Transformers and Induction Machines Exam Prep
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Transformers and Induction Machines Exam Prep
English 28 leveled MCQs Free
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10 marks Long answer
With neat circuit diagram, equations and waveform, Derive an expression for Emf induced in a single phase transformer.
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The EMF induced is given by E = 4.44 f φm N.
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10 marks Long answer
Determine the regulation while supplying full load on 0.8 leading PF and the secondary terminal voltage on full load & 0.8 leading PF of transformer 200/125, 10 KVA, it has primary resistance is 0.6 ohms and reactance of 0.8 ohms. The secondary resistance is 0.07 ohms & reactance of 0.3 ohms.
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Regulation can be calculated using the formula: Regulation = (V_no_load - V_full_load) / V_full_load * 100.
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10 marks Long answer
Explain the characteristics of ideal transformers. Briefly write a short note on losses in a single phase transformer.
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Ideal transformers have no losses, perfect coupling, and maintain the same frequency.
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10 marks Long answer
A 100KVA transformer has its maximum efficiency of 0.98 at full load with UPF. During the day it is loaded as follows: For 12hours → 20KW at 0.5 PF lag, For 6hours → 45KW at 0.9 PF lag, For 6hours → 80KW at 0.8 PF lag. Determine its all-day efficiency.
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All-day efficiency can be calculated by considering the total energy output and input over the day.
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10 marks Long answer
With a neat circuit diagram explain auto transformers. Mention the advantages and disadvantages of auto transformers.
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Auto transformers use a single winding for both primary and secondary, offering reduced size and cost but lack electrical isolation.
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10 marks Long answer
A 3 phase step down transformer is connected to 6600V mains and takes a 10A. Calculate the secondary line voltage, line current and output for the following connections star-delta ii) delta-star.
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Use transformation ratios to determine secondary voltage and current based on connection type.
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10 marks Long answer
For a 6-pole 3-phase, 50Hz Induction motor ratio of stator to rotor turns is 3. On a certain load, its speed is observed to be 995rpm, when connected to 415V supply calculate, (i) Frequency of rotor EMF in running condition. (ii) Magnitude of induced EMF in the rotor at standstill.
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Rotor frequency can be calculated as fr = (Ns - N) * P / 120.
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10 marks Long answer
Suggest and explain with neat diagram the type of Induction Motor which has low starting torque.
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Squirrel cage induction motors typically have low starting torque.
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10 marks Long answer
Explain the need of starters. Explain with a neat circuit diagram Direct On Line starter(DOL).
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Starters are necessary to limit inrush current and DOL starters connect the motor directly to the supply.
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10 marks Long answer
Draw the circle diagram for a 10HP, 50HZ, 3-phase delta connected induction motor with the following data. Determine the starting current and full load torque from the circle diagram.
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Circle diagrams are used to analyze motor performance and determine operational parameters.
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10 marks Long answer
Briefly explain the steps to conduct no load and blocked rotor test in induction motor.
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No load test measures core losses while blocked rotor test measures copper losses.
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10 marks Long answer
Draw the circle diagram for a 10HP, 50HZ, 3-phase star connected induction motor with the following data. Determine the line current and efficiency for full load condition from the circle diagram.
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Circle diagrams help in calculating line current and efficiency based on operational data.
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10 marks Long answer
With a neat sketch explain double revolving theory and principle of operation of single phase induction motor.
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Double revolving theory explains the operation of single-phase motors by considering two rotating fields.
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10 marks Long answer
Suggest a suitable type of induction motor which is used in refrigerators and air conditioners, explain with a neat diagram. Justify.
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Split-phase induction motors are commonly used in refrigerators and air conditioners for their starting torque.
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6 marks Long answer
Describe the behavior of an ideal transformer on no load.
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On no load, primary current is magnetizing current, which is small and lags the supply voltage by 90°.
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5 marks Long answer
Draw the phasor diagram of a practical transformer on no load and explain.
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The phasor diagram shows the relationship between the primary voltage, no load current, and magnetising current.
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5 marks Long answer
Explain the concept of load component in a transformer.
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The load component is the additional current drawn by the primary due to load, which is in antiphase with secondary current.
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4 marks Medium answer
Explain the efficiency of a transformer.
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Efficiency η = Pout/(Pout + Pi + Pcu), expressed as a percentage.
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4 marks Medium answer
List the features of an ideal transformer.
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An ideal transformer has no losses, zero resistance windings, zero leakage flux, and high core permeability.
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4 marks Medium answer
The no load current of a transformer is 10 A at a power factor of 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate: c) Maximum value of flux in the core. Assume primary winding turns as 500.
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φm = 3.6036 mWb
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3 marks Medium answer
What is the relationship between copper losses and kVA rating?
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Copper losses are proportional to the square of the kVA rating.
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3 marks Medium answer
What condition must be met for maximum efficiency in a transformer?
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For maximum efficiency, variable losses (Pcu) must equal constant losses (Pi).
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3 marks Medium answer
The no load current of a transformer is 10 A at a power factor of 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate: a) Magnetising component of no load current.
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Im = 9.6824 A
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3 marks Medium answer
The no load current of a transformer is 10 A at a power factor of 0.25 lagging, when connected to 400 V, 50 Hz supply. Calculate: b) Iron loss.
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W0 = 1000 W
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2 marks Short answer
Define core or iron losses in a transformer.
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Core losses are constant losses in a transformer due to hysteresis and eddy currents.
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2 marks Short answer
What are copper losses in a transformer?
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Copper losses are power wasted as I^2R losses in the transformer windings.
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1 marks Short answer
State the formula for hysteresis loss.
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Hysteresis loss = Kh Bm^1.6 f v watts.
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1 marks Short answer
State the formula for eddy current loss.
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Eddy current loss = Ke Bm^2 f^2 t^2 watts/unit volume.
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