# Electric Drives Questions and Answers – Motors – Heating Effects

This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Motors – Heating Effects”.

1. Iron losses occur in the machine are variable losses.
a) True
b) False

Explanation: Iron losses consist of eddy current and hysteresis losses. These are constant losses. Iron losses are independent of the load current. The copper losses are variable losses.

2. Which of the following are the classes of insulating material in electrical machines?
a) Γ, A, E, F, H, C
b) Γ, E, B, F, H, C
c) A, E, B, F, H, C
d) Γ, A, E, F, C

Explanation: Depending upon the temperature limits, insulating materials are classified into seven classes Γ, A, E, B, F, H, C. The insulation has the lowest temperature limit.

3. Cooling air travels 10 m distance in 5 sec from point A to point B. Calculate the velocity of the cooling air.
a) 2 m/s
b) 4 m/s
c) 6 m/s
d) 10 m/s

Explanation: The velocity of the cooling air can be calculated using the ratio of displacement and time. The value of the displacement is 10 m. The value of the velocity is 10÷5=2 sec.

4. Calculate the heating time constant of the machine if the thermal capacity (c) of the machine is 24 watts/℃ and heat dissipation constant value (D) is 3 watts/℃.
a) 8 sec
b) 7 sec
c) 6 sec
d) 9 sec

Explanation: The heating time constant of the machine is the ratio of the thermal capacity of the machine and heat dissipation constant value. Τ = c÷D=24÷3 = 8 sec.

5. Inadequate rating of the equipment can ___________
a) Increase the speed
b) Decrease the speed
c) Damage the winding
d) Increase the efficiency

Explanation: The inadequate rating of the equipment can damage the winding and stops the machine operation. The windings can burn out.
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6. What is heating time constant of the machine?
a) The ratio of the thermal capacity and heat dissipation constant value
b) The ratio of the thermal capacity and heat dissipation constant value
c) The ratio of the thermal capacity and heat dissipation constant value
d) The ratio of the thermal capacity and heat dissipation constant value

Explanation: The heating time constant of the machine is the ratio of the thermal capacity of the machine and heat dissipation constant value. Τ = c÷D. It has a unit of second.

7. Insufficient rating of the machine can damage the windings.
a) True
b) False

Explanation: The inadequate rating of the equipment can damage the winding and halt the machine operation. The windings can burn out and lead to the breakdown of the machine.

8. A circuit consists of two 7 F capacitors connected in series and 12 H inductor. Determine the order of the circuit.
a) 1
b) 2
c) 0
d) 3

Explanation: The order of the circuit is the number of memory/storing elements which are non-separable present in the circuit. In mathematics, the order is defined as the highest order derivate in the differential equation. The order of the circuit is 2 because two 7 F capacitors combine to form a single capacitor.

9. Calculate the steady state temperature value for T(t)=1.5(1-e-17t).
a) 1.5
b) 4.8
c) 3.9
d) 2.4

Explanation: The steady state temperature value is obtained at t=∞. The value of T(t) at t=∞ is 1.5(1-e-∞)=1.5(1-0)=1.5. The term e-17t is an exponentially decaying function.

10. Calculate the quality factor for the R-C circuit connected in the cooling system if R=1 Ω and C=10 F.
a) 0.1
b) 0.2
c) 0.3
d) 0.4

Explanation: The quality factor is defined as the ratio of the reactive power to the active power consumed. The resistor always absorbs active power and capacitor absorbs the reactive power. Quality factor=1÷RC=1÷10=.1.

11. Calculate the heat absorbed by a 3 Ω resistor when 6 V is applied across it.
a) 12 W
b) 10 W
c) 8 W
d) 11 W

Explanation: The resistor always absorbs active power and dissipates in the form of the heat. Power absorbed by a 3 Ω resistor is 6×6÷3=12 W.

12. What is the unit of power?
a) Joule/sec
b) Joule/sec2
c) Joule/sec3
d) Joule/sec4

Explanation: The power is defined as the rate at which work is done. The ratio of energy and time is power. It is expressed in Watt or Joule/sec.

13. Convert 20 ℃ into Fahrenheit.
a) 68 ℉
b) 58 ℉
c) 78 ℉
d) 48 ℉

Explanation: Temperature can be expressed in Celsius and Fahrenheit. The Standard unit of the temperature is the Kelvin(K). (20℃ × 9/5) + 32 = 68℉.

14. Convert 50℉ into Celsius.
a) 20 ℃
b) 10 ℃
c) 30 ℃
d) 40 ℃

Explanation: Temperature can be expressed in Celsius and Fahrenheit. The basic unit of temperature in SI is the Kelvin. The formula used is (50℉ − 32) × 5/9 = 10℃.

Sanfoundry Global Education & Learning Series – Electric Drives.

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