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A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center. The time-average poynting vector at 50 km is
- a)6.36ur μW/m2
- b)4.78ur μW/m2
- c)9.55ur μW/m2
- d)12.73ur μW/m2
Correct answer is option 'C'. Can you explain this answer?
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The average power transmitted by the antenna is 150 kW. This power is radiated uniformly over the surface of a hemisphere with the antenna at the center. We are required to find the time-average Poynting vector at a distance of 50 km from the antenna.
To solve this problem, we can use the formula for the Poynting vector:
S = (P r^2) / (2πr^2)
where S is the time-average Poynting vector, P is the average power transmitted by the antenna, and r is the distance from the antenna.
Let's calculate the time-average Poynting vector at a distance of 50 km:
P = 150 kW = 150,000 W
r = 50 km = 50,000 m
Substituting these values into the formula, we get:
S = (150,000 × 50,000^2) / (2π × 50,000^2)
Simplifying the equation, we have:
S = 150,000 / (2π)
To find the value of S, we can use the approximation of π as 3.14:
S ≈ 150,000 / (2 × 3.14)
S ≈ 23,885.35 W/m^2
Therefore, the time-average Poynting vector at a distance of 50 km is approximately 23,885.35 W/m^2.
Comparing this result with the given options, we can see that none of the options match. It seems that there may be an error in the options provided.
Please check the options again or provide additional information if available.
To solve this problem, we can use the formula for the Poynting vector:
S = (P r^2) / (2πr^2)
where S is the time-average Poynting vector, P is the average power transmitted by the antenna, and r is the distance from the antenna.
Let's calculate the time-average Poynting vector at a distance of 50 km:
P = 150 kW = 150,000 W
r = 50 km = 50,000 m
Substituting these values into the formula, we get:
S = (150,000 × 50,000^2) / (2π × 50,000^2)
Simplifying the equation, we have:
S = 150,000 / (2π)
To find the value of S, we can use the approximation of π as 3.14:
S ≈ 150,000 / (2 × 3.14)
S ≈ 23,885.35 W/m^2
Therefore, the time-average Poynting vector at a distance of 50 km is approximately 23,885.35 W/m^2.
Comparing this result with the given options, we can see that none of the options match. It seems that there may be an error in the options provided.
Please check the options again or provide additional information if available.
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A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer?
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A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer?.
A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A antenna located on the surface of a flat earth transmit an average power of 150 kW. Assume that all the power is radiated uniformly over the surface of hemisphere with the antenna at the center.The time-average poynting vector at 50 km isa)6.36ur μW/m2b)4.78ur μW/m2c)9.55ur μW/m2d)12.73ur μW/m2Correct answer is option 'C'. Can you explain this answer?.
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