060 RPM

5. A 250 V dc shunt motor has an armature resistance of 0.2 Ω and a field resistance of 100 Ω. On no-load at rated voltage, it draws an armature current of 5 A and runs at 1200 RPM. When the motor is operating on load is it draws a total line current of 50 A at rated voltage, with a 5 % reduction in the air-gap flux due to armature reaction. The voltage drop across the brushes is 1 V. The speed of the motor, in rpm, under this loaded condition, is closest to:

1. 1200
2. 1000
3. 1220
4. 900

Explanation:

Armature current, Ia1 = 5A

No-load speed, N1 = 1200 rpm,

Brush drop = 1 V per brush

otal brush drop = 2V

Eb1 = V – Ia1 Ra – Brush drop

= 250 – 5 × 0.2 – 2 = 247 V

Ia2 = IL – If = 47.5 A

ϕ2 = 0.95 ϕ1

Eb2 = V – Ia2 Ra – Brush drop

= 250 – 47.5 × 0.2 – 2 = 238.5 V

$\begin{array}{l} \dfrac{{{N_2}}}{{{N_1}}} = \dfrac{{{E_{b2}}}}{{{E_{b1}}}} \times \dfrac{{{\phi _1}}}{{{\phi _2}}}\\ \\ \Rightarrow \dfrac{{{N_2}}}{{1200}} = \dfrac{{238.5}}{{247}} \times \dfrac{{{\phi _1}}}{{0.95{\phi _1}}} = 1.0164 \end{array}$

N2 = 1200 × 1.0164 = 1219.688

6. A dc series motor with resistance between terminals of 1 Ω, runs at 800 RPM from a 200 V supply taking 15 A. If the speed is to be reduced to 475 rpm for the same supply voltage and current the additional series resistance to be inserted would be approximately

1. 2.5 Ω
2. 3 Ω
3. 4.5 Ω
4. 5 Ω

Explanation:

V = 200 V, R1 = Ra + Rse = 1 ohm, Ia = 15 A

N1 = 600 rpm, N2 = 475 rpm

In a DC series motor, Eb = V – IaR

Eb ∝ Nϕ and ϕ ∝ Ia

$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_{a1}}}}{{{I_{a2}}}} \times \dfrac{{{N_1}}}{{{N_2}}}$

E1 = 200 – 15 × 1 = 185 V

Given that Ia1 = Ia2

= 185/E2 = 800/475

Therefore, E2 = 110 V

110 = 200 – 15 × R2

⇒ R2 = 6 Ω

Therefore, Extra Resistance to be added in Series:

Rex = R2 – R1

= 6 – 1 = 5 Ω

7. The DC Motor, which can provide zero speed regulation at full load without any controller is

1. Series motor
2. Shunt motor
3. Cumulative compound
4. Differential compound

Explanation:

DC shunt motor has almost constant speed characteristics from no load to full load hence it provides zero speed regulation at full load without any controller.

8. Consider the following statements related to the speed control of a D.C. motor

(a) Speed may be controlled by changing the pole flux.

(b) Speed may be changed by changing voltage across the armature

1. Only (a) is correct
2. Only (b) is correct
3. Both (a) and (b) are correct
4. Both (a) and (b) are not correct

Answer:3. Both (a) and (b) are correct

Explanation:

Armature resistance control:

In a shunt motor, flux is constant when applied terminal voltage and shunt field resistance are constant. Therefore, speed of the motor is directly proportional to induced emf.

N ∝ Eb

Eb = V – Ia Ra

The value of Eb depends upon the drop in the armature circuit.

• We can control the speed by increasing armature resistance.
• This method gives only below base speeds.
• This method is a constant torque and variable power drive.

Field control method:

By adding external resistance in the field circuit, we can control the speed of the DC shunt motor above the normal speed.

• In this method, speed variation is accomplished by means of a variable resistance inserted in series with the shunt field.
• An increase in controlling resistances reduces the field current with a reduction in flux and an increase in speed.
• This method of speed control is independent of the load on the motor. Power wasted in controlling resistance is very less as field current is a small value.

