

MCQOPTIONS
Saved Bookmarks
This section includes 18 Mcqs, each offering curated multiple-choice questions to sharpen your Power Electronics knowledge and support exam preparation. Choose a topic below to get started.
1. |
Determine the distortion factor (μ) for a full bridge inverter with supply Vs = 60 V. |
A. | 0.8 |
B. | 0.7 |
C. | 0.9 |
D. | 1 |
Answer» D. 1 | |
2. |
A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the fundamental frequency output power. |
A. | 1200 W |
B. | 856 W |
C. | 972 W |
D. | 760 W |
Answer» D. 760 W | |
3. |
A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the rms output voltage and the peak reverse blocking voltage of each transistor. |
A. | 30 V, 60 V |
B. | 30 V, 30 V |
C. | 60 V, 60 V |
D. | 60 V, 30 V |
Answer» D. 60 V, 30 V | |
4. |
Find the distortion factor (μ), for a single phase half wave bridge inverter with dc source Vs = 1 kV. |
A. | 0.87 |
B. | 1 |
C. | 0.9 |
D. | 0.7 |
Answer» D. 0.7 | |
5. |
A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the average current through each SCR inverter switch. |
A. | 11.5 A |
B. | 5.75 A |
C. | 23 A |
D. | none of the mentioned |
Answer» C. 23 A | |
6. |
A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the fundamental power delivered to the load. |
A. | 1.07 W |
B. | 107.2 W |
C. | 1.07 kW |
D. | 107.2 kW |
Answer» D. 107.2 kW | |
7. |
The distortion factor (μ) is the ratio of |
A. | total rms output voltage to fundamental rms output voltage |
B. | fundamental rms output voltage to fundamental average output voltage |
C. | total rms output voltage to rms value of all the harmonic components |
D. | fundamental rms output voltage to total rms output voltage |
Answer» E. | |
8. |
If Vr is the rms value of the inverter output voltage and V1 is the rms value of the fundamental component, then the total harmonic distortion (THD) is given by |
A. | Vr/V1 |
B. | (Vr + V1r |
C. | [ (Vr/V1)2 – 1 ]1/2 |
D. | [ (Vr/V1)1/2 + 1 ]2 |
Answer» D. [ (Vr/V1)1/2 + 1 ]2 | |
9. |
A_SINGLE_PHASE_FULL_BRIDGE_INVERTER_USING_TRANSISTORS_AND_DIODES_IS_FEEDING_A_R_LOAD_OF_3_‚ÄÖ√Ñ√∂‚ÀÖ√´¬¨‚ÀÇ_WITH_THE_DC_INPUT_VOLTAGE_OF_60_V._FIND_THE_FUNDAMENTAL_FREQUENCY_OUTPUT_POWER.?$# |
A. | 1200 W |
B. | 856 W |
C. | 972 W |
D. | 760 W |
Answer» D. 760 W | |
10. |
Determine_the_distortion_factor_(μ)_for_a_full_bridge_inverter_with_supply_Vs_=_60_V.$# |
A. | 0.8 |
B. | 0.7 |
C. | 0.9 |
D. | 1 |
Answer» D. 1 | |
11. |
A single phase full bridge inverter using transistors and diodes is feeding a R load of 3 Ω with the dc input voltage of 60 V. Find the rms output voltage and the peak reverse blocking voltage of each transistor?# |
A. | 30 V, 60 V |
B. | 30 V, 30 V |
C. | 60 V, 60 V |
D. | 60 V, 30 V |
Answer» D. 60 V, 30 V | |
12. |
What would be the harmonic factor of lowest order harmonic in case of a half wave bridge inverter? |
A. | 1/1 |
B. | 1/3 |
C. | 1/2 |
D. | Insufficient data |
Answer» C. 1/2 | |
13. |
A single phase inverter gives rms value of output voltage as 115 V and the fundamental output voltage of as 103.5 V. Find the THD (Total Harmonic Distortion). |
A. | 0.4 % |
B. | 40.8 % |
C. | 48.3 % |
D. | 4.83 % |
Answer» D. 4.83 % | |
14. |
Find the distortion factor (μ), for a single phase half wave bridge inverter with dc source Vs = 1 kV.$ |
A. | 0.87 |
B. | 1 |
C. | 0.9 |
D. | 0.7 |
Answer» D. 0.7 | |
15. |
A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the average current through each SCR inverter switch.$ |
A. | 11.5 A |
B. | 5.75 A |
C. | 23 A |
D. | none of the mentioned |
Answer» C. 23 A | |
16. |
A single-phase half bridge inverter is connected to a 230 V dc source which is feeding a R load of 10 Ω. Determine the fundamental power delivered to the load.$ |
A. | 1.07 W |
B. | 107.2 W |
C. | 1.07 kW |
D. | 107.2 kW |
Answer» D. 107.2 kW | |
17. |
The distortion factor (μ) is the ratio of$ |
A. | total rms output voltage to fundamental rms output voltage |
B. | fundamental rms output voltage to fundamental average output voltage |
C. | total rms output voltage to rms value of all the harmonic components |
D. | fundamental rms output voltage to total rms output voltage |
Answer» E. | |
18. |
If Vr is the rms value of the inverter output voltage and V1 is the rms value of the fundamental component, then the total harmonic distortion (THD) is given by |
A. | V<sub>r</sub>/V<sub>1</sub> |
B. | (V<sub>r</sub> + V<sub>1r</sub> |
C. | [ (V<sub>r</sub>/V<sub>1</sub>)<sup>2</sup> – 1 ]<sup>1/2</sup> |
D. | [ (V<sub>r</sub>/V<sub>1</sub>)<sup>1/2</sup> + 1 ]<sup>2</sup> |
Answer» D. [ (V<sub>r</sub>/V<sub>1</sub>)<sup>1/2</sup> + 1 ]<sup>2</sup> | |