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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 power in the Y phase.
- a)1000
- b)2000
- c)3000
- d)4000
Correct answer is option 'B'. 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. The total impedance in the circuit is the sum of the individual impedances:
Total impedance = Z1 + Z2 + Z3
Total impedance = 20 Ω + 30 Ω + 40 Ω
Total impedance = 90 Ω
Therefore, the total impedance in the circuit is 90 Ω when the impedances Z1= 20 Ω, Z2=30 Ω, and Z3=40 Ω are connected in series.
Total impedance = Z1 + Z2 + Z3
Total impedance = 20 Ω + 30 Ω + 40 Ω
Total impedance = 90 Ω
Therefore, the total impedance in the circuit is 90 Ω when the impedances Z1= 20 Ω, Z2=30 Ω, and Z3=40 Ω are connected in series.
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The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 102 x 20 = 2000W.
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