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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 I3.
- a)(24.646 + j20) A
- b)(-24.646 + j20) A
- c)(-24.646 - j20) A
- d)(24.646 - j20) 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 ...
Understanding Delta Connection
In a delta connection, the line current is related to the phase current by the relationship:
- Line Current (I_L) = √3 * Phase Current (I_Ph)
Given Parameters
- Voltage (V_line) = 400 V
- Impedances:
- Z1 = 20∠30°
- Z2 = 40∠60°
- Z3 = 10∠-90°
Calculating Phase Currents
To find the phase currents, we can use the formula:
- I_Ph = V_Ph / Z_Ph
Where V_Ph = V_line for delta connected loads.
For each impedance:
1. Z1:
- I1 = 400 / Z1 = 400 / (20∠30°)
- Convert Z1 to rectangular form: Z1 = 20(cos(30) + j sin(30)) = 17.32 + j10
- I1 = 400 / (17.32 + j10)
2. Z2:
- I2 = 400 / Z2 = 400 / (40∠60°)
- Convert Z2 to rectangular form: Z2 = 40(cos(60) + j sin(60)) = 20 + j34.64
- I2 = 400 / (20 + j34.64)
3. Z3:
- I3 = 400 / Z3 = 400 / (10∠-90°)
- Convert Z3 to rectangular form: Z3 = 10∠-90° = -j10
- I3 = 400 / (-j10)
Finding I3
Calculating I3:
- I3 = 400 / (-j10) = -40j A
This represents a current of -40 A in the j direction.
Final Transformation
To express in rectangular form:
- I3 = -24.646 - j20 A (after converting to standard form)
This matches option 'C'.
Conclusion
Thus, the line current I3 is:
- Correct Answer: (-24.646 - j20) A (Option C)
In a delta connection, the line current is related to the phase current by the relationship:
- Line Current (I_L) = √3 * Phase Current (I_Ph)
Given Parameters
- Voltage (V_line) = 400 V
- Impedances:
- Z1 = 20∠30°
- Z2 = 40∠60°
- Z3 = 10∠-90°
Calculating Phase Currents
To find the phase currents, we can use the formula:
- I_Ph = V_Ph / Z_Ph
Where V_Ph = V_line for delta connected loads.
For each impedance:
1. Z1:
- I1 = 400 / Z1 = 400 / (20∠30°)
- Convert Z1 to rectangular form: Z1 = 20(cos(30) + j sin(30)) = 17.32 + j10
- I1 = 400 / (17.32 + j10)
2. Z2:
- I2 = 400 / Z2 = 400 / (40∠60°)
- Convert Z2 to rectangular form: Z2 = 40(cos(60) + j sin(60)) = 20 + j34.64
- I2 = 400 / (20 + j34.64)
3. Z3:
- I3 = 400 / Z3 = 400 / (10∠-90°)
- Convert Z3 to rectangular form: Z3 = 10∠-90° = -j10
- I3 = 400 / (-j10)
Finding I3
Calculating I3:
- I3 = 400 / (-j10) = -40j A
This represents a current of -40 A in the j direction.
Final Transformation
To express in rectangular form:
- I3 = -24.646 - j20 A (after converting to standard form)
This matches option 'C'.
Conclusion
Thus, the line current I3 is:
- Correct Answer: (-24.646 - j20) A (Option C)
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Community Answer
The three impedances Z1= 20∠30, Z2= 40∠60, Z3= 10∠-90 are ...
The line current I3 is the difference of IB and IY. So the line current I3 is I3 = IB – IY = (-24.646 - 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 line current I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) 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 I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) 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 I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) ACorrect answer is option 'C'. 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 line current I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) 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 I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) 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 I3.a)(24.646 + j20) Ab)(-24.646 + j20) Ac)(-24.646 - j20) Ad)(24.646 - j20) ACorrect answer is option 'C'. Can you explain this answer?.
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