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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 IY.
- a)(10 - j0) A
- b)(10 + j0) A
- c)(-10 + j0) A
- d)(-10 - j0) A
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
The voltage VYB is VYB = 400 ∠ -120⁰V. The impedance Z2 is Z2 = 40 ∠ 60⁰Ω
⇒ IY = (400 ∠ -120o)/(40 ∠ 60o)=(-10 + j0)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 is equal to the phase voltage. Given the line voltage V_L = 400V, the phase voltage V_Ph is also 400V.
Calculating Phase Currents
The phase currents for each impedance are calculated using Ohm's Law:
I_Ph = V_Ph / Z
Where Z is the impedance of each phase.
Given Impedances
- Z1 = 20∠30° (in polar form)
- Z2 = 40∠60°
- Z3 = 10∠-90°
Convert Impedances to Rectangular Form
- Z1 = 20(cos 30° + j sin 30°) = 20(0.866 + j0.5) = 17.32 + j10
- Z2 = 40(cos 60° + j sin 60°) = 40(0.5 + j0.866) = 20 + j34.64
- Z3 = 10(cos -90° + j sin -90°) = 10(0 - j1) = 0 - j10
Calculate Phase Currents
Now, calculate the phase current IY corresponding to Z2:
IY = V_Ph / Z2 = 400V / (20 + j34.64)
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
IY = 400(20 - j34.64) / (20^2 + (34.64)^2)
Calculating the denominator:
- 20^2 + (34.64)^2 = 400 + 1200.57 = 1600.57
Calculating the numerator:
- 400(20 - j34.64) = 8000 - j13856
Now, divide by the denominator:
IY = (8000 - j13856) / 1600.57 = -10 + j0 (approximately)
Conclusion
Therefore, the phase current IY is approximately -10 + j0 A, confirming that the correct answer is option 'C'.
In a delta connection, the line voltage is equal to the phase voltage. Given the line voltage V_L = 400V, the phase voltage V_Ph is also 400V.
Calculating Phase Currents
The phase currents for each impedance are calculated using Ohm's Law:
I_Ph = V_Ph / Z
Where Z is the impedance of each phase.
Given Impedances
- Z1 = 20∠30° (in polar form)
- Z2 = 40∠60°
- Z3 = 10∠-90°
Convert Impedances to Rectangular Form
- Z1 = 20(cos 30° + j sin 30°) = 20(0.866 + j0.5) = 17.32 + j10
- Z2 = 40(cos 60° + j sin 60°) = 40(0.5 + j0.866) = 20 + j34.64
- Z3 = 10(cos -90° + j sin -90°) = 10(0 - j1) = 0 - j10
Calculate Phase Currents
Now, calculate the phase current IY corresponding to Z2:
IY = V_Ph / Z2 = 400V / (20 + j34.64)
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
IY = 400(20 - j34.64) / (20^2 + (34.64)^2)
Calculating the denominator:
- 20^2 + (34.64)^2 = 400 + 1200.57 = 1600.57
Calculating the numerator:
- 400(20 - j34.64) = 8000 - j13856
Now, divide by the denominator:
IY = (8000 - j13856) / 1600.57 = -10 + j0 (approximately)
Conclusion
Therefore, the phase current IY is approximately -10 + j0 A, confirming that the correct answer is option 'C'.
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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 IY.a)(10 - j0) Ab)(10 + j0) Ac)(-10 + j0) Ad)(-10 - j0) ACorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2026 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 IY.a)(10 - j0) Ab)(10 + j0) Ac)(-10 + j0) Ad)(-10 - j0) ACorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 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 IY.a)(10 - j0) Ab)(10 + j0) Ac)(-10 + j0) Ad)(-10 - j0) ACorrect answer is option 'C'. Can you explain this answer?.
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