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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 line current I1.
  • a)
    (-51.96 - j10) A
  • b)
    (-51.96 + j10) A
  • c)
    (51.96 + j10) A
  • d)
    (51.96 - j10) 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 ...
Ω, Z2= 30 Ω, and Z3= 40 Ω are connected in series.

To find the total impedance, we simply add the individual impedances together:

Total impedance = Z1 + Z2 + Z3
Total impedance = 20 Ω + 30 Ω + 40 Ω
Total impedance = 90 Ω

Therefore, the total impedance of the three impedances connected in series is 90 Ω.
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Community Answer
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
The line current I1 is the difference of IR and IB. So the line current I1 is I1 = IR – IB = (51.96 + j10) 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 line current I1.a)(-51.96 - j10) Ab)(-51.96 + j10) Ac)(51.96 + j10) Ad)(51.96 - j10) 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 line current I1.a)(-51.96 - j10) Ab)(-51.96 + j10) Ac)(51.96 + j10) Ad)(51.96 - j10) 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 line current I1.a)(-51.96 - j10) Ab)(-51.96 + j10) Ac)(51.96 + j10) Ad)(51.96 - j10) ACorrect answer is option 'C'. Can you explain this answer?.
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