DC Motors Operating Characteristics MCQ || Operating Characteristics of DC Motor Questions and Answers

1. When a DC motor generates maximum power, the ratio of applied voltage to back emf is

  1. 1:1
  2. 2:3
  3. 2:1
  4. √2:1

Answer:3. 2:1


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

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


V = Applied voltage

Eb = Back emf

Ia = Armature current

Ra = 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


2. A DC series motor running at the speed of 800 rpm draws 18 A from the supply. If the load is changed such that the current drawn by the motor is 55 A, calculate the speed of the motor (in rad/sec on the new load. Armature and field winding resistances are 0.2 and 0.4 Ω, respectively. Assume a supply voltage of 230 V and also that the flux produced is proportional to the current.

  1. 25 rad/s
  2. 20 rad/s
  3. 15 rad/s
  4. 10 rad/s

Answer:1. 25 rad/s


Calculating  the constant K of DC Motor

$K. = \frac{{V – {I_a}\left( {{R_a} + {R_a}} \right)}}{{{I_a}.\omega }}$

Next, calculate the speed of the motor at the current of Ia = 55. $\omega = \frac{{230 – 33}}{{55.0.1454}} = 24.6\;rad/s$


3. A separately excited dc motor runs at 1500rpm under no load with 200V applied to the armature. The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of  5 Nm is 1400 rpm. The rotational and armature reaction is neglected. The armature resistance of the motor is

  1. 2 Ω
  2. 3.4 Ω
  3. 4.4 Ω
  4. 7.7 Ω

Answer:2. 3.4 Ω


$\frac{{{E_1}}}{{{E_2}}} = \frac{{{N_1}}}{{{N_2}}}$ E1 is 200V itself since Ia 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

$\begin{array}{l} 186.67 = 200 – 3.92{R_a}\\ {R_a} = 3.4\ {\rm{\Omega }} \end{array}$


4. A dc shunt motor is connected to the source through a 3-point starter. Suddenly if the starter handle is moved fastly from off to on position, then the ______

  1. Motor will draw a large current
  2. Motor will not start
  3. Motor will burn
  4. All of the mentioned

Answer: 1. Motor will draw a large current


When the starter handle is moved rapidly from ‘OFF’ to ‘ON’ position, the resistance connected in series with the armature is being cut out at a faster rate and the speed of the motor is increased at a slow rate. Because of this, the armature current will increase to a larger value and the ratio of upper and lower values of currents will not be maintained.


5. A DC motor is used in a conveyor belt. The belt has been wrapped around the rollers, and one of the rollers is connected to the motor. The series-connected motor rotates at 1200 rpm with the torque of 12 N-m and the field current of 40 A. Find output voltage, assuming RA = Rf = 0.

  1. 37.69 V
  2. 25.48 V
  3. 45.24 V
  4. 40.44 V

Answer: 1. 37.69 V


In order to find the output voltage

${E_a} = \frac{{T.\omega }}{{{I_a}}}$

Before solving, first, find ω.

$\omega = \frac{{2\pi N}}{{60}} = \frac{{2\pi .1200}}{{60}} = 1215.66\;rad/sec$

Thus, we can solve for the voltage:

$V = \frac{{12 \times 125.66}}{{40}} = 37.69\;V$


6. With an armature voltage of 100 V and rated field winding voltage, the speed of a separately excited DC motor driving a fan is 1000 rpm, and its armature current is 10 A. The armature resistance is 1 Ω. The load torque of the fan load is proportional to the square of the rotor. Neglecting rotational losses, the value of the armature voltage (in Volt) which will reduce the rotor speed to 500 rpm is

  1. 50 V
  2. 47.5 V
  3. 35.5 V
  4. 30 V

Answer: 2. 47.5 V



Equivalent circuit of separately excited D.C motor,

V = Eb + IaRa



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. Ia

T ∝ ϕ Ia

T ∝ Ia {∴ 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


$E_{b_1}=V_1-I{a_1}R_a=100-10=90$ 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



⇒ V2 = 45 + 2.5 × 1

⇒ V2 = 47.5 V

Hence, The armature voltage is 47.5 V


7. For a 100 hp 250 V, compound dc motor with compensating winding has a field current of 5 A to produce a voltage of 250 V at 1200 rpm. What will be the shunt field current of this machine at no load?

  1. 5 A
  2. 5.6 A
  3. 4 A
  4. 0 A

Answer: 1. 5 A


At no load condition, load on the motor is zero hence armature current(Ia) is also zero,  so the internally generated voltage is 250 V. So 5 A will be the shunt field current at no load.


8. A 250-V DC shunt motor has a shunt field resistance of 150 Ω and an armature resistance of 0.3 Ω. For a given load. The motor runs at 1400 rpm and draws 20 A current. If the resistance of 75 Ω is added in series with the field, find the new armature current. Assume the load torque remains constant.

