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In a system an input resistance of 20 kΩ causes a noise voltage of 15μV. If two such input resistances each of 20 kΩ are used in parallel, the total noise voltage will be about
- a)15 μV
- b)30 μV
- c)7.5 μV
- d)10.6 μV
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
In a system an input resistance of 20 kΩ causes a noise voltage of 15μ...
Overall resistance = 10 kW. Noise voltage = (15)
= 10.6 μV.
= 10.6 μV.Most Upvoted Answer
In a system an input resistance of 20 kΩ causes a noise voltage of 15μ...
Given:
Input resistance of a system = 20 kΩ
Noise voltage for a single input resistance = 15 μV
To find:
Total noise voltage when two input resistances are used in parallel
Solution:
Explanation:
When resistances are connected in parallel, the total resistance is given by the formula:
1/R_total = 1/R_1 + 1/R_2 + 1/R_3 + ... + 1/R_n
In this case, we have two input resistances of 20 kΩ each connected in parallel. So, the total resistance is:
1/R_total = 1/20 kΩ + 1/20 kΩ
1/R_total = 2/20 kΩ
1/R_total = 1/10 kΩ
R_total = 10 kΩ
Now, the total noise voltage can be calculated using the formula:
V_total = √(V_1^2 + V_2^2 + V_3^2 + ... + V_n^2)
In this case, we have two input resistances connected in parallel, so the total noise voltage becomes:
V_total = √(V_1^2 + V_2^2)
Substituting the given values:
V_total = √((15 μV)^2 + (15 μV)^2)
V_total = √(225 μV^2 + 225 μV^2)
V_total = √(450 μV^2)
V_total = √450 μV
V_total = 21.21 μV (approximately)
Therefore, the total noise voltage when two input resistances of 20 kΩ each are used in parallel is approximately 21.21 μV.
Answer:
The correct option is (d) 21.21 μV.
Input resistance of a system = 20 kΩ
Noise voltage for a single input resistance = 15 μV
To find:
Total noise voltage when two input resistances are used in parallel
Solution:
Explanation:
When resistances are connected in parallel, the total resistance is given by the formula:
1/R_total = 1/R_1 + 1/R_2 + 1/R_3 + ... + 1/R_n
In this case, we have two input resistances of 20 kΩ each connected in parallel. So, the total resistance is:
1/R_total = 1/20 kΩ + 1/20 kΩ
1/R_total = 2/20 kΩ
1/R_total = 1/10 kΩ
R_total = 10 kΩ
Now, the total noise voltage can be calculated using the formula:
V_total = √(V_1^2 + V_2^2 + V_3^2 + ... + V_n^2)
In this case, we have two input resistances connected in parallel, so the total noise voltage becomes:
V_total = √(V_1^2 + V_2^2)
Substituting the given values:
V_total = √((15 μV)^2 + (15 μV)^2)
V_total = √(225 μV^2 + 225 μV^2)
V_total = √(450 μV^2)
V_total = √450 μV
V_total = 21.21 μV (approximately)
Therefore, the total noise voltage when two input resistances of 20 kΩ each are used in parallel is approximately 21.21 μV.
Answer:
The correct option is (d) 21.21 μV.
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In a system an input resistance of 20 kΩ causes a noise voltage of 15μV. If two such input resistances each of 20 kΩ are used in parallel, the total noise voltage will be abouta)15 μVb)30 μVc)7.5 μVd)10.6 μVCorrect answer is option 'D'. Can you explain this answer?
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