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A 230 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hours at unity at unity power factor. The meter disc makes 2208 revolutions during this period. If the number of revolutions made by the meter are 1472 when operating at 230 V and 5 A for 4 hours, then power factor of the load would be
  • a)
    0.6 lag
  • b)
    0.707 lag
  • c)
    0.866 lag
  • d)
    0.8 lag
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A 230 V, single phase, watt hour meter has a constant load of 4 A pass...
Energy supplied = VI cosφ) x t x 10-3 kWh
= 230 x 4 x 1 x 6 x 10-3
= 5.52 kWh
Energy meter constant,

Energy consumed when the meter makes 1472 revolutions

Now, energy consumed
 VI cos φ x f x 10-3 kWh = 230 x 5 x cost}) x 4 x 10-3 kWh
= 3.68 kWh (As obtained above)
or, 
Hence, power factor of the load = 0.8
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Most Upvoted Answer
A 230 V, single phase, watt hour meter has a constant load of 4 A pass...
Solution:

Given data:

For the first case,

Voltage (V) = 230 V

Load current (I) = 4 A

Time (t) = 6 hours

Number of revolutions made by the meter = 2208

For the second case,

Voltage (V) = 230 V

Load current (I) = 5 A

Time (t) = 4 hours

Number of revolutions made by the meter = 1472

To find: Power factor of the load in the second case

Formula used:

The formula used to calculate the power consumed by a load in a single-phase circuit is given by,

P = V x I x PF

where,

P = Power consumed by the load in watts (W)

V = Voltage across the load in volts (V)

I = Current passing through the load in amperes (A)

PF = Power factor of the load

Calculation:

From the first case,

Number of revolutions made by the meter in 1 hour = 2208/6 = 368

Power consumed by the load in the first case = V x I x PF

= 230 x 4 x 1 = 920 W

Energy consumed by the load in the first case = Power x time

= 920 x 6 = 5520 Wh

From the second case,

Number of revolutions made by the meter in 1 hour = 1472/4 = 368

Energy consumed by the load in the second case = Power x time

= V x I x PF x t

=> PF = Energy consumed by the load / (V x I x t)

=> PF = 1472 / (230 x 5 x 4)

=> PF = 0.8 lag

Therefore, the power factor of the load in the second case is 0.8 lag.
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A 230 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hours at unity at unity power factor. The meter disc makes 2208 revolutions during this period. If the number of revolutions made by the meter are 1472 when operating at 230 V and 5 A for 4 hours, then power factor of the load would bea)0.6 lagb)0.707 lagc)0.866 lagd)0.8 lagCorrect answer is option 'D'. Can you explain this answer?
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A 230 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hours at unity at unity power factor. The meter disc makes 2208 revolutions during this period. If the number of revolutions made by the meter are 1472 when operating at 230 V and 5 A for 4 hours, then power factor of the load would bea)0.6 lagb)0.707 lagc)0.866 lagd)0.8 lagCorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 230 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hours at unity at unity power factor. The meter disc makes 2208 revolutions during this period. If the number of revolutions made by the meter are 1472 when operating at 230 V and 5 A for 4 hours, then power factor of the load would bea)0.6 lagb)0.707 lagc)0.866 lagd)0.8 lagCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 230 V, single phase, watt hour meter has a constant load of 4 A passing through it for 6 hours at unity at unity power factor. The meter disc makes 2208 revolutions during this period. If the number of revolutions made by the meter are 1472 when operating at 230 V and 5 A for 4 hours, then power factor of the load would bea)0.6 lagb)0.707 lagc)0.866 lagd)0.8 lagCorrect answer is option 'D'. Can you explain this answer?.
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