NIELIT 2017 July Scientist B (CS) - Section B: 31

Question

At a room temperature of $300K$, calculate the thermal noise generated by two resistors of $10K\Omega$ and $20K\Omega$ when the bandwidth is $10KHz$.

• A. $1.2868\times10^{-6}V, 1.819\times10^{-6}V$
• B. $6.08\times10^{-6}V, 15.77\times10^{-6}V$
• C. $16.66\times10^{-6}V, 2.356\times10^{-6}V$
• D. $1.66\times10^{-6}V, 0.23\times10^{-6}V$

Answer 1

Option A. $1.2868 × 10^{-6}V, 1.819 × 10^{-6}V$

Noise voltage Vn = √(4R KTB)

Where, $K = 1.38×10^{-23} J/K$, the Boltzmann constant

T is the absolute temperature in kelvins,

R is the resistance

B is the bandwidth at which the power $P_{n}$ is delivered.

Noise voltage by individual resistors

$V_{n1}= √(4R_{1} KTB) = √(4×10×10^{3} ×1.38×10^{-23} ×300× 10×10^{3}) $

$= √(1.656 × 10^{-12}) = 1.2868 × 10^{-6}V$

$V_{n2}= √(4R_{1} KTB) = √(4×20×10^{3} ×1.38×10^{-23} ×300× 10×10^{3}) $

$= √(3.312 × 10^{-12}) = 1.819 × 10^{-6}V$