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- Question
A 12 kΩ resistor is in series with a 0.02 µF capacitor across a 1.2 kHz ac source. If the current is expressed in polar form as I = 0.3∠0° mA, what is the source voltage expressed in polar form?
Options- A. 45.6 V
- B. 411.3 V
- C. 0.411 V
- D. 4.11 V
- Correct Answer
- 4.11 V
- 1. If a load is purely inductive and the reactive power is 12 VAR, the apparent power is
Options- A. 0 VA
- B. 12 VA
- C. 6 VA
- D. 24 VA Discuss
- 2. When these numbers are multiplied, (6 × 103) (5 × 105), the result is
Options- A. 3 × 108
- B. 30 × 108
- C. 300 × 109
- D. 3,000 × 107 Discuss
- 3. Referring to Problem 21, the filter's bandwidth is
Options- A. 11.5 Hz
- B. 530 Hz
- C. 53 Hz
- D. 106 Hz Discuss
- 4. If you used 400 W of power for 30 h, you have used
Options- A. 1.3 kWh
- B. 13.3 kWh
- C. 1.2 kWh
- D. 12 kWh Discuss
- 5. A dc component is the peak value of a pulse waveform.
Options- A. True
- B. False Discuss
- 6. A 120 Ω load is connected across an ideal voltage source with VS = 12 V. The voltage across the load
Options- A. is 0 V
- B. is 12 V
- C. is 120 V
- D. cannot be determined Discuss
- 7. The number of microamperes in 2 milliamperes is
Options- A. 2 µA
- B. 20 µA
- C. 200 µA
- D. 2,000 µA Discuss
- 8. The total resistance of eight 5.6 kΩ resistors in series is
Options- A. 8 Ω
- B. 5,600 Ω
- C. 4,480 Ω
- D. 44,800 Ω Discuss
- 9. A capacitor of 0.02 µF is larger than
Options- A. 0.000020 F
- B. 200,000 pF
- C. 2,000 pF
- D. all of the above Discuss
- 10. Four resistors are connected in parallel. Fifteen mA flows through resistor R. If the second resistor is 2R, the third resistor 3R, and the fourth resistor 4R, the total current in the circuit is
Options- A. 60 mA
- B. 15 mA
- C. 135 mA
- D. 31.25 mA Discuss
More questions
Correct Answer: 12 VA
Correct Answer: 30 × 108
Correct Answer: 53 Hz
Correct Answer: 12 kWh
Correct Answer: False
Correct Answer: is 12 V
Correct Answer: 2,000 µA
Correct Answer: 44,800 Ω
Correct Answer: 2,000 pF
Correct Answer: 31.25 mA
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