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This section includes 227 Mcqs, each offering curated multiple-choice questions to sharpen your Electrical Engineering knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Calculate the output voltage of the Buck converter if the supply voltage is 789 V and duty cycle value is .9. |
| A. | 711.1 V |
| B. | 710.1 V |
| C. | 722.2 V |
| D. | 713.2 V |
| Answer» C. 722.2 V | |
| 52. |
Calculate the output voltage of the Boost converter if the supply voltage is 156 V and duty cycle value is .4. |
| A. | 260 V |
| B. | 264 V |
| C. | 261 V |
| D. | 268 V |
| Answer» B. 264 V | |
| 53. |
In a single phase full wave regulator, the firing angles in the positive and negative half cycles are generally |
| A. | Equal |
| B. | Different |
| C. | equal or different |
| D. | different but sometimes equal |
| Answer» B. Different | |
| 54. |
A single phase half wave converter is used to charge a battery of 50 V. The thyristor is continuously fired by dc signal. Input voltage is v = Vm sin t. The thyristor will conduct |
| A. | from 0 to 180 |
| B. | when v > 50 V |
| C. | from 0 to 90 |
| D. | When v < 50 V |
| Answer» C. from 0 to 90 | |
| 55. |
Thyristors connected in series need static and dynamic stabilizing circuits. |
| A. | True |
| B. | False |
| C. | none |
| D. | all |
| Answer» B. False | |
| 56. |
In a 3 phase full converter, the instantaneous output voltage can have negative part if |
| A. | firing angle is less than 60 |
| B. | firing angle is less than 90 |
| C. | firing angle is more than 60 |
| D. | firing angle is more than 90 |
| Answer» D. firing angle is more than 90 | |
| 57. |
In a 3 phase full converter the average load current is 150 A. The average thyristor current is |
| A. | 150 A |
| B. | 75A |
| C. | 50 A |
| D. | 25 A |
| Answer» D. 25 A | |
| 58. |
When thyristors are connected in series and parallel, it may be necessary to have : |
| A. | current derating |
| B. | voltage derating |
| C. | both current and voltage derating |
| D. | none of the above |
| Answer» D. none of the above | |
| 59. |
Assertion (A): A TRIAC is a bidirectional SCR. Reason (R): A TRIAC is a four layers 3 terminal device and can conduct in both directions. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» B. Both A and R correct but R is not correct explanation of A | |
| 60. |
For dynamic equalizing circuit used for series connected SCRs, the choice of C is based on: |
| A. | Reverse recovery characteristics |
| B. | Turn-on characteristics |
| C. | Turn-off characteristics |
| D. | Rise time characteristics |
| Answer» B. Turn-on characteristics | |
| 61. |
A combination of synchronized leading edge and trailing edge modulation has also been used to control a____________? |
| A. | Boost single – phase power factor converter |
| B. | A buck dc – dc converter to reduce ripple in the intermediate dc bus capacitor |
| C. | Both A. and B. |
| D. | None f these |
| Answer» D. None f these | |
| 62. |
A carrier based PWM technique in CSI is composed of____________? |
| A. | A switching pulse generator and a shorting pulse generator |
| B. | A shorting pulse distributor |
| C. | A switching and shorting pulse combination |
| D. | All of these |
| Answer» E. | |
| 63. |
A capacitive load in voltage source inverters generates ________________? |
| A. | Small current spikes and can be reduced by using an inductive filter |
| B. | Large current spikes and can be increased by using an inductive filter |
| C. | Small current spikes and can be increased by using an inductive filter |
| D. | Large current spikes and can be reduced by using an inductive filter |
| Answer» E. | |
| 64. |
60 thyrsistors are connected in series and parallel to form a 10 KV and 5.5 KA switch. Each thyristor is rated for 1.2 KV, 1 KA. The no. of parallel path are 6. The efficiency of the switch is______________________? |
| A. | 76.3 % |
| B. | 91.6 % |
| C. | 83.3 % |
| D. | 90.9 % |
| Answer» B. 91.6 % | |
| 65. |
4 thyristors rated 200 V in series. The operating voltage of the string is 600 V. Derating factor of the string is_____________________? |
| A. | 0.75 |
| B. | 0.7 |
| C. | 0.2 |
| D. | 0.25 |
| Answer» E. | |
| 66. |
As the firing of a thyristor is delayed, the output voltage of a controlled rectifier decreases. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 67. |
