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

Answer:3. 2:1

Explanation: 

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

Answer:1. 25 rad/s

Explanation: 

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 I_a = 55. $\omega = \frac{{230 – 33}}{{55.0.1454}} = 24.6\;rad/s$

Answer:2. 3.4 Ω

Explanation: 

$\frac{{{E_1}}}{{{E_2}}} = \frac{{{N_1}}}{{{N_2}}}$ 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

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

Answer: 1. Motor will draw a large current

Explanation: 

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.

Answer: 1. 37.69 V

Explanation: 

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:

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

Answer: 2. 47.5 V

Explanation: 

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

Answer: 1. 5 A

Explanation: 

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.

Answer: 4. 27.42 A

Explanation: 

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 = I_L – I_f1 = 20 – 1.66 = 18.34 Amps

Where, Ia1 = Armature current,

I_L = Current drawing by motor,

I_f1 = Field current,

Now, by adding 75Ω in the field circuit,

The field current becomes I_f2

Hence

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

as the load torque is constant (T_L= k_aϕ I_a)

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

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

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

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

The new armature current (I_a2) = 27.42 A.

Answer:2. 1500 rpm

Explanation: 

E = kϕN

V = E + I_ar_a

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

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

Answer: 1. 47.76 N

Explanation:

Given

Power P = 5 kW = 5000 Watt

Revolution = 1000 RPM

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

Calculation

Power = 2πNT

T = P/2πN

= 5000/(2 ×π × 50/3)

= 47.76 N.

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

Explanation: 

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

Important:

• 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.

Answer: 1. non-linear

Explanation: 

• 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.

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

Explanation: 

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

Answer: 1. V = Eb + IaRa

Explanation: 

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.

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.

Answer:1. Field control method

Explanation: 

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

In field control method

N ∝ E_b/φ

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.

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

Explanation: 

In a DC shunt motor E_b ∝ 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.

Answer:3. 4 point starter

Explanation: 

• 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.

Answer:1. field winding

Explanation: 

• 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.

Answer:3. Both shunt and compound motor

Explanation: 

The back emf (E_b) 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.