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51.	Calculate the PIV for the Mid-point configuration of Full-wave rectifier if the peak value of the supply voltage is 311.(a) 622 V(b) 620 V(c) 624 V(d) 626 VI have been asked this question in an internship interview.This intriguing question originated from Solid State Controlled Drives topic in chapter Introduction to Solid State Controlled Drives of Electric Drives
Answer» Correct choice is (a) 622 V Explanation: The peak INVERSE VOLTAGE for the Mid-point configuration of Full wave rectifier is 2Vm=2×311=622 V. The peak inverse is the MAXIMUM NEGATIVE voltage ACROSS the thyristor.

52.	In single phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (Vs)r.m.s=220 V, f=40 Hz, R=1 Ω, E=90 V, β=230°.(a) -328.33 V(b) -325.48 V(c) -254.85 V(d) -284.48 VThis question was addressed to me in unit test.This interesting question is from Solid State Controlled Drives topic in chapter Introduction to Solid State Controlled Drives of Electric Drives
Answer» RIGHT answer is (a) -328.33 V To explain I would SAY: In single phase RLE LOAD the voltage across the THYRISTOR when current decays to ZERO VT=Vmsin(β)-E=220×√2sin(230°)-90=-328.33 V.

53.	Calculate the displacement factor if the fundamental voltage is 24sin(140πt-240°) and fundamental current is 47sin(140πt-120°).(a) -0.5(b) -0.7(c) 0.9(d) 0.4This question was posed to me in unit test.Query is from Solid State Controlled Drives topic in section Introduction to Solid State Controlled Drives of Electric Drives
Answer» Correct option is (a) -0.5 The explanation is: The DISPLACEMENT factor is the COSINE of the ANGLE DIFFERENCE between the fundamental VOLTAGE and fundamental current. D.F=cos(120°)=-0.5.

54.	RLE load is also known as DC motor load.(a) True(b) FalseI have been asked this question during an internship interview.My doubt is from Solid State Controlled Drives in division Introduction to Solid State Controlled Drives of Electric Drives
Answer» Right choice is (a) True Explanation: RLE load is also known as DC MOTOR load because the ARMATURE circuit consists of BACK e.m.f, inductive COILS, and armature resistance.

55.	Calculate the conduction angle in purely inductive load if the firing angle is π÷2.(a) 205°(b) 175°(c) 180°(d) 195°The question was posed to me by my school principal while I was bunking the class.The doubt is from Solid State Controlled Drives in section Introduction to Solid State Controlled Drives of Electric Drives
Answer» The correct CHOICE is (C) 180° For explanation I would SAY: The conduction angle in the purely inductive load is β-α=2(π-∝)=2(180°-90°). The conduction angle is the angle for which the current exists in the circuit. The average VALUE of the voltage in a purely inductive load is zero.

56.	Calculate the extinction angle in purely inductive load if the firing angle is π÷4.(a) 315°(b) 145°(c) 345°(d) 285°The question was asked in an interview for job.This key question is from Solid State Controlled Drives topic in chapter Introduction to Solid State Controlled Drives of Electric Drives
Answer» Correct CHOICE is (a) 315° Explanation: The extinction ANGLE in the purely inductive load is 2π-∝=360°-45°=315°. The extinction angle is the angle at which CURRENT in the circuit becomes zero. The average value of the VOLTAGE in a purely inductive load is zero.

57.	In Half-wave controlled rectifier calculate the average value of the voltage if the supply is 10sin(50t) and firing angle is 30°.(a) 2.32 V(b) 2.97 V(c) 4.26 V(d) 5.64 VI had been asked this question during an online exam.My question is taken from Solid State Controlled Drives in chapter Introduction to Solid State Controlled Drives of Electric Drives
Answer» The CORRECT ANSWER is (B) 2.97 V To explain I would say: In Half-wave controlled rectifier, the average value of the VOLTAGE is VM(1+cos(∝))÷2π=10(1+cos(30°)÷6.28=2.97 V. The thyristor will conduct from ∝ to π.

58.	In Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 23sin(50t).(a) 7.32 V(b) 8.32 V(c) 9.32 V(d) 7.60 VI had been asked this question during an online exam.This is a very interesting question from Solid State Controlled Drives in portion Introduction to Solid State Controlled Drives of Electric Drives
Answer» RIGHT ANSWER is (a) 7.32 V To EXPLAIN: In HALF-wave uncontrolled rectifier, the average value of the voltage is Vm÷π=23÷π=7.32 V. The diode will CONDUCT only for the positive half cycle. The conduction period of the diode is π.

59.	RLE load is a voltage stiff load.(a) True(b) FalseThe question was posed to me by my school principal while I was bunking the class.This question is from Solid State Controlled Drives topic in portion Introduction to Solid State Controlled Drives of Electric Drives
Answer» Right choice is (B) False Easiest explanation: RLE LOAD is a current STIFF load because of the presence of the inductor. The load current does not suddenly with a change in voltage. Inductor OPPOSES the change of current.