  1. 32.32 A
  2. 18.84 A
  3. 16.60 A
  4. 27.42 A

Answer: 4. 27.42 A


The field current corresponds to the given circuit of dc shunt motor is

${I_{f1}} = \frac{{{V_t}}}{{{R_f}}} = \frac{{250}}{{150}} = 1.66$ A.

Ia1 = IL – If1 = 20 – 1.66 = 18.34 Amps

Where, Ia1 = Armature current,

IL = Current drawing by motor,

If1 = Field current,

Now, by adding 75Ω in the field circuit,

The field current becomes If2


${I_{f2}} = \frac{{{V_t}}}{{{R_f} + {R_{ext}}}} = \frac{{250}}{{150 + 75}} = 1.11$ A

as the load torque is constant (TL= kaϕ Ia)

$\phi \propto \frac{1}{{{I_a}}}$

${I_f} \propto \frac{1}{{{I_a}}}$

$\therefore \;\frac{{{I_{f1}}}}{{{I_{f2}}}} = \frac{{{I_{a1}}}}{{{I_{a2}}}}$

${I_{a2}} = \frac{{1.66 \times 18.34}}{{1.11}}$= 27.42 A.

The new armature current (Ia2) = 27.42 A.


9. A DC shunt motor when connected to 250 V DC supply runs at 1500 rpm. When the same motor is connected to a 300 V DC supply, then what will be the speed of the motor neglecting losses and saturation?

  1. 1200 rpm
  2. 1500 rpm
  3. 1800 rpm
  4. 750 rpm

Answer:2. 1500 rpm


E = kϕN

V = E + Iara

$V{I_a} = E{I_a} + I_a^2{r_a}$

Now $I_a^2{r_a} \approx 0$ (neglecting losses)

So V = E

⇒ V = kϕN

250 = Kϕ(1500)

$K\phi = \frac{{250}}{{1500}} = \frac{1}{6}$

When 300 V is applied

300 = Kϕ’ (N’)

Where N’ = speed at 300 V

Φ’ = flux at 300 V

Now, for shunt motor, ϕα V (applied voltage)

Φ α 250, ϕ’ α 300

$\phi ‘ = \frac{{300\phi }}{{250}}$

∴ 300 = K ϕ’ N’

$= K \times \frac{{300\phi }}{{250}} \times N’$

$300 = \frac{1}{6} \times \frac{{300}}{{250}} \times N’$

N’ = 1500 rpm


10. For a dc shunt motor of 5 kW, running at 1000 RPM the induced torque will be _______

  1. 47.76 N
  2. 57.76 N
  3. 35.76 N
  4. 37.76 N

Answer: 1. 47.76 N



Power P = 5 kW = 5000 Watt

Revolution = 1000 RPM

Speed N = revoution/time = 1000/60 = 50/3


Power = 2πNT

T = P/2πN

= 5000/(2 ×π × 50/3)

= 47.76 N.


11. It is advised not to run dc series motor with no load Why?

  1. Because zero torque at no load will make speed infinite
  2. Because zero torque as no-load will not let machine start
  3. Because infinite torque will be produced
  4. None of the mentioned

Answer: 1. Because zero torque at no load will make speed infinite


A DC series motor should always be started with load because at no load it will rotate with dangerously high speed.


  • When the motor is connected across the supply mains without load, it draws small current from the supply mains
  • This current will flow through the series field and armature, the speed tends to increase so that back emf may approach the applied voltage in magnitude
  • The increase in back emf weakens the armature current and hence the field current
  • This cause again increases the speed so the back emf
  • Thus, the field continues to weaken and speed continues to increase until the armature produced such centrifugal force that it is coming out from its shaft and gets damaged.


12. The flux and the internally generated voltage of a dc machine is a ________ function of its magneto-motive force.

  1. non-linear
  2. linear
  3. constant
  4. inverse

Answer: 1. non-linear


  • When the DC machine is put into its first operation after its construction or when the machine is completely demagnetized prior to this operation, the initial reading of E will be zero when If =0.
  • Then on increasing field current, one can observe almost a linear variation of E as long as the machine is in unsaturated condition. Further, an increase in field current causes the machine to get saturated and it requires more magnetizing force for the same amount of increase in E. Hence the curve becomes non-linear.
  • The flux and induced emf are non-linear functions of its MMF.


13. The basic difference between 4 point starter when compared to 3 point starter is:

  1. The holding coil is removed from the shunt field circuit
  2. The overcurrent release is connected in the supply line
  3. Field circuit is completed through starting resistance
  4. The starting resistance is gradually cut out till when the armature reaches the running position

Answer: 1. The holding coil is removed from the shunt field circuit


Four-point starter:

The four-point starter works as a current controlling device in the deficiency of back EMF while it starts running off the DC motor. A four-point starter also works as a protecting device.

