Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 4 (Q.No. 45)
45.
A device has two resistances 300 Ω and 200 Ω respectively. The noise voltages due to these two resistance calculated separately are 5.38 μV and 4.38 μV respectively. When both of them are used together, the noise voltage will be
Answer: Option
Explanation:
.
Discussion:
3 comments Page 1 of 1.
Asha said:
4 years ago
Vn^2 = vn1^2+vn2^2.
Gives 6.937microvolt.
Gives 6.937microvolt.
Manu david said:
5 years ago
Can anyone please explain this?
RUSHIKESH DESHMUKH said:
6 years ago
But it depends on whether it is connected I series or in the shunt.
For series, it would be (vn1)^2+(vn2)^2.
Where vn1 is noise voltage due to R1 and vn2 is the noise voltage due to R2.
For series, it would be (vn1)^2+(vn2)^2.
Where vn1 is noise voltage due to R1 and vn2 is the noise voltage due to R2.
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