MCQOPTIONS
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MCQOPTIONS
This section includes 10 Mcqs, each offering curated multiple-choice questions to sharpen your Electronic Devices Circuits knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
There is no collector resistance in a simplified CC amplifier configuration. |
| A. | True |
| B. | False |
| Answer» B. False | |
| 2. |
What is the approximate value of the voltage gain of a simplified CC amplifier configuration? |
| A. | 1 |
| B. | β |
| C. | ∞ |
| D. | 0 |
| Answer» B. β | |
| 3. |
What is the voltage gain of a simplified CC configuration amplifier if RE = 25kΩ and re‘ = 12kΩ? |
| A. | 0.68 |
| B. | 0.59 |
| C. | 0.48 |
| D. | 0.086 |
| Answer» B. 0.59 | |
| 4. |
How do we compute the output impedance of a simplified CC amplifier? |
| A. | RE || (re‘ – (R1 || R2) / (β + 1)) |
| B. | RE || (re‘ – (R1 || R2) / (β – 1)) |
| C. | RE || (re‘ + (R1 || R2) / (β + 1)) |
| D. | RE || (re‘ + (R1 || R2) / (β – 1)) |
| Answer» D. RE || (re‘ + (R1 || R2) / (β – 1)) | |
| 5. |
How do we compute the input impedance of a simplified CC amplifier? |
| A. | β × (Re – re‘) |
| B. | β × (Re + re‘) |
| C. | RB || β × (Re – re‘) |
| D. | RB || β × (Re + re‘) |
| Answer» E. | |
| 6. |
How do we compute the base impedance of a simplified CC amplifier? |
| A. | β × (Re – re‘) |
| B. | β × (Re + re‘) |
| C. | RB || β × (Re – re‘) |
| D. | RB || β × (Re + re‘) |
| Answer» C. RB || β × (Re – re‘) | |
| 7. |
What is the value of voltage VB of a CC amplifier configuration if VCC = 12V, R1 = 5kΩ and R2 = 8kΩ? |
| A. | 6.54V |
| B. | 0.7V |
| C. | 4.6V |
| D. | 7.38V |
| Answer» E. | |
| 8. |
How do we compute the voltage gain of a simplified CC amplifier model using hybrid parameters? |
| A. | 1 + hie × Ri |
| B. | 1 – hie × Ri |
| C. | 1 + hie / Ri |
| D. | 1 – hie / Ri |
| Answer» E. | |
| 9. |
How do we compute the input resistance of a simplified CC amplifier model using hybrid parameters? |
| A. | hie – (1 + hfe) × RL |
| B. | hie + (1 + hfe) × RL |
| C. | hie – (1 + hfe) / RL |
| D. | hie + (1 + hf |
| E. | × RLb) hie + (1 + hfe) × RLc) hie – (1 + hfe) / RLd) hie + (1 + hfe) / RL |
| Answer» C. hie – (1 + hfe) / RL | |
| 10. |
How do we computethe current gain of a simplified CC amplifier model using hybrid parameters? |
| A. | 1 + hoe |
| B. | 1 – hoe |
| C. | 1 + hfe |
| D. | 1 – hfe |
| Answer» D. 1 – hfe | |