Discussion
Home ‣ Electrical Engineering ‣ Energy and Power Comments
- Question
A 15 V source is connected across a 12 Ω resistor. How much energy is used in three minutes?
Options- A. 938 Wh
- B. 0.938 Wh
- C. 56.25 Wh
- D. 5.6 Wh
- Correct Answer
- 0.938 Wh
- 1. How many watt-hours represent 65 W used for 18 h?
Options- A. 11.7 Wh
- B. 1,170 Wh
- C. 11,700 Wh
- D. 117,000 Wh Discuss
- 2. In 0.025 W, there are
Options- A. 25 kW
- B. 0.00025 mW
- C. 2,500 µW
- D. 25 mW Discuss
- 3. A 75 Ω load uses 2 W of power. The output voltage of the power supply is approximately
Options- A. 120 V
- B. 1.2 V
- C. 12 V
- D. 6 V Discuss
- 4. A power supply produces a 0.6 W output with an input of 0.7 W. Its percentage of efficiency is
Options- A. 8.57%
- B. 42.85%
- C. 4.28%
- D. 85.7% Discuss
- 5. A 220 Ω resistor dissipates 3 W. The voltage is
Options- A. 73.3 V
- B. 2.5 V
- C. 25.7 V
- D. 257 V Discuss
- 6. A 120 Ω resistor must carry a maximum current of 25 mA. Its rating should be at least
Options- A. 4.8 W
- B. 150 mW
- C. 15 mW
- D. 480 mW Discuss
- 7. A 68 Ω resistor is connected across the terminals of a 3 V battery. The power dissipation of the resistor is
Options- A. 132 mW
- B. 13.2 mW
- C. 22.6 mW
- D. 226 mW Discuss
- 8. A 3.3 kΩ resistor dissipates 0.25 W. The current is
Options- A. 8.7 mA
- B. 87 mA
- C. 8.7 µA
- D. 8.7 A Discuss
- 9. A given power supply is capable of providing 6 A for 3.5 h. Its ampere-hour rating is
Options- A. 0.58 Ah
- B. 2.1 Ah
- C. 21 Ah
- D. 58 Ah Discuss
- 10. At the end of a 14 day period, your utility bill shows that you have used 18 kWh. What is your average daily power?
Options- A. 1.286 kWh
- B. 12.85 kWh
- C. 535 kWh
- D. 252 kWh Discuss
Energy and Power problems
Search Results
Correct Answer: 1,170 Wh
Correct Answer: 25 mW
Correct Answer: 12 V
Correct Answer: 85.7%
Correct Answer: 25.7 V
Correct Answer: 150 mW
Correct Answer: 132 mW
Correct Answer: 8.7 mA
Explanation:
W = I2R
I = square root(W/R)
= square root( 0.25/(3.3x1000) )
= 0.0087038828
= 8.7 mA
I = square root(W/R)
= square root( 0.25/(3.3x1000) )
= 0.0087038828
= 8.7 mA
Correct Answer: 21 Ah
Correct Answer: 1.286 kWh
Comments
There are no comments.More in Electrical Engineering:
Programming
Copyright İCuriousTab. All rights reserved.