Half bridge and full bridge inverters of one type use different commutation methods. |
| A. | True |
| B. | False |
| Answer» C. | |
| 68. |
In 180° mode of operation of a 3 phase bridge inverter, two thyristors conduct at one time. |
| A. | True |
| B. | False |
| Answer» C. | |
| 69. |
Assertion (A): Thyristors can be used in controlled heating, excitation systems of alternators, speed control of motors and HVDCReason (R): A static var system using thyristors is very commonly used in high voltage ac systems. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» C. A is correct but R is wrong | |
| 70. |
Assertion (A): The circuit of figure thyristor will conduct for 180° during positive half cycle if it is continuously fired. Reason (R): A thyristor can conduct only when it is forward biased. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» E. | |
| 71. |
Assertion (A): A transistor has lower voltage drop during conduction as compared to thyristorReason (R): Transistors are manufactured in very high voltage and current ratings. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» D. A is wrong but R is correct | |
| 72. |
Assertion (A): A transistor requires a continuous base signal for conduction but a thyristor requires a gate pulse.Reason (R): Transistor find widespread application in power electronic circuits. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» D. A is wrong but R is correct | |
| 73. |
Assertion (A): SUS is a PUT and avalanche diode connected between gate and cathodeReason (R): SBS has two SUS inbuilt together and connected in anti-parallel |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» C. A is correct but R is wrong | |
| 74. |
Assertion (A): Typical value of thermal resistance from source to sink of an SCR is about 0.3° C/WReason (R): Heat sinks of thyristors are generally made of aluminium. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» C. A is correct but R is wrong | |
| 75. |
Assertion (A): A TRIAC is a bidirectional SCR.Reason (R): A TRIAC is a four layers 3 terminal device and can conduct in both directions. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» B. Both A and R correct but R is not correct explanation of A | |
| 76. |
Assertion (A): When an SCR is conducting, the voltage drop across is about 1 V. Reason (R): When an SCR is conducting, the junction temperature is about 200°C. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» D. A is wrong but R is correct | |
| 77. |
Assertion (A): Light triggering is very suitable for HVDC transmission.Reason (R): Light triggering has the advantage of complete electrical isolation of gate circuit. |
| A. | Both A and R are correct and R is correct explanation of A |
| B. | Both A and R correct but R is not correct explanation of A |
| C. | A is correct but R is wrong |
| D. | A is wrong but R is correct |
| Answer» B. Both A and R correct but R is not correct explanation of A | |
| 78. |
If peak value of input voltage is Vm and firing angle is a, the average output voltage for single phase half wave and single phase full wave regulators are |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 79. |
In figure, thyristor Th is in off state. When thyristor is turned on, the peak thyristor current can be |
| A. | 5 A |
| B. | 10 A |
| C. | 15 A |
| D. | 20 A |
| Answer» D. 20 A | |
| 80. |
A capacitor C farads is charged to a voltage V. It discharges through a thyristor, a diode and an inductance L. The peak current in the circuit is |
| A. | VCL |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 81. |
In the circuit of figure, the load current when thyristor is off |
| A. | is constant |
| B. | increases exponentially |
| C. | decreases exponentially |
| D. | zero |
| Answer» D. zero | |
| 82. |
In the circuit of figure, the waveshape of load current when thyristor conducting is |
| A. | constant |
| B. | increasing exponentially |
| C. | decreasing exponentially |
| D. | either (b) or (c) |
| Answer» C. decreasing exponentially | |
| 83. |
In a step down chopper using time ratio control, the input voltage is V and duty cycle is a. The load is purely resistive having resistance R. The average and rms load currents are |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 84. |
