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A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core is
- a)1.2 Wb/m2
- b)0.94 Wb/m2
- c)2.1 Wb/m2
- d)0.25 Wb/m2
Correct answer is option 'B'. Can you explain this answer?
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A single phase transformer has 400 primary and 1000 secondary turns. T...
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A single phase transformer has 400 primary and 1000 secondary turns. T...
Given data:
- Primary turns, N1 = 400
- Secondary turns, N2 = 1000
- Cross-sectional area of the core, A = 60 cm² = 0.006 m²
- Voltage of the primary source, V1 = 500 V
- Frequency of the primary source, f = 50 Hz
To find: Peak value of flux density in the core
Formula used:
- EMF equation of transformer: V1/V2 = N1/N2
- Magnetic flux, Φ = B × A
- RMS value of voltage, Vrms = Vm/√2
- RMS value of current, Irms = Im/√2
- Magnetic flux density, B = μ0μr(N/L)I
- Peak value of magnetic flux density, Bm = √2B
where μ0 = permeability of free space = 4π × 10^-7 H/m, μr = relative permeability of the core material, N = number of turns, L = mean length of the magnetic path, and I = current.
Solution:
1. RMS value of secondary voltage:
Using the EMF equation of the transformer,
V1/V2 = N1/N2
500/V2 = 400/1000
V2 = 500 × 1000/400
V2 = 1250 V
The RMS value of secondary voltage is:
V2rms = V2/√2 = 1250/√2 = 884.5 V
2. RMS value of secondary current:
The power in the primary and secondary circuits is the same, neglecting losses. Therefore,
V1I1 = V2I2
I2 = (V1/V2)I1 = (500/1250)I1 = 0.4I1
The RMS value of secondary current is:
I2rms = I2/√2 = 0.4I1/√2
3. Magnetic flux density:
The mean length of the magnetic path can be estimated as follows:
L ≈ 2 × (mean length of the core limb) + (mean length of the air gap)
Assuming the core limb to be rectangular in shape, its length can be taken as the length of the core, and its width can be taken as the cross-sectional area divided by the length. The length of the air gap can be taken as 0.2 mm, which is a typical value for power transformers.
Therefore,
L = 2 × (0.15 m) + (0.0002 m) = 0.3004 m
The current in the core is the secondary current divided by the turns ratio:
I = I2/N2 = (0.4I1/√2)/1000 = (0.2I1/√2)/500
The magnetic flux density is:
B = μ0μr(N/L)I
B = (4π × 10^-7) × 2000 × (400/0.3004) × (0.2I1/√2)/500
B = 0.94 × 0.2I1/√2
B = 0.133I1
4. Peak value of magnetic flux density:
The peak value of magnetic flux density is
- Primary turns, N1 = 400
- Secondary turns, N2 = 1000
- Cross-sectional area of the core, A = 60 cm² = 0.006 m²
- Voltage of the primary source, V1 = 500 V
- Frequency of the primary source, f = 50 Hz
To find: Peak value of flux density in the core
Formula used:
- EMF equation of transformer: V1/V2 = N1/N2
- Magnetic flux, Φ = B × A
- RMS value of voltage, Vrms = Vm/√2
- RMS value of current, Irms = Im/√2
- Magnetic flux density, B = μ0μr(N/L)I
- Peak value of magnetic flux density, Bm = √2B
where μ0 = permeability of free space = 4π × 10^-7 H/m, μr = relative permeability of the core material, N = number of turns, L = mean length of the magnetic path, and I = current.
Solution:
1. RMS value of secondary voltage:
Using the EMF equation of the transformer,
V1/V2 = N1/N2
500/V2 = 400/1000
V2 = 500 × 1000/400
V2 = 1250 V
The RMS value of secondary voltage is:
V2rms = V2/√2 = 1250/√2 = 884.5 V
2. RMS value of secondary current:
The power in the primary and secondary circuits is the same, neglecting losses. Therefore,
V1I1 = V2I2
I2 = (V1/V2)I1 = (500/1250)I1 = 0.4I1
The RMS value of secondary current is:
I2rms = I2/√2 = 0.4I1/√2
3. Magnetic flux density:
The mean length of the magnetic path can be estimated as follows:
L ≈ 2 × (mean length of the core limb) + (mean length of the air gap)
Assuming the core limb to be rectangular in shape, its length can be taken as the length of the core, and its width can be taken as the cross-sectional area divided by the length. The length of the air gap can be taken as 0.2 mm, which is a typical value for power transformers.
Therefore,
L = 2 × (0.15 m) + (0.0002 m) = 0.3004 m
The current in the core is the secondary current divided by the turns ratio:
I = I2/N2 = (0.4I1/√2)/1000 = (0.2I1/√2)/500
The magnetic flux density is:
B = μ0μr(N/L)I
B = (4π × 10^-7) × 2000 × (400/0.3004) × (0.2I1/√2)/500
B = 0.94 × 0.2I1/√2
B = 0.133I1
4. Peak value of magnetic flux density:
The peak value of magnetic flux density is
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A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core isa)1.2 Wb/m2b)0.94Wb/m2c)2.1 Wb/m2d)0.25 Wb/m2Correct answer is option 'B'. Can you explain this answer?
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A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core isa)1.2 Wb/m2b)0.94Wb/m2c)2.1 Wb/m2d)0.25 Wb/m2Correct answer is option 'B'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core isa)1.2 Wb/m2b)0.94Wb/m2c)2.1 Wb/m2d)0.25 Wb/m2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core isa)1.2 Wb/m2b)0.94Wb/m2c)2.1 Wb/m2d)0.25 Wb/m2Correct answer is option 'B'. Can you explain this answer?.
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ample number of questions to practice A single phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm2 If the primary is connected to 500 V, 50 Hz source, the peak value of flux density in the core isa)1.2 Wb/m2b)0.94Wb/m2c)2.1 Wb/m2d)0.25 Wb/m2Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice SSC tests.
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