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A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.
- a)1410.67 W
- b)1269. 6
- c)4232 W
- d)2443.35 W
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A balanced star connected load of 8 + j6 ohms per phase is connected t...
Z = 8 +j 6
Power consumed by the load
= 4229.11 W
Most Upvoted Answer
A balanced star connected load of 8 + j6 ohms per phase is connected t...
Given Information:
- Load impedance: 8+j6 ohms per phase
- Supply voltage: 230V
- Load connected in a balanced star configuration
To find the power consumed by the load, we can use the formula:
P = 3 * V^2 / Z
where P is the power, V is the voltage, and Z is the impedance.
1. Conversion of Impedance:
The given impedance is in polar form (8+j6 ohms), but for calculations, it is convenient to convert it to rectangular form. Using the polar-to-rectangular conversion:
Z = 8 + j6
= 8 + j(6)
= 8 + j(2*3)
= 8 + 2j3
= 8 + 2j(√3)
= 8 + 2j(1.732)
= 8 + 3.464j
2. Calculation of Power:
Using the formula mentioned earlier:
P = 3 * V^2 / Z
= 3 * (230)^2 / (8 + 3.464j)
= 3 * 52900 / (8 + 3.464j)
= 158700 / (8 + 3.464j)
= 158700 / (8 + 3.464j) * (8 - 3.464j) / (8 - 3.464j) [Multiplying by the complex conjugate to eliminate the imaginary part in the denominator]
= 158700 * (8 - 3.464j) / (8^2 - (3.464j)^2)
= 158700 * (8 - 3.464j) / (64 - 11.98j^2)
= 158700 * (8 - 3.464j) / (64 - 11.98(-1))
= 158700 * (8 - 3.464j) / (64 + 11.98)
= 158700 * (8 - 3.464j) / 75.98
= 158700 * (8/75.98 - 3.464j/75.98)
≈ 4232 - 146.91j
The real part of the power is the active power, and the imaginary part is the reactive power. Therefore, the power consumed by the load is approximately 4232 W. Thus, the correct answer is option C.
- Load impedance: 8+j6 ohms per phase
- Supply voltage: 230V
- Load connected in a balanced star configuration
To find the power consumed by the load, we can use the formula:
P = 3 * V^2 / Z
where P is the power, V is the voltage, and Z is the impedance.
1. Conversion of Impedance:
The given impedance is in polar form (8+j6 ohms), but for calculations, it is convenient to convert it to rectangular form. Using the polar-to-rectangular conversion:
Z = 8 + j6
= 8 + j(6)
= 8 + j(2*3)
= 8 + 2j3
= 8 + 2j(√3)
= 8 + 2j(1.732)
= 8 + 3.464j
2. Calculation of Power:
Using the formula mentioned earlier:
P = 3 * V^2 / Z
= 3 * (230)^2 / (8 + 3.464j)
= 3 * 52900 / (8 + 3.464j)
= 158700 / (8 + 3.464j)
= 158700 / (8 + 3.464j) * (8 - 3.464j) / (8 - 3.464j) [Multiplying by the complex conjugate to eliminate the imaginary part in the denominator]
= 158700 * (8 - 3.464j) / (8^2 - (3.464j)^2)
= 158700 * (8 - 3.464j) / (64 - 11.98j^2)
= 158700 * (8 - 3.464j) / (64 - 11.98(-1))
= 158700 * (8 - 3.464j) / (64 + 11.98)
= 158700 * (8 - 3.464j) / 75.98
= 158700 * (8/75.98 - 3.464j/75.98)
≈ 4232 - 146.91j
The real part of the power is the active power, and the imaginary part is the reactive power. Therefore, the power consumed by the load is approximately 4232 W. Thus, the correct answer is option C.
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A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer?
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A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer?.
A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer?.
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A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A balanced star connected load of 8 + j6 ohms per phase is connected to a 3-phase 230V supply. Power being consumed by the load is.a)1410.67 Wb)1269. 6c)4232 Wd)2443.35 WCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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