Discussion
A meter marked 1200 rev/kWh makes 20 revolutions in 30 seconds. The load is
Answer: C
The energy is \( 20/1200 = 0.01667 \) kWh, and dividing by the period expressed in hours, \( 30/3600 = 0.00833 \) h, gives 2 kW. The general form is \( P = 3600n/(Kt) \) with n revolutions in t seconds and K in rev/kWh.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development