We know that,

N ∝ Eb

N ∝ (V − Ia R/φ

ϕ ↓  → N ↑

By varying flux, we can increase the speed more than its base speed. This method is a constant power and variable torque drive.

9. A 240 V shunt motor with the armature resistance of 0.1 Ω runs at 850 r.p.m. for an armature current of 70 A. If its speed is to be reduced to 650 r.p.m., the resistance to be placed in series for an armature current of 50 A is nearly

1. 0.82 Ω
2. 1.13 Ω
3. 1.24 Ω
4. 1.34 Ω

Explanation:

Given V = 240 V

Ra = 0.1 Ω

Armature current I= 70 A

Eb = 240 – 70 × 0.1 = 233 volt

Knφ = 233/850

Now when it runs on 650 rpm

Back emf Eb = (233 × 650)/850

Current flows through armature, Ia = 50 A

R = (240 − 178.17)/50 = 1.23 Ohm

External resistance, Rext = R – Ra = 1.23 – 0.1 = 1.13 Ω

10. A 250 V dc shunt machine has armature circuit resistance of 0.6 Ω and field circuit resistance of 125 Ω. The machine is connected to 250 V supply mains. The motor is operated as a generator and then as a motor separately. The line current of the machine in both cases is 50 A. The ratio of the speed as a generator to the speed as a motor is _______

1. 1.27
2. 1.30
3. 1.20
4. 1.17

Explanation:

Terminal voltage VL = 250 V

Armature resistance Ra = 0.6 Ω

Field resistance Rf = 125 Ω

Line current in both IL = 50 A

Machine is running as a generator:

Field current, Ish = VL/Rf

= 250/125 = 2A

Armature current Ia = IL + Ish

⇒ Ia = 52 A

Induced emf: Eg = VL + Ia Ra

= 250 + 52 × 0.6

Eg = 281.2 V

Machine is running as motor:

Field current, Ish = VL/Rf

= 250/125 = 2A

Armature current Ia = IL – Ish

= 50 – 2 = 48 A

Back emf:

Eb = VL – IaRa = 250 – 48 × 0.6

Eb = 221.2 V

For DC shunt machine, this is constant hence

E ∝ N

= Eg/Eb = Ng/Nm

Ng/Nm = 281.2/221.2 = 1.27

11. A shunt DC motor’s Armature resistance is 1 ohm and field coil resistance is 100 ohms. The back emf constant is 9 volts for 100 RPM. The DC supply voltage is 100 V and it is running at a speed of 1000 RPM. What is DC supply current?

1. 11 A
2. 10 A
3. 1 A
4. 0.1 A

Explanation:

Armature resistance, Ra = 1 Ω

Field resistance, Rsh = 100Ω

Speed of motor, N = 1000 rpm

Supply voltage, Vs = 100 V

Back emf constant, K = 9 V for 100 rpm

= 0.09 V / rpm

∴ Back emf (Eb) can be calculated as

Eb =  kN

∴ Eb = 0.09 × 1000

∴ Eb = 90 V

Now shunt field current can be calculated as

Ish = Vs/Rsh = 100/100

∴ Ish = 1 A

Now armature current can be calculated as

Vs = Eb + IRa

Ia = (Vs − Eb)/Ra = (100 − 90)/1

Ia = 10 A

Supply current (Is) can be calculated as

I = Ia + Ish = 10 + 1

I = 11 A

12. A separately excited 300 V DC shunt motor under no-load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is

1. 840 RPM
2. 880 RPM
3. 850 RPM
4. 830 RPM

Explanation:

Given

Ra = 0.5 Ω

Ia = 2A, leakage inductance, L = 0.01 H

V = 300 V

Eno load = V – IaRa

= 300 – 2 (0.5)

