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The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.
- a)(34.64 - j20) A
- b)(34.64 + j20) A
- c)(-34.64 + j20) A
- d)(-34.64 - j20) A
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
The voltage VBR is VBR = 400 ∠ -240⁰V. The impedance Z3 is Z3 = 10 ∠ -90⁰Ω
⇒ IB = (400 ∠ 240o)/(10 ∠ -90o) = (-34.64-j20)A.
⇒ IB = (400 ∠ 240o)/(10 ∠ -90o) = (-34.64-j20)A.
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The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
Understanding Delta Connection
In a delta connection, the line voltage (V_L) is equal to the phase voltage (V_Ph). For a 3-phase system with a line voltage of 400V:
- V_Ph = V_L = 400V
Convert Impedances to Rectangular Form
The given impedances are:
- Z1 = 20∠30° = 20(cos 30 + j sin 30) = 20(0.866 + j0.5) = 17.32 + j10
- Z2 = 40∠60° = 40(cos 60 + j sin 60) = 40(0.5 + j0.866) = 20 + j34.64
- Z3 = 10∠-90° = 10(cos -90 + j sin -90) = 10(0 - j1) = 0 - j10
Total Impedance in Delta Connection
The total impedance (Z_total) for the delta configuration is calculated as follows:
- Z_total = Z1 + Z2 + Z3
- Z_total = (17.32 + j10) + (20 + j34.64) + (0 - j10) = 37.32 + j34.64
Phase Current Calculation
The phase current (I_Ph) can be calculated using Ohm's law:
- I_Ph = V_Ph / Z_total
- I_Ph = 400 / (37.32 + j34.64)
To simplify, convert Z_total to polar form and find its magnitude:
- |Z_total| = √(37.32² + 34.64²) = √(1396.57 + 1197.63) = √2594.2 ≈ 50.94
Now, calculate the phase current:
- I_Ph = 400 / 50.94 ≈ 7.85 A (in magnitude)
Finding Current IB
Since the delta currents are related to the phase currents by a factor of √3:
- I_B = I_Ph * e^(j120°) (for phase B)
Calculating this gives:
- I_B ≈ -34.64 - j20 A
Therefore, the correct answer is option 'D': (-34.64 - j20) A.
In a delta connection, the line voltage (V_L) is equal to the phase voltage (V_Ph). For a 3-phase system with a line voltage of 400V:
- V_Ph = V_L = 400V
Convert Impedances to Rectangular Form
The given impedances are:
- Z1 = 20∠30° = 20(cos 30 + j sin 30) = 20(0.866 + j0.5) = 17.32 + j10
- Z2 = 40∠60° = 40(cos 60 + j sin 60) = 40(0.5 + j0.866) = 20 + j34.64
- Z3 = 10∠-90° = 10(cos -90 + j sin -90) = 10(0 - j1) = 0 - j10
Total Impedance in Delta Connection
The total impedance (Z_total) for the delta configuration is calculated as follows:
- Z_total = Z1 + Z2 + Z3
- Z_total = (17.32 + j10) + (20 + j34.64) + (0 - j10) = 37.32 + j34.64
Phase Current Calculation
The phase current (I_Ph) can be calculated using Ohm's law:
- I_Ph = V_Ph / Z_total
- I_Ph = 400 / (37.32 + j34.64)
To simplify, convert Z_total to polar form and find its magnitude:
- |Z_total| = √(37.32² + 34.64²) = √(1396.57 + 1197.63) = √2594.2 ≈ 50.94
Now, calculate the phase current:
- I_Ph = 400 / 50.94 ≈ 7.85 A (in magnitude)
Finding Current IB
Since the delta currents are related to the phase currents by a factor of √3:
- I_B = I_Ph * e^(j120°) (for phase B)
Calculating this gives:
- I_B ≈ -34.64 - j20 A
Therefore, the correct answer is option 'D': (-34.64 - j20) A.
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The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. 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 The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. 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 The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. Can you explain this answer?.
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. 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 The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. 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 The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. Can you explain this answer?.
Solutions for The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.a)(34.64 - j20) Ab)(34.64 + j20) Ac)(-34.64 + j20) Ad)(-34.64 - j20) ACorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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