Due to constant speed operation, DC shunt motors are used for driving line shaft, machine tools, lathe, woodworking machines, milling machines, weaving machines, small printing press, etc.

2. As the load is increased the speed of the DC shunt motor will ___________

Reduce slightly

Increase slightly

Increase proportionately

Reduce rapidly

Answer:1. Reduce slightly

Explanation:

As the load increases, to maintain the voltage at a constant value, current decreases, due to which the strength of the magnetic field produced by the Internal winding of the DC motor decreases. This results in a decreased interaction of the two magnetic fields and thus a decrease in the speed of the motor. The change is so negligible that in many cases it is assumed that the speed of the DC motor remains constant. Hence, characteristic is also called shunt characteristic.

3. A 15 kW, 230 V dc shunt motor has armature circuit, resistance of 0.4 Ω and field circuit resistance of 230 Ω. At no load and rated voltage, the motor runs at 1440 rpm and the line current drawn by the motor is 5 A. At full load, the motor draws a line current of 70 A. Neglect armature reaction. The full load speed of the motor in rpm

1000

1080

1276

1340

Answer:3. 1276

Explanation:

The circuit for the given question is

Current through field resistance is

I_{f} = V/R_{f} = 230/230 = 1 A

Hence, we have the armature current

Ia = I − I_{f} = 4 − 1 = 3 A

From the given circuit, we have

V = E_{f1} + I_{a}R_{s} = 230 = E_{f1} + 4 × 0.4

E_{f1} = 202.4 V

As emf ∝ flux × speed

for DC shunt motor flux is constant. So,

E_{f} ∝ ω or E_{f1}/E_{f2} = ω_{1}/ω_{2}

228.4/202.4 = 1440/ω_{2}

ω_{2} = 1276 RPM

4. A 220V 15KW 1000rpm shunt motor with armature resistance of 0.25 has a rated line current of 68 A and a rated field current of 2.2A. The change in field flux required to obtain a speed of 1600 rpm while drawing a line current of 52.8 A and a field current of 1.8A is

18.18% increase

18.18% decrease

36.36% increase

36.36% decrease

Answer: 4. 36.36% decrease

Explanation:

Given Data

Supply Voltage V = 220 V Speed N_{1} = 1000 RPM Armature Resistance R_{a} = 0.25 Ω Line Current I_{L} = 68A & 52.8A Field current I_{F}= 2.2 A & 1.8 A Speed N_{2} = 1600 RPM

The Line current of DC shunt motor is the sum of Armature current and Shunt field current

I_{L} = I_{a} + I_{sh}

∴ I_{a} = I_{L} − I_{sh}

I_{a1} = 68 − 2.2 = 65.8 A

From the voltage equation, the Back EMF will be

E_{b1} = V − I_{a1}R_{a}

= 220 × 65.8 × 0.25

E_{b1} = 203.55

Similarly for the line current of 52.8 A and a field current of 1.8 A the armature current will be

I_{a2} = 52.8 − 1.8 = 51 A

Hence E_{b2} = V − I_{a2}R_{a}

= 220 × 51 × 0.25

E_{b2} = 207.25

Let the on load speed be N. As we know that back EMF of DC motor is directly proportional to the flux and speed.

5. In a DC shunt motor, speed is related to armature current as _____________

Directly proportional to the armature current

Proportional to the square of the current

Independent of armature current

Inversely proportional to the armature current

Answer: 4. Inversely proportional to the armature current

Explanation:

When an external load is applied to the shunt motor it tends to slow down slightly. The slight decrease in speed causes a corresponding decrease in counter emf. If the armature resistance is low, the resulting increase in armature current and torque will be relatively large. Therefore, the torque is increased until it matches the resisting torque of the load. The speed of the motor will then remain constant at the new value as long as the load is constant.

Conversely, if the load on the shunt motor is reduced, the motor tends to speed up slightly, The increased speed causes a corresponding increase in counter emf and a relatively large decrease in armature current and torque.

So when the speed of the shunt motor increases the armature current reduces and when the speed of the shunt motor decreases the armature current Increase. Hence the speed of the shunt motor is inversely proportional to the armature current.

6. The armature torque of the DC shunt motor is proportional to

Field flux only

Armature current only

Field flux and armature current

Field current

Answer: 2. Armature current only

Explanation:

When an external load is applied to the shunt motor it tends to slow down slightly. The slight decrease in speed causes a corresponding decrease in counter emf.

If the armature resistance is low, the resulting increase in armature current and torque will be relatively large. Therefore, the torque is increased until it matches the resisting torque of the load. The speed of the motor will then remain constant at the new value as long as the load is constant.

Conversely, if the load on the shunt motor is reduced, the motor tends to speed up slightly, The increased speed causes a corresponding increase in counter emf and a relatively large decrease in armature current and torque.

So when the armature torque of the shunt motor increases the armature current Increase and when the armature torque of the shunt motor decreases the armature current reduces. Hence the speed of the shunt motor is directly proportional to the armature current.

7. The speed of ______ motor is practically constant

Shunt motor

Series motor

Cumulatively compounded

Differentially compounded

Answer: 1. Shunt motor

Explanation:

In case of the DC shunt motor, flux is practically constant as the field is parallel to the armature.

If the load is increased, then the speed of the motor will remain almost constant because the field current remains almost constant.

So that it is also known as a constant speed motor.

Hence, the highest speed attained by the DC shunt motor at rated flux is equal to no-load speed.

DC shunt motor is used to drive the constant speed line shafting, lathes, blower, fan, etc.

8. A DC shunt motor of 200 V, 10.5 A, 2000 rpm has an armature resistance of 0.5 Ω and field resistance of 400 Ω. It drives a load whose torque is constant at rated motor torque. What is the value of armature current if the source voltage drops to 175 V?

9.7 A

12.4 A

11.4 A

10.7 A

Answer: 3. 11.4 A

Explanation:

Given V_{1} = 200 V

Armature resistance (R = 0.5 Ω

Field winding resistance (R_{sh}) = 400 Ω

Field current = 200/400 = 0.5 A

Load current = 10.5 A

Armature current = 10.5 – 0.5 = 10 A

Given that load torque is constant.

I_{a2} = 200 × 10/175 = 11.4 A

9. In a DC shunt motor for zero armature current we get speed ______

Non-zero and minimum

Not Practically applicable

Non-zero and maximum

Doesn’t depend on armature current

Answer: 2. Not Practically applicable

Explanation:

DC shunt motor characteristic does not have a point of zero armature current because a small current (no-load current) is necessary to maintain rotation of the motor at no-load.

10. If a DC shunt motor is working at full load and if shunt field circuit suddenly opens ________

Will make armature to take heavy current, possibly burning it

Will result in excessive speed, possibly destroying armature due to excessive centrifugal stresses

Nothing will happen to motor

Motor will come to stop

Answer: 1. Will make armature to take heavy current, possibly burning it

Explanation:

If the field circuit is opened accidentally, the field flux will suddenly decrease to its relatively small residual value. If the armature circuit is not opened immediately, the motor speed will increase to dangerously high values and the motor will destroy itself in a few seconds either by the windings being forced from the slots or the commutator segments being thrown out by cen- centrifugal force.

Since the sudden decrease in the field flux reduces the counter voltage to a very small amount, the armature current of the motor will increase to a very high value. This will take place before the motor starts to rotate at a high speed.

The field circuit of a shunt motor must never be opened if the motor is running. Otherwise, the motor will “run away” and will destroy itself in a few seconds!