Root-mean-square thermal noise associated with a resister should be calculated. The fate of the thermal noise when the bandwidth is reduced to 100 Hz should be predicted. Concept introduction: Data given: Temperature (T) = Room temperature = 298 K Bandwidth ( Δ f ) = 1 MHz Resistance of the resistor (R) = 1.0 M Ω Thermal noise for a resistive circuit element can be given as υ ¯ r m s = 4 k T R Δ f Where, k = Boltzmann's constant = 1 .38 × 10 − 23 J/K T = Temperature in K R = Resistance in Ω Δ f = Bandwidth

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Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213
BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

Solutions

Chapter 5, Problem 5.9QAP
Interpretation Introduction

Interpretation:

Root-mean-square thermal noise associated with a resister should be calculated. The fate of the thermal noise when the bandwidth is reduced to 100 Hz should be predicted.

Concept introduction:

Data given:

Temperature (T) = Room temperature = 298 K

Bandwidth ( Δf ) = 1 MHz

Resistance of the resistor (R) = 1.0 MΩ

Thermal noise for a resistive circuit element can be given as

υ¯rms=4kTRΔfWhere,k = Boltzmann's constant = 1.38×1023 J/KT = Temperature in KR = Resistance in ΩΔf = Bandwidth

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