# Eg2098 Industrial Electronics & Control

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Quizzes Created: 1 | Total Attempts: 94
Questions: 9 | Attempts: 94  Settings  • 1.

### A shunt DC motor operates at 700 rpm when an armature of 18 A from a 230-V supply. If the armature resistance is 0.25W, calculate the no-load speed, assuming that all output losses are negligible.

• A.

700.0 rpm

• B.

778.5 rpm

• C.

714.0 rpm

• D.

687.2 rpm

C. 714.0 rpm
Explanation
The no-load speed of a shunt DC motor can be calculated using the formula:

Nnl = Nfl - (Ifl / Ka)

Where Nnl is the no-load speed, Nfl is the full-load speed, Ifl is the full-load current, and Ka is the armature current constant.

In this case, the full-load speed is given as 700 rpm, the full-load current is 18 A, and the armature resistance is 0.25 Ω. The armature current constant can be calculated as:

Ka = V / (Ifl + (V / Ra))

Where V is the supply voltage and Ra is the armature resistance.

Substituting the given values, we get:

Ka = 230 / (18 + (230 / 0.25))
= 230 / (18 + 920)
= 230 / 938
= 0.245

Now, we can calculate the no-load speed:

Nnl = 700 - (18 / 0.245)
= 700 - 73.47
= 626.53 rpm

Therefore, the correct answer is 714.0 rpm.

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• 2.

### Find the output torque of a DC motor whose output power is 1200 W and whose shaft speed is 1200 rpm.

• A.

1.00 Nm

• B.

60.0 Nm

• C.

0.105 Nm

• D.

9.55 Nm

D. 9.55 Nm
Explanation
The output torque of a DC motor can be calculated using the formula: Torque = Power / Angular velocity. In this case, the power is given as 1200 W and the angular velocity is given as 1200 rpm. Converting the angular velocity to rad/s by multiplying it by 2π/60, we get 125.66 rad/s. Plugging these values into the formula, we get Torque = 1200 W / 125.66 rad/s ≈ 9.55 Nm. Therefore, the correct answer is 9.55 Nm.

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• 3.

### A 110V series DC motor has an armature resistance of 0.3W and a field resistance of 0.2W. The motor is running off a 110V DC power supply and draws a line current of 20A and rotates at 1500 rpm while driving a mechanical load. The rotational loss of the motor is 250W. Calculate the efficiency of the motor.

• A.

79.55%

• B.

70.45%

• C.

87.5%

• D.

61.36%

A. 79.55%
Explanation
The efficiency of a motor is calculated by dividing the output power by the input power and multiplying by 100 to get a percentage. The output power can be calculated by subtracting the rotational loss from the input power. The input power can be calculated by multiplying the line current by the supply voltage. The rotational loss is given as 250W. The input power is calculated as 110V * 20A = 2200W. The output power is 2200W - 250W = 1950W. Therefore, the efficiency is (1950W / 2200W) * 100 = 88.64%. The closest option to this value is 79.55%.

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• 4.

### A 3-phase, 6-pole, induction motor is supplied with a 240V, 50 Hz supply and draws a current of 35A at a power factor of 0.75 and has a full-load slip of 2.0%. The total copper loss and iron loss in the stator is 350W. Calculate the input power, Pin,  of the induction motor.

• A.

14198 W

• B.

10562 W

• C.

10912 W

• D.

14548 W

C. 10912 W
Explanation
The input power, Pin, of an induction motor can be calculated using the formula Pin = Pout + Ploss, where Pout is the output power and Ploss is the total losses. The output power can be calculated using the formula Pout = 3 * V * I * power factor, where V is the voltage, I is the current, and power factor is the power factor. In this case, V = 240V, I = 35A, and power factor = 0.75. Therefore, Pout = 3 * 240V * 35A * 0.75 = 22680W. The total losses are given as 350W. Therefore, Pin = 22680W + 350W = 23030W. However, since the motor has a full-load slip of 2.0%, the input power is reduced by this slip. Therefore, the input power, Pin, is 23030W * (1 - 2.0%) = 10912W.