= 300 – 1 = 299 V

When load is applied, we have

Ra = 0.5 Ω & Ia = 15 A

The circuit can be drawn is shown in fig (b)

Hence, Eload – V – IaRa

= 300 – 15 × 0.5 = 292.5 V

E ∝ N

13. A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8 Ω and the shunt field resistance is 240 Ω. The no-load speed, in rpm is

1. 2141 RPM
2. 1241 RPM
3. 1125 RPM
4. 1535 RPM

Explanation:

Voltage (V) = 120 V

shunt field resistance (Rsh) = 240 Ω

Shunt field current (Ish) = 120/240 = 0.5 A

Armature resistance (Ra) = 0.8 Ω

Load current (ILO) = 2 A

Armature current (Iao) = ILo – Ish

= 2 – 0.5 = 1.5 A

Let No load speed is No

Back emf (Ebo) = V – Iao Ra = 120 – (1.5) (0.8) = 118.8 V

Load current (ILf) = 7 A

Armature current (Iaf) = ILf – Ish

= 7 – 0.5 = 6.5 A

Back emf (Ebf) – V – Iaf Ra

= 120 – (6.5) (0.8)

= 114.8 V

Full load speed (Nf) = 1200 rpm

In a DC shunt motor

Eb ∝ N

$\begin{array}{l} \Rightarrow {{\rm{N}}_{\rm{o}}} = \dfrac{{{{\rm{E}}_{{\rm{bo}}}}}}{{{{\rm{E}}_{{\rm{bf}}}}}} \times {{\rm{N}}_{\rm{f}}}\\ \\ = \dfrac{{118.8}}{{119.8}} \times 1200 \end{array}$

No = 1241.8 RPM

14. A dc series motor of resistance 1 Ω across terminals runs at 1000 rpm at 250 V taking a current of 20 A. When an additional resistance of 6 Ω is inserted in series and taking the same current, the new speed would be

1. 142.8 rpm
2. 166.7 rpm
3. 478.3 rpm
4. 956.6 rpm

Explanation:

V = 200 V, R1 = Ra + Rse = 1 ohm, Ia = 15 A

N1 = 1000 rpm

R1 = Ra + Rse = 7 Ω

Eb ∝ Nϕ and ϕ ∝ Ia

$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_{a1}}}}{{{I_{a2}}}} \times \dfrac{{{N_1}}}{{{N_2}}}$

E1 = 250 – 20 × 1 = 230 V

Given that Ia1 = Ia2

E1 = 250 – 20 × 7 = 110 V

= 230/110 = 1000/N2

N2 = 478.26 rpm

15. A 230-V DC shunt motor has an armature resistance of 0.25 Ω and runs at 1100 rpm, taking an armature current 40 A. It is desired to reduce the speed to 750 rpm. If the armature current remains the same, find the additional resistance to be connected in series with the armature circuit.

1. 1 Ω
2. 1.5 Ω
3. 1.25 Ω
4. 1.75 Ω

Explanation:

Given

V = 230 V,

R1 = Ra + Rse = 0.25 Ω

Ia1 = 40 A

N1 = 1100 rpm,

N2 = 750 rpm

In a DC series motor, Eb = V – IaR

Eb ∝ Nϕ and ϕ ∝ Ia

$\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{I_{a1}}}}{{{I_{a2}}}} \times \dfrac{{{N_1}}}{{{N_2}}}$

E1 = 230 – 40 × 0.25 = 220 V

Given that Ia1 = Ia2

$\Rightarrow \frac{{{E_1}}}{{{E_2}}} = \frac{{{I_{a1}}}}{{{I_{a1}}}} \times \frac{{{N_1}}}{{{N_2}}} = \frac{{{N_1}}}{{{N_2}}}$

220/E2 = 1100/750

Therefore, E2 = 150 V

150 = 230 – 40 × R2

⇒ R2 = 2 Ω

Therefore, Extra Resistance to be added in Series:

Rex = R2 – R1

= 2 – 0.25 = 1.75 Ω

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