DC Motor Torque MCQ || DC Motor Torque Questions and Answers - Page 2 of 2

11. A DC motor runs at 200 rpm speed. If the power developed by the machine is 400 W, what is the Torque developed?

25.5 Nm
12.2 Nm
33.33 Nm
19.1 Nm

Answer: 4. 19.1 Nm

Explanation:

Given that, N = 200 rpm, P_m = 400 W

ω = (2 × π × 200)/60 = 20.94

Torque = power / speed

T = 400/20.94

T = 19.1 Nm

12. A 240 V, 4-pole wave wound DC series motor has 600 conductors on its armature; it has armature and series field resistance of 0.5 ohms. The motor takes a current of 30 A. Estimate its gross torque developed if it has a flux per pole of 25 mWb.

160 N-m
155.3 N-m
150 N-m
143.2 N-m

Answer: 4. 143.2 N-m

Explanation:

Given,

P = 4

ϕ = 25 mwb

I_L = Ia = 30 A (Since it is DC series motor)

Z = 600

For Wave Winding, A = 2

From the above concept,

Ta = 0.159ZφIa × (P/A)

Ta = 0.159 × 600 × 25 × 10⁻³ × 30 × 4/2

∴ Gross Torque = 143.2 Nm

13. A separately excited dc generator rotating at 3000 rpm produces an emf of 157 V and delivers a current of 20 A. The braking torque exerted by the armature is

17 N-m
10 N-m
12 N-m
12.5 N-m

Answer: 2. 10 N-m

Explanation:

Given sped of the generator N = 3000 rpm

Terminal voltage V_t = 157 V

Delivered load current I_L = 20 A

Power developed by generator P = V_t × I_L

P = 157 × 20 = 3140 W

Torque exerted by armature T = P / ω

where ω = (2 π N) / 60

ω = (2 π × 3000) / 60 = 314 rad / sec

∴ T = 3140 / 314 = 10 N-m

14. The torque of a given motor can be varied by

Changing flux
Changing armature current
Changing flux and armature current both
Changing number of parallel paths

Answer: 3. Changing flux and armature current both

Explanation:

In a DC machine, torque (T) is given as

T = P_out/ω

Where,

P_out = Output power in kW

ω = Angular speed in radian per second = 2π N

E_b = Back emf

I_a = Armature current

N = Speed in rpm

Z = Number of conductors

A = Number of parallel paths

P = Number of poles

Output power of DC motor is given by

Pout = E_bI_a ……..(2)

Back EMF of DC motor is given by

E_b = (ϕZNP)/(60 A) ——-(3)

Where,

ϕ = flux/pole

From equations (1), (2) and (3), we get

$T = \dfrac{{{E_b}{I_a}}}{{2\pi N}} = \dfrac{{ZP\phi ({I_a})}}{{2\pi \times 60A}}$

For any DC machine, (ZP)/(60 A)

remains constant

∴ T ∝ ϕ Ia

So that for DC Machine, torque is directly proportional to armature current and flux.

Hence torque of a given motor can be varied by changing flux and armature current.

15. A dc motor develops an electromagnetic torque 150 N-m in certain operating conduction. from this operating conduction, a 10% reduction in field flux and 50% increase in armature current is made. What will be the new value of electromagnetic torque?

225 N-m
202.5 N-m
22.5 N-m
20.25 N-m

Answer: 2. 202.5 N-m

Explanation:

Given ϕ₂ = 0.9ϕ₁ and I_a2= 1.5 I_a1

T ∝ ϕ I_a

$\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{\phi _1}}}{{{\phi _2}}} \times \dfrac{{{I_{a1}}}}{{{I_{a2}}}}$

${T_2} = 150 \times \dfrac{{0.9{\phi _1}}}{{{\phi _1}}} \times \dfrac{{1.5{I_{a1}}}}{{{I_{a1}}}} = 202.5\:Nm$

16. In a DC shunt motor, the torque produced is proportional to:

Armature current
(Armature current)²
Ampere-turn
(Ampere-turn)²

Answer: 1. Armature current

Explanation:

For DC Machine, torque is directly proportional to armature current and flux.

T ∝ ϕ Ia

In dc shunt motors, flux is constant (not dependent on armature current). Hence torque is proportional to armature current.

T ∝ Ia

In dc series motors, flux is directly proportional to armature current. Hence torque is directly proportional to square of the armature current.

Ta ∝ Ia2

In dc series motor at saturation condition, flux is constant. Hence torque is proportional to armature current.