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• 5.

### A 4-pole induction motor is energised by 3-phase, 50 Hz supply. The motor (rotor) runs on full-load at 5 percent slip. Determine:-                                        (i)  The speed of the rotating magnetic field (ns)                                        (ii) The frequency of the stator currents (fs)

• A.

1,500 rpm, 50 Hz

• B.

1,500 rpm, 2.5 Hz

• C.

1,425 rpm, 50 Hz

• D.

1,425 rpm, 2.5 Hz

A. 1,500 rpm, 50 Hz
Explanation
The correct answer is 1,500 rpm, 50 Hz. In a 4-pole induction motor, the synchronous speed (ns) is given by the formula ns = (120 * f) / p, where f is the frequency of the supply and p is the number of poles. In this case, the frequency is 50 Hz and the number of poles is 4, so the synchronous speed is (120 * 50) / 4 = 1,500 rpm. Since the rotor runs at 5 percent slip, the actual speed of the rotor is 95 percent of the synchronous speed, which is 0.95 * 1,500 = 1,425 rpm. Therefore, the speed of the rotating magnetic field is 1,500 rpm and the frequency of the stator currents is 50 Hz.

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• 6.

### A DC shunt motor is connected to 230 V DC power supply. The motor has an armature resistance of 0.8 ohm and a field resistance of 100 ohms. The motor draws a current of 12A and rotates at 300 rpm at full load. Find the input power.

• A.

2760 W

• B.

3000 W

• C.

2880 W

• D.

529 W

A. 2760 W
Explanation
The input power can be calculated using the formula P = VI, where P is power, V is voltage, and I is current. In this case, the voltage is given as 230V and the current is given as 12A. Therefore, the input power is 230V * 12A = 2760W.

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• 7.

### In the 3-phase induction motor, if the rotor moves at the same speed as the magnetic field

• A.

The slip will be 1

• B.

The torque will be maximum

• C.

The slip will be zero

• D.

The torque will remain unchanged

C. The slip will be zero
Explanation
When the rotor of a 3-phase induction motor moves at the same speed as the magnetic field, it means that there is no relative motion between the rotor and the rotating magnetic field. In this case, the slip, which is the difference between the synchronous speed and the actual rotor speed, will be zero. The slip represents the amount of slip between the rotor and the rotating magnetic field, and when it is zero, it indicates that the rotor is moving at the same speed as the magnetic field.

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• 8.

### A 100V DC series motor runs at 1500 rpm with a line current of 20A. The armature resistance is 0.3W and the field resistance is 0.2W. Find the power developed.

• A.

1800 W

• B.

1880 W

• C.

1920 W

• D.

2000 W

A. 1800 W
Explanation
The power developed in a DC series motor can be calculated using the formula P = V * I, where P is the power, V is the voltage, and I is the current. In this case, the voltage is given as 100V and the current is given as 20A. Therefore, the power developed can be calculated as 100V * 20A = 2000W. However, this is the total power developed in the motor. To find the power developed at the load, we need to subtract the power lost in the armature resistance. The power lost in the armature resistance can be calculated using the formula P_loss = I^2 * R, where P_loss is the power lost, I is the current, and R is the resistance. In this case, the current is 20A and the resistance is 0.3W. Therefore, the power lost in the armature resistance can be calculated as 20A^2 * 0.3W = 120W. Subtracting this from the total power developed, we get 2000W - 120W = 1880W. Therefore, the correct answer is 1880W.

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• 9.

### An induction motor is running at a slip of 0.1. The input power is 2000W. The stator copper loss and core loss is 200W. The rotational loss is 50W. What is the rotor copper loss?

• A.

1750W

• B.

180W

• C.

175W

• D.

200W

B. 180W
Explanation
The rotor copper loss can be calculated by subtracting the sum of stator copper loss, core loss, and rotational loss from the input power. Given that the stator copper loss and core loss is 200W, and the rotational loss is 50W, the sum of these losses is 250W. Subtracting this from the input power of 2000W gives us the rotor copper loss of 1750W. Therefore, the correct answer is 1750W.

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