In a certain inverter circuit each thyristor carries 26.67 A for 60°, 13.33 A for 120° and zero A for 180° of the cycle. The rms current of thyristor is |
| A. | 20 A |
| B. | 13.33 A |
| C. | 6.67 A |
| D. | 3.33 A |
| Answer» C. 6.67 A | |
| 85. |
A 3 phase bridge inverter is fed by 400 V battery. The load is star connected and has a resistance of 10 Ω per phase. The peak value of thyristor current is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 86. |
A 3 phase bridge inverter is fed by 400 V battery. The load is star connected and has a resistance of 10 Ω per phase. The peak value of load current is |
| A. | 40 A |
| B. | 20 A |
| C. | 10 A |
| D. | 5 A |
| Answer» C. 10 A | |
| 87. |
In a 3 phase bridge inverter with 120° mode of operation the. number of thyristors conducting at one time are |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» C. 3 | |
| 88. |
In a 3 phase bridge inverter with 180° mode of operation the numbers of thyristors conducting at one time are |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| 89. |
In a full bridge inverter fed by a battery of voltage V, the rms value of fundamental component of output voltage is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 90. |
In a half bridge inverter circuit fed by a battery of voltage V, the rms value of fundamental component of output voltage is |
| A. | 2 V |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» C. [C]. | |
| 91. |
If 0.5 T is the time period of oscillations and Toff is the time between turn off of one thyristor and turn on of second thyristor, the frequency of output f of a series inverter is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 92. |
In a 3 phase full converter, the firing angle is less than 60°. The instantaneous output voltage |
| A. | will have positive part only |
| B. | will have both positive and negative parts |
| C. | may have both positive and negative parts |
| D. | will have negative part if load is inductive |
| Answer» B. will have both positive and negative parts | |
| 93. |
In a 3 phase full converter, the instantaneous output voltage can have negative part if |
| A. | firing angle is less than 60° |
| B. | firing angle is less than 90° |
| C. | firing angle is more than 60° |
| D. | firing angle is more than 90° |
| Answer» D. firing angle is more than 90° | |
| 94. |
A 3 phase fully controlled bridge converter is fed by a 3 phase system having phase voltage v = Vm sin ωt. The firing angle is a. The dc output voltage is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» B. [B]. | |
| 95. |
Two single phase semi-converters are connected in series to form a series converter. The input is v = Vm sin ωt and a1, a2 are the firing angles. If a1 = 0, the dc output voltage is |
| A. | (1 + cos a2) |
| B. | (2 + cos a2) |
| C. | (3 + cos a2) |
| D. | (4 + cos a2) |
| Answer» D. (4 + cos a2) | |
| 96. |
Two semiconverters are connected in series to form a single phase series converter. The input v = Vm sin ωt and a1 = a2 are the firing angles of the two converters. If a1 = a2 = 0, the dc output voltage is |
| A. | [A]. |
| B. | [B]. |
| C. | [C]. |
| D. | [D]. |
| Answer» D. [D]. | |
| 97. |
For a single phase half wave converter feeding a resistive load, the firing angle a. The waveshape in below figure is for |
| A. | output voltage |
| B. | input voltage |
| C. | voltage drop across thyristor |
| D. | none of the above |
| Answer» E. | |
| 98. |
A fully controlled bridge converter is feeding an RLE load. For some circuit conditions firing angle is 47°. If connections of battery are reversed, the firing angle is likely to be |
| A. | 47° |
| B. | less than 47° |
| C. | more than 47° |
| D. | either more or less than 47° |
| Answer» D. either more or less than 47° | |
| 99. |
A fully controlled bridge converter is feeding an RLE load. The load current is I0. The input voltage is Vm sin ωt. The KVL equation to find firing angle a is |
| A. | cos a = E + I0R |
| B. | cos a = E + I0R |
| C. | cos a = E - I0R |
| D. | cos a = E - I0R |
| Answer» B. cos a = E + I0R | |
| 100. |
In the below figure the rms voltage across each half of secondary is 200 V. The peak inverse voltage across each of thyristor is |
| A. | 200 V |
| B. | 2(200) V |
| C. | 2(400) V |
| D. | 800 V |
| Answer» D. 800 V | |