DC Machines Multiple Choice Questions on “Electromagnetic Power and Circuit Models”.
1. Product of torque and mechanical angular velocity ω is_____
a) Ea/ω
b) Ea*Ia
c) ω/Ea
d) Can’t tell
Answer: b
Clarification: According to statement of energy conversion, electrical and mechanical power of the machine must balance in a machine. Ea*Ia is referred to as electromagnetic power. Thus, torque = (Electromagnetic power)/ ω.
2. Condition for linear magnetization is______
a) φ α If
b) φ α Ia
c) φ α 1/If
d) φ α 1/Ia
Answer: a
Clarification: For a linear magnestization to take place, flux produced by the current flowing through particaluar flux producing coil must vay in direct proportion. Here, φ is produced due to field winding which carried current If.
3. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. Here, motor is drawing 25 A current at 1500 rpm.
What is the available torque (N-m) at shaft?
a) 29.9
b) 59.8
c) 14.95
d) 44.85
Answer: a
Clarification: Gross mechanical power developed is equal to Ea*Ia. Ea*Ia is referred to as electromagnetic power. Thus, torque = (Electromagnetic power)/ ω. Substituting all the values we get torque = 29.9 Nm.
4.When Ea>Vt machine is said to be operating in which of the following mode?
a) Depends on the Shaft speed
b) Generating
c) Motoring
d) Inducing
Answer: b
Clarification: The machine operates in generating mode (puts out electrical power) when Ia is in the direction of induced emf Ea. For the given armature circuit Vt = armature terminal voltage= Ea – Ia * Ra; which implies Vt < Ea.
5. When Ea
b) Generating
c) Motoring
d) Inducing
Answer: c
Clarification: The machine operates in motoring mode (puts in electrical power) when Ia is in the direction opposite of induced emf Ea. For the given armature circuit Vt = armature terminal voltage= Ea + Ia * Ra; which implies Vt> Ea.
6. Simple equation of DC machine operating in generating mode, with non-zero Ra value is________
a) Vt = Ea – Ia/Ra
b) Vt = Ea + Ia*Ra
c) Vt = – Ea + Ia*Ra
d) Vt = Ea – Ia*Ra
Answer: d
Clarification: The machine operates in generating mode (puts out electrical power) when Ia is in the direction of induced emf Ea, also Ea>Vt. For the given armature circuit, with non-zero value of Ra, Vt= armature terminal voltage= Ea- Ia*Ra.
7. Simple equation of DC machine operating in motoring mode, with non-zero Ra value is________ 8. When torque of the electromagnetic origin is in the opposite direction of rotation of armature, machine is said to be operating in which of the following mode?
a) Vt = Ea – Ia/Ra
b) Vt = Ea + Ia*Ra
c) Vt = – Ea + Ia*Ra
d) Vt = Ea – Ia*Ra
Answer: b
Clarification: The machine operates in motoring mode (puts in electrical power) when Ia is in the direction opposite of induced emf Ea, also Ea
a) Depends on other parameters
b) Generating
c) Motoring
d) Inducing
Answer: b
Clarification: In a generating mode, torque of the electromagnetic origin is in the opposite direction of rotation of armature, implying mechanical power is absorbed and a prime mover is needed to run the machine. Ea and Ia are in the same directions.
9. When torque of the electromagnetic origin is in the direction of rotation of armature, machine is said to be operating in which of the following mode?
a) Depends on other parameters
b) Generating
c) Motoring
d) Inducing
Answer: c
Clarification: In a motoring mode, torque of the electromagnetic origin is in the direction of rotation of armature, implying electrical power is absorbed and a prime mover is not needed to run the machine. Ea and Ia are in the opposite directions.
10. What is the armature current for a DC motor if terminal voltage is 255V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?
a) 10A
b) 100A
c) 1KA
d) 1A
Answer: b
Clarification: Open circuit voltage means the voltage at armature current equal to 0, i.e. Vt=Ea= 250V. Actual Vt at loaded condition is 255V. For motoring mode, Vt – Ea = Ia*Ra. By substituting all the given values, we get Ia= 100A.
11. For a DC generator feeding 100kW power into 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____
a) 436.8 A
b) 434.8 a
c) 432.8 A
d) Data insufficient
Answer: a
Clarification: Field current is equal to 230/115 = 2A, by Ohm’s law. When running as a generator line current IL= 100k/230= 434.8A. Since power is supplied to 230V mains, Ia=Il+If = 434.8+2 =436.8A.
For motoring mode Ia= IL- If.
12. What is the armature current for a DC generator if terminal voltage is observed to be 245V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?
a) 10A
b) 100A
c) 1KA
d) 1A
Answer: b
Clarification: Open circuit voltage means the voltage at armature current equal to 0, i.e. Vt=Ea= 250V. Actual Vt at loaded condition is 245V. For motoring mode, Ea – Vt= Ia*Ra. By substituting all the given values, we get Ia= 100A.
13. For a DC motor taking 10kW power from 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____
a) 45.68 A
b) 43.48 a
c) 41.48 A
d) Data insufficient
Answer: c
Clarification: Field current is equal to 230/115 = 2A, by Ohm’s law. When running as a motor line current IL= 10k/230= 43.48A. Since power is taken from 230V mains, Ia=IL- If = 43.48-2 =41.48A. For generating mode Ia= IL+ If.