# DC Motor Torque MCQ || DC Motor Torque Questions and Answers

1. A dc shunt motor has the following characteristics, Ra = 0.5 Ω, Rf = 200 Ω, base speed = 1000 rpm, rated voltage = 250 V. On no-load, it draws a current = 5 A. At what speed will this run while delivering a torque of 150 N.m?

1. 881 rpm
2. 920 rpm
3. 950 rpm
4. 990 rpm

Explanation:

In motor back emf is given by

Eb = V – Iara

And

Eb = KnϕN

Where N = speed in rpm

Kn in V/rpm

ϕ = flux per pole

Torque T = KaϕIa

Ka=30πKn

Calculation:

Given V = 250 volt

Ia = 5 A

If = 250/200 = 1.25 A

Ia = 5 – 1.25 = 3.75 A

Eb = 250 – 3.75 × 0.5 = 248.125 volt

Knφ = 248.125/1000 = 0.248125

$T = \frac{{30}}{\pi }{K_n}\phi {I_a}$

Ia = (150 × π)/(0.248 × 30) = 63.33 A

Eb = 250 – 63.33 × 0.5 = 218.335

N = 218.335/0.248125 = 881 RPM

2. A 4-pole dc motor takes a 50 A armature current. The armature has lap-connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the armature of the motor.

1. 52.6 N-m
2. 39.5 N-m
3. 76.4 N-m
4. 15.2 N-m

Explanation:

Given,

Armature current Ia = 50 A

Number of poles p = 4

Lap wound A = p = 4

Number of conductors Z = 480

The flux per pole ϕ = 20 mWb

$T = \dfrac{{\phi Zp{I_a}}}{{2\pi A}} = \dfrac{{20 \times {{10}^{ – 3}} \times 480 \times 50 \times 4}}{{4 \times 2\pi }}$

T = 76.4 N-m

3. 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. 45.5 V
3. 30 V
4. 47.5 V

Explanation:

Let,

Torque T1 at N1 = 1000 rpm

Torque T= at N2 = 500 rpm

∴ T ∝ N2 {Given}

T2/T1 = (N2/N1)2

T2/T1 = (500/1000)2

T2/T1 = 1/4

Load torque can be expressed as,

T = k ϕ. Ia

T ∝ ϕ Ia

T ∝ Ia {∴ ϕ = constant}

T2/T1 = Ia2/Ia1

Ia2/Ia1 = 1/4

Ia2 = 10/4 = 2.5 A

Therefore,

Eb1 = V − Ia1Ra = (100 − 10) = 90 V

${E_b} = \frac{{NP\phi Z}}{{60{\rm{\;}}A}}$

Eb1/Eb2 = N1/N2

Eb2 = 90 × 500/1000

Eb2 = 45 V

Therefore,

V2 = Eb2 + Ia2Ra

⇒ V2 = 45 + 2.5 × 1

⇒ V2 = 47.5 V

Hence, The armature voltage is 47.5 V

4. A DC shunt machine develops an AC EMF of 250 V at 1500 rpm. Find the torque developed for an armature current of 50 A.

1. 59.6 N-m
2. 79.6 N-m
3. 69.6 N-m
4. 49.6 N-m

Explanation:

Speed is in radians per second (ω).

ω = (2 × π × N)/60

ω = (2 × π × 1500)/60

Power = generated emf × armature current

= 250 × 50

= 12500 watts.

Torque = power/speed.