T ∝ Ia

Where, T = torque, Ia = armature current, ϕ = field flux

17. A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra = 0.02 Ω. When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is

34.2 A
30 A
22 A
4.84 A

Answer: 2. 30 A

Explanation:

Voltage (V) = 220 V

Torque (τ) = 70 N-m

Speed (N) = 900 rpm

Armature resistance (R_a) = 0.02 Ω

Power (P) = 10 kW

⇒ E_bI_a = 10 × 10³

⇒ (V – I_aR_a) I_a = 10,000

⇒ [220 – I_a(0.02)] I_a = 10,000

⇒ I_a = 45.75 A

P = τω

$\begin{array}{l} 10 \times {10^3} = \tau \left( {\dfrac{{2\pi {\rm{N}}}}{{60}}} \right)\\ \\ \tau = \dfrac{{10 \times {{10}^3} \times 60}}{{2\pi \times 900}} = 106.1\:{\rm{N}} – {\rm{m}} \end{array}$

Torque α ϕ I_a

For separately excited motor

T α I_a

T₁ = 106.1 N-m, I_a1 = 45.75 A

T₂ = 70 N-m

Ia₂ = (T₂Ia₁)/T₁

Ia₂ = (70 × 45.75)/106.1 = 30.18 A

18. The no-load speed of a 230 V separately excited dc motor is 1400 rpm. The armature resistance drop and the brush drop are neglected. The field current is kept constant at the rated value. The torque of the motor in Nm for an armature current of 8 A is _______

12.5 Nm
10.5 Nm
14.5 Nm
7.5 Nm

Answer:1. 12.5 Nm

Explanation:

The torque on the motor is

T = P/ω = VIa/ω

ω = 2πNs/60

I_a = Armature current, V = Excitation voltage

Calculation:

No load speed, N = 1400 rpm

I_a = 8A, V = 230 V

$T = \frac{{V{I_a}}}{\omega } = \dfrac{{V{I_a}}}{{\dfrac{{2\pi \: \times \:1400}}{{60}}}} = \dfrac{{230\: \times \:8}}{{\dfrac{{2\: \times \:\pi \: \times \:1400}}{{60}}}}$

T = (9.55 × 230 × 8)/1400

T = 12.554 Nm

19. Two dc series motor run at the speeds of 360 rpm and 420 rpm respectively and separately when drawing 50 A at 400 V.These armature resistance are 0.5 Ω and 0.7 Ω respectively. What is the speed of the combination is rpm if they get mechanically coupled while still drawing same current are same voltage?

180.94
177.94
175.90
170.90

Answer: 2. 177.94

Explanation:

For motor 1, back emf ${E_{{b_1}}} = 400 – \left( {0.5} \right) \times 50$ = 375 V

For motor 2, back emf ${E_{{b_2}}} = 400 – \left( {0.7} \right) \times 50$

= 365 V

Now E_b ∝ Nf

∴ E_b = KNf

where K is constant

$\begin{array}{*{20}{l}} \begin{array}{l} {E_{{b_1}}} = K{N_1}{\emptyset _1}\\ \end{array}\\ {K{\emptyset _1} = \frac{{{E_{{b_1}}}}}{{{N_1}}} = \frac{{375}}{{360}}}\\ \begin{array}{l} \\ {E_{{b_2}}} = K{N_2}{\emptyset _2} \end{array}\\ \begin{array}{l} \\ K{\emptyset _2} = \frac{{{E_{{b_2}}}}}{{{N_2}}} = \frac{{365}}{{420}} \end{array} \end{array}$

When motor are coupled, then

E_b = 400 – 50 × (1.2) = 240 V

Now, if N is the common speed of the coupled motors, then

${E_b} = {E_{{b_1}}} + {E_{{b_2}}}$

= Kf₁N + Kf₂N

$\Rightarrow 340 = \frac{{375}}{{360}}N + \frac{{365}}{{420}}N$

N = 177.94 rpm

20. Two series motors are separately running at speeds of 450 rpm and 500 rpm respectively while drawing 60 A at 450 V. The armature resistance of slower motor is 0.6 Ω and that of faster one is 0.8 Ω. If they are mechanically coupled and connected in series while the combination is still drawing 60 A at 450 V. what is the speed of the combination?

212.3 rpm
300 rpm
480.6 rpm
512.7 rpm

Answer: 1. 212.3 rpm

Explanation:

Eb₁ = (450 −(60 × 0.6) = 414 V

Eb₂ = (450 −(60 × 0.8) = 402 V

Eb = KNφ

Eb₁ = KN₁φ₁

Kφ₁ = Eb₁/N₁ = 414/450

Eb₂ = KN₂φ₂

Kφ₂ = Eb₂/N₂ = 402/500

If N is the common speed of the coupled motors, then

Eb = Eb₁ + Eb₂

= Kφ₁ N + Kφ₂N

= 366 = (414/450) + (402/500)

N = 212.3 RPM

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