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A 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circuit. The magnetising component of no-load primary current is
- a)0.150 A
- b)0.477 A
- c)0.795 A
- d)0.352 A
Correct answer is option 'B'. Can you explain this answer?
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A 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circui...
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A 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circui...
Given data:
Transformer rating (S) = 2200 VA
Primary voltage (V1) = 250 V
Primary current at no-load with a power factor of 0.3 (I0) = 0.5 A
To find: Magnetizing component of no-load current
Calculations:
Apparent power (S) = V1 × I0
= 250 × 0.5
= 125 VA
The real power (P) consumed by the transformer is zero as it is on open-circuit. Hence, the reactive power (Q) consumed by the transformer is equal to the apparent power (S).
Q = S = 125 VAR
The power factor (cosφ) of the transformer is given as 0.3. Therefore, the reactive power (Q) and the active power (P) can be related using the power factor as:
cosφ = P/S
0.3 = P/125
P = 37.5 W
The magnetizing component of the no-load current, Im can be calculated as:
Im = √(I0² - Ic²)
where Ic = I0 × cosφ
Ic = 0.5 × 0.3
= 0.15 A
Im = √(0.5² - 0.15²)
= 0.477 A
Therefore, the magnetizing component of no-load primary current is 0.477 A, which is closest to option (B).
Transformer rating (S) = 2200 VA
Primary voltage (V1) = 250 V
Primary current at no-load with a power factor of 0.3 (I0) = 0.5 A
To find: Magnetizing component of no-load current
Calculations:
Apparent power (S) = V1 × I0
= 250 × 0.5
= 125 VA
The real power (P) consumed by the transformer is zero as it is on open-circuit. Hence, the reactive power (Q) consumed by the transformer is equal to the apparent power (S).
Q = S = 125 VAR
The power factor (cosφ) of the transformer is given as 0.3. Therefore, the reactive power (Q) and the active power (P) can be related using the power factor as:
cosφ = P/S
0.3 = P/125
P = 37.5 W
The magnetizing component of the no-load current, Im can be calculated as:
Im = √(I0² - Ic²)
where Ic = I0 × cosφ
Ic = 0.5 × 0.3
= 0.15 A
Im = √(0.5² - 0.15²)
= 0.477 A
Therefore, the magnetizing component of no-load primary current is 0.477 A, which is closest to option (B).
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A 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circuit. The magnetising component of no-load primary current isa)0.150 Ab)0.477 Ac)0.795 Ad)0.352 ACorrect answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circuit. The magnetising component of no-load primary current isa)0.150 Ab)0.477 Ac)0.795 Ad)0.352 ACorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 2200 / 250 V transformer takes 0.5 A at a p.f. of 0.3 on open-circuit. The magnetising component of no-load primary current isa)0.150 Ab)0.477 Ac)0.795 Ad)0.352 ACorrect answer is option 'B'. Can you explain this answer?.
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