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
9.76 V
1 μV
23.56 μV
6.94 μV
Answer: Option
Explanation:

.

Discussion:
3 comments Page 1 of 1.

Asha said:   4 years ago
Vn^2 = vn1^2+vn2^2.

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.

Post your comments here:

Your comments will be displayed after verification.