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An industrial consumer has a load pattern of 2000 KW 0.8 lag for 12 hours and 1000 KW unity power factor for 12 hours. The load factor is _____
  • a)
    0.5
  • b)
    0.75
  • c)
    0.6
  • d)
    2
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
An industrial consumer has a load pattern of 2000 KW 0.8 lag for 12 h...
Hence, the correct option is (C)
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Most Upvoted Answer
An industrial consumer has a load pattern of 2000 KW 0.8 lag for 12 h...
Given:
- Load pattern: 2000 KW 0.8 lag for 12 hours and 1000 KW unity power factor for 12 hours.

To find: Load factor.

Formula:
Load factor = Average load / Maximum demand

Calculation:
1. Maximum demand:
- Maximum demand occurs when the load is 2000 KW at 0.8 lag.
- Apparent power = Real power / Power factor = 2000 / 0.8 = 2500 KVA
- Maximum demand = 2500 KVA

2. Average load:
- Average load = (2000 KW x 0.8 x 12 hours) + (1000 KW x 1 x 12 hours) / 24 hours
- Average load = (19200 KWh) + (12000 KWh) / 24
- Average load = 31200 / 24
- Average load = 1300 KW

3. Load factor:
- Load factor = Average load / Maximum demand
- Load factor = 1300 / 2500
- Load factor = 0.6

Therefore, the load factor is 0.6.
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An industrial consumer has a load pattern of 2000 KW 0.8 lag for 12 hours and 1000 KW unity power factor for 12 hours. The load factor is _____a)0.5b)0.75c)0.6d)2Correct answer is option 'C'. Can you explain this answer?
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