The main difference between a 4-point starter compared to a 3-point starter is, the holding coil is detached from the shunt-field circuit.

The 4-point starter uses four terminals for speeding up the motor. These four terminals namely, armature terminal (A), field terminal (F), and the line terminal (L).

  • NVC (No Volt Coil): The connection of a four-point starter can be done in parallel with the field coil
  • The line terminal (L) is connected to a positive supply
  • The armature terminal (A) is connected to the winding of an armature
  • The field terminal (F) is connected to the field winding
  • It is provided as not to affect the current flowing through ‘Hold on’ coil even when the field current changes


14. Voltage equation of a dc motor is

  1. V = Eb + IaRa
  2. V = Eb – IaRa
  3. V = EbIa – Ra
  4. V = EbIa + Ra

Answer: 1. V = Eb + IaRa


In the case of the d.c motor, supply voltage V has to overcome back e.m f. Eb which is opposing V and also various drops as armature resistance drop la Ra brush drop etc. Infect the electrical work done in overcoming the back EMF gets converted into Be mechanical energy developed in the armature. Hence the voltage equation of a d.c motor can be written as

V = Eb + IaRa

The back EMF is always less than the supply voltage otherwise it will act as a generator.


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

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

Answer:2. Half


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.


16. For speed control of a DC motor, which technique provides speed above base speed?

  1. Field control method
  2. Supply current control method
  3. Supply voltage control method
  4. Armature resistance control method

Answer:1. Field control method


Field control method of a DC motor provides speed above base speed.

In field control method

N ∝ Eb

N ∝ (V − IaRa)/φ

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


17. If the field of a DC shunt motor is opened while running, what will happen to the speed?

  1. Speed of the motor will remain constant
  2. The motor will get locked
  3. Speed of the motor will reduce
  4. Speed of the motor will become dangerously high

Answer:4. Speed of the motor will become dangerously high


In a DC shunt motor Eb ∝ Nφ

For a constant back emf, flux is inversely proportional to the speed of the motor.

If field winding is disconnected accidentally, the speed would dangerously increase in order to maintain the back emf of the motor. For a constant back emf, flux is inversely proportional to the speed of the motor.

In the case of a series machine if field winding is disconnected, then the motor circuit is open and no current will pass through the armature, for the operation of the machine interaction of two fluxes is necessary thus the machine won’t operate.


18. High-speed protection of DC shunt motor is not provided by:

  1. 3 point starter
  2. 2 point starter
  3. 4 point starter
  4. 5 point starter

Answer:3. 4 point starter


  • The four-point starter works as a current controlling device in the deficiency of back EMF while it starts running off the DC motor.
  • A four-point starter also works as a protecting device.
  • The main difference between a 4-point starter compared to a 3-point starter is, the holding coil is detached from the shunt-field circuit.
  • The only limitation or drawback of the 4 point starter is that it cannot limit or control the high current speed of the motor.
  • If the field winding of the motor gets opened under the running condition, the field current automatically reduces to zero.
  • But as some of the residual flux is still present in the motor, we know that the flux is directly proportional to the speed of the motor.
  • Therefore, the speed of the motor increases drastically, which is dangerous, and thus High-speed protection of DC shunt motor is not possible.
  • This sudden increase in the speed of the motor is known as the High-Speed Action of the Motor.


19. The connection of NVC in DC 3 point starter, is connected across DC supply in series with:

  1. field winding
  2. handle
  3. overload relay
  4. starting resistor

Answer:1. field winding


  • The no-volt trip coil (NVC) is connected in series with the field winding of the motor.
  • In the event of switching off, or when the supply voltage falls below a predetermined value, or the complete failure of supply while the motor is running NVC is de-energized.
  • This results in the release of the handle, which is then pulled back to the OFF position by the action of the spring.
  • The current to the motor is cut off, and the motor is not restarted without resistance R in the armature circuit.
  • The NVC also provides protection against an open circuit in the field winding.
  • The NVC is called no-volt or Undervoltage protection of the motor. Without this protection, the supply voltage might be restored with the handle in the RUN position. Consequently, full line voltage may be applied directly to the armature resulting in a very large current.


20. Three-point starter can be used for

  1. Series motor only
  2. Shunt motor only
  3. Both shunt and compound motor
  4. Compound motor only

Answer:3. Both shunt and compound motor


The back emf (Eb) is developed as motor armature starts to rotate in presence of the magnetic field. Initially, back emf is zero, starting current is high.

To limit this starting current, we use a starter. A three-point starter helps in starting and running the shunt-wound motor or compound wound DC motor.

A three-point starter is used to reduce the starting current and it has three terminals

i) ‘L’ Line terminal (Connected to positive supply).

ii) ‘A’ Armature terminal (Connected to armature winding).

iii) ‘F’ Field terminal (Connected to field winding).

The no-volt coil is connected in series with the shunt field as shown in the figure of 3 point starter.

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