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A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, is
- a)1
- b)4
- c)100
- d)400
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A 100 W light source emits uniformly in all directions. A photodetecto...
The whole power of 100 W will not fall on the active area
Intensity of light at a distance ‘d’ from an isotropic source is given by
Here the photodetector is at 1m distance
So, d = 1m
Power falling on photodetector = Intensity x active area
Output current = Power on photodetector x Responsivity
Hence, the photo-current generated in the detector is 1 mA
Intensity of light at a distance ‘d’ from an isotropic source is given by
Here the photodetector is at 1m distance
So, d = 1m
Power falling on photodetector = Intensity x active area
Output current = Power on photodetector x Responsivity
Hence, the photo-current generated in the detector is 1 mA
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Community Answer
A 100 W light source emits uniformly in all directions. A photodetecto...
Given:
- Power of the light source = 100 W
- Diameter of the photodetector = 2 cm = 0.02 m
- Distance between the source and the photodetector = 1 m
- Responsivity of the photodetector = 0.4 A/W
To find:
The photo-current generated in the detector in units of mA.
Formula:
The photo-current generated in the photodetector can be calculated using the formula:
I = P * R
Where:
I = Photo-current generated in the photodetector (in Amperes)
P = Power incident on the photodetector (in Watts)
R = Responsivity of the photodetector (in Ampere/Watts)
Calculation:
1. Area of the circular active area of the photodetector:
The area of a circle can be calculated using the formula:
A = π * r^2
Where:
A = Area of the circle
π = Pi (approximately 3.14159)
r = Radius of the circle
Given that the diameter of the circular active area is 2 cm, the radius can be calculated as:
r = d/2 = 0.02/2 = 0.01 m
Substituting the values in the formula, the area can be calculated as:
A = π * (0.01)^2 ≈ 0.000314 m^2
2. Power incident on the photodetector:
The power incident on the photodetector can be calculated using the formula:
P = P_source * A_detector / A_source
Where:
P = Power incident on the photodetector (in Watts)
P_source = Power of the light source (in Watts)
A_detector = Area of the photodetector (in square meters)
A_source = Area of the sphere surrounding the light source (in square meters)
The area of the sphere surrounding the light source can be calculated using the formula:
A_source = 4 * π * r^2
Substituting the values, the area can be calculated as:
A_source = 4 * π * (1)^2 ≈ 12.57 m^2
Substituting the values in the formula, the power incident on the photodetector can be calculated as:
P = 100 * 0.000314 / 12.57 ≈ 0.0025 W
3. Photo-current generated in the photodetector:
Using the formula mentioned earlier:
I = P * R
Substituting the values, the photo-current can be calculated as:
I = 0.0025 * 0.4 ≈ 0.001 A
Converting the photo-current to milliamperes:
1 A = 1000 mA
Therefore, the photo-current generated in the detector is approximately 1 mA.
Hence, the correct answer is option 'A' (1 mA).
- Power of the light source = 100 W
- Diameter of the photodetector = 2 cm = 0.02 m
- Distance between the source and the photodetector = 1 m
- Responsivity of the photodetector = 0.4 A/W
To find:
The photo-current generated in the detector in units of mA.
Formula:
The photo-current generated in the photodetector can be calculated using the formula:
I = P * R
Where:
I = Photo-current generated in the photodetector (in Amperes)
P = Power incident on the photodetector (in Watts)
R = Responsivity of the photodetector (in Ampere/Watts)
Calculation:
1. Area of the circular active area of the photodetector:
The area of a circle can be calculated using the formula:
A = π * r^2
Where:
A = Area of the circle
π = Pi (approximately 3.14159)
r = Radius of the circle
Given that the diameter of the circular active area is 2 cm, the radius can be calculated as:
r = d/2 = 0.02/2 = 0.01 m
Substituting the values in the formula, the area can be calculated as:
A = π * (0.01)^2 ≈ 0.000314 m^2
2. Power incident on the photodetector:
The power incident on the photodetector can be calculated using the formula:
P = P_source * A_detector / A_source
Where:
P = Power incident on the photodetector (in Watts)
P_source = Power of the light source (in Watts)
A_detector = Area of the photodetector (in square meters)
A_source = Area of the sphere surrounding the light source (in square meters)
The area of the sphere surrounding the light source can be calculated using the formula:
A_source = 4 * π * r^2
Substituting the values, the area can be calculated as:
A_source = 4 * π * (1)^2 ≈ 12.57 m^2
Substituting the values in the formula, the power incident on the photodetector can be calculated as:
P = 100 * 0.000314 / 12.57 ≈ 0.0025 W
3. Photo-current generated in the photodetector:
Using the formula mentioned earlier:
I = P * R
Substituting the values, the photo-current can be calculated as:
I = 0.0025 * 0.4 ≈ 0.001 A
Converting the photo-current to milliamperes:
1 A = 1000 mA
Therefore, the photo-current generated in the detector is approximately 1 mA.
Hence, the correct answer is option 'A' (1 mA).
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A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. Can you explain this answer?
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A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. 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 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. 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 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. Can you explain this answer?.
A 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. 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 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. 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 100 W light source emits uniformly in all directions. A photodetector having a circular active area whose diameter is 2 cm is placed 1 m away from the source, normal to the incident light. If the responsivity of the photodetector is 0.4 A/W, the photo-current generated in the detector, in units of mA, isa)1b)4c)100d)400Correct answer is option 'A'. Can you explain this answer?.
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