T = 12500/157.08

T = 79.577 Nm

5. An Electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/s2, the moment of inertia of the system is (neglecting viscous and Coulomb friction)

1. 0.25 kg m2
2. 0.25 Nm2
3. 4 kg m2
4. 4 Nm2

Explanation:

Accelerating torque in an electric motor is given as

Ta = Tm – T

Ta = J α

Where, Tm = Mechanical torque or starting torque

Te = Electrical torque or load torque

J = Moment of inertia

α = Angular acceleration

Calculation:

Given-

T= 15 Nm, Te = 7 Nm, α = 2 rad /sec

J = (15 – 7) / 2

J = 4 Nm2

6. Neglecting saturation, if the current taken by a series motor is increased from 10 A to 12 A, the percentage increase in its torque will be:

1. 44%
2. 40%
3. 20%
4. 16%

Explanation:

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

Ta ∝ Ia      (∵ Flux saturation is neglected)

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

Calculation:

Let at I = 10 A, the torque is T1 and at I = 12 A torque is T2

T1/T2 = (10/12)2

T2 = 1.44 T1

i.e. T2 increases by 44%

7. A DC motor has the following specifications: 10 hp, 37.5 A, 230 V; flux / pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature resistance = 0.267Ω. The armature reaction is negligible and rotational losses are 600W. The motor operates from a 230V DC supply. If the motor runs at 1000 rpm, the output torque produced in (in Nm) is _______

1. 60.34 Nm
2. 57.78 Nm
3. 51.34 Nm
4. 40.23 Nm

Explanation:

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

Eb = (0.01 × 666 × 4 × 1000)/60 × 2 = 222 V

Ia = (V – E) / Ra = (230 – 222) / 0.267

Ia = 29.96 A

Internal power ( Pd) = EIa = 222 × 29.96 = 6651.685 Watt

Shaft Power output,

Pout = P– Rotational losses

Pout = 6651.685 − 600 = 6051.685

${T = \dfrac{{{P_{out}}}}{\omega } = \frac{{6051.685}}{{2\pi \times \dfrac{{1000}}{{60}}}} = 57.78\:Nm}$

8. A DC Motor develops a torque of 150 N-m. A 10 percent reduction in the field flux causes a 50 percent increase in armature current. The new value of torque is:

1. 102.5 N-m
2. 202.5 N-m
3. 172.5 N-m
4. 232.5 N-m

Explanation:

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

For any DC machine ZP/60 A Remains constant

∴ T ∝ ϕ Ia

Calculation:

Given,

Torque T1 = 150 N-m

There is 10% reduction in field flux

⇒ ϕ2 = 90% ϕ1 = 0.9 ϕ1

Also armature current increased by 50%

⇒ Ia2 = 150% Ia1 = 1.5 Ia1

As T ∝ ϕ Ia

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

$\dfrac{{{T_2}}}{{150}} = \dfrac{{0.9{\phi _2}}}{{{\phi _1}}} \times \dfrac{{1.5{I_{a2}}}}{{{I_{a1}}}} = 1.35$

T2 = 202.5 N-m

Therefore the new value of torque is 202.5 N-m

9. A shunt–connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radians/second. At its rated torque of 500 Nm, its speed is 180 radian/second, The motor is used to directly drive a load whose load torque TL depends on its rotational speed (in radian/second), such that TL = 2.78 × ωT. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load is __________

Explanation:

Given,

Rated torque (Tr) = 500 Nm

Rated Speed (ωr) = 180 rad/sec

Load torque (TL) = 2.78 ωr

Developed torque can be expressed as,

T = (200 − ωr) × 500/20

10. Which of the given options is correctly represented by the following expression? 9.55 (output / N), where N is in rpm?

1. Back EMF
2. Maximum efficiency
3. Armature torque
4. Shaft torque

Explanation:

The torque that is available at the motor shaft for doing useful work is known as shaft torque. It is represented by Tsh.

The total or gross torque (Ta) developed in the armature of a motor is not available at the shaft because a part of it is lost in overcoming the iron and frictional losses in the motor.

Therefore, shaft torque Tsh is less than the armature torque Ta.

The difference Ta – Tsh is called lost torque.

We know that, Power (P) = Torque (T) × Angular Speed (ω)

Here, We have concerns with Shaft Torque,

Tsh = P/ω

If N is the speed in RPM then,

Angular speed (ω) = 2πN/60

From equation (1) & (2)

Tsh = (P × 60)/2πN

60/2π = 9.55

Hence, Tsh = 9.55 (output / N)

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