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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. Determine the phase current IR.
- a)(17.32 - j10) A
- b)(-17.32 - j10) A
- c)(17.32 + j10) A
- d)(-17.32 + j10) A
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
Impedances Calculation:
- Given impedances: Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90
- Convert delta impedances to star impedances: Zs = Zd/3
- Zs1 = 20/3∠30, Zs2 = 40/3∠60, Zs3 = 10/3∠-90
Equivalent Star Impedance Calculation:
- Calculate equivalent impedance in star connection: Zs = Zs1 + Zs2 + Zs3
- Zs = (20/3∠30) + (40/3∠60) + (10/3∠-90)
Calculate Phase Current:
- Phase voltage (Vph) = Line voltage (Vline) / √3
- Vph = 400 / √3
- Calculate phase current (I) using Ohm's Law: I = Vph / Zs
Therefore, the phase current IR is 17.32∠-10A which is equivalent to (17.32 - j10) A.
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The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
Taking VRY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have VRY = 400∠0⁰V Z1 = 20∠30⁰Ω = (17.32 + j10)Ω IR = (400∠0o)/(20∠30o) = (17.32 - j10) A.
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