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A 415 V, three-phase star-connected alternator supplies a delta-connected load, each phase of which has an impedance of (86∠54.46°) Ω. Calculate kVA rating of the alternator, neglecting losses in the line between the alternator and load.
- a)6.002 kVA
- b)3.465 kVA
- c)4 KVA
- d)8 KVA
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A 415 V, three-phase star-connected alternator supplies a delta-connec...
+ j63) ohms. Determine the line current and power factor of the load.
To find the line current, we need to convert the delta-connected load to an equivalent star-connected load.
In a delta-connected load, the line current is equal to the phase current. In a star-connected load, the line current is equal to the square root of 3 times the phase current.
First, let's convert the impedance of each phase of the delta-connected load to the equivalent impedance of the star-connected load:
Z_star = Z_delta / sqrt(3)
Z_star = (86 + j63) / sqrt(3)
To simplify the calculation, we can convert the impedance to polar form:
Z_star = 102.7 ∠ 36.87 degrees ohms
Now, we can find the line current:
I_line = I_phase * sqrt(3)
To find the phase current, we can use Ohm's Law:
Z = V / I
I_phase = V / Z
I_phase = 415 V / (102.7 ∠ 36.87 degrees ohms)
I_phase = 4.04 ∠ -36.87 degrees A
Finally, we can find the line current:
I_line = 4.04 ∠ -36.87 degrees A * sqrt(3)
I_line = 7 A ∠ -36.87 degrees A
The line current is 7 A with a phase angle of -36.87 degrees.
To find the power factor, we can use the cosine of the phase angle:
power factor = cos(-36.87 degrees)
power factor = 0.8
The power factor of the load is 0.8.
To find the line current, we need to convert the delta-connected load to an equivalent star-connected load.
In a delta-connected load, the line current is equal to the phase current. In a star-connected load, the line current is equal to the square root of 3 times the phase current.
First, let's convert the impedance of each phase of the delta-connected load to the equivalent impedance of the star-connected load:
Z_star = Z_delta / sqrt(3)
Z_star = (86 + j63) / sqrt(3)
To simplify the calculation, we can convert the impedance to polar form:
Z_star = 102.7 ∠ 36.87 degrees ohms
Now, we can find the line current:
I_line = I_phase * sqrt(3)
To find the phase current, we can use Ohm's Law:
Z = V / I
I_phase = V / Z
I_phase = 415 V / (102.7 ∠ 36.87 degrees ohms)
I_phase = 4.04 ∠ -36.87 degrees A
Finally, we can find the line current:
I_line = 4.04 ∠ -36.87 degrees A * sqrt(3)
I_line = 7 A ∠ -36.87 degrees A
The line current is 7 A with a phase angle of -36.87 degrees.
To find the power factor, we can use the cosine of the phase angle:
power factor = cos(-36.87 degrees)
power factor = 0.8
The power factor of the load is 0.8.
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Community Answer
A 415 V, three-phase star-connected alternator supplies a delta-connec...
Concept
The kVA rating for a delta-connected load is given by:
S = 3VPIP
where, VP = Phase Voltage
IP = Phase Current
The phase current is given by: IP = VP/ZP
Calculation:
Given, VP = 415 V
ZP = (86∠54A6°)Ω
IP = 415 / (86∠54.46)
IP = 4.82 ∠-54.46
S = 3 x 415 x 4.82
S = 6.002 kVA
The kVA rating for a delta-connected load is given by:
S = 3VPIP
where, VP = Phase Voltage
IP = Phase Current
The phase current is given by: IP = VP/ZP
Calculation:
Given, VP = 415 V
ZP = (86∠54A6°)Ω
IP = 415 / (86∠54.46)
IP = 4.82 ∠-54.46
S = 3 x 415 x 4.82
S = 6.002 kVA
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A 415 V, three-phase star-connected alternator supplies a delta-connected load, each phase of which has an impedance of (86∠54.46°) Ω. Calculate kVA rating of the alternator, neglecting losses in the line between the alternator and load.a)6.002 kVAb)3.465 kVAc)4 KVAd)8 KVACorrect answer is option 'A'. Can you explain this answer?
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