Problem Statement
A resistor $R=1\ \text{k}\Omega$ is at $T=300\ \text{K}$. Find the rms thermal voltage noise in a bandwidth $\Delta f=10\ \text{kHz}$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Johnson-Nyquist theorem: the spectral power density of thermal noise in a resistor:
Step 2 — Apply the relevant physical law or equation: $$S_V(f) = 4k_BTR \quad \text{(V}^2\text{/Hz)}$$
Step 3 — Solve algebraically for the unknown: RMS voltage in bandwidth $\Delta f$:
Step 4 — Substitute numerical values with units: $$V_{rms} = \sqrt{4k_BTR\,\Delta f}$$
$$= \sqrt{4\times1.38\times10^{-23}\times300\times1000\times10^4}$$
$$= \sqrt{4\times1.38\times10^{-23}\times3\times10^9}$$
$$= \sqrt{1.656\times10^{-13}} = 1.29\times10^{-7}\ \text{V} = 0.129\ \mu\text{V}$$
Step 5 — Compute and check the result: This fundamental noise sets the limit for sensitive measurements. It is unavoidable (a consequence of the fluctuation-dissipation theorem).
Step 6: Result: $V_{rms} \approx 0.13\ \mu\text{V}$.
Worked Calculation
$$S_V(f) = 4k_BTR \quad \text{(V}^2\text{/Hz)}$$
$$V_{rms} = \sqrt{4k_BTR\,\Delta f}$$
$$= \sqrt{4\times1.38\times10^{-23}\times300\times1000\times10^4}$$
Johnson-Nyquist theorem: the spectral power density of thermal noise in a resistor:
$$S_V(f) = 4k_BTR \quad \text{(V}^2\text{/Hz)}$$
RMS voltage in bandwidth $\Delta f$:
$$V_{rms} = \sqrt{4k_BTR\,\Delta f}$$
$$= \sqrt{4\times1.38\times10^{-23}\times300\times1000\times10^4}$$
$$= \sqrt{4\times1.38\times10^{-23}\times3\times10^9}$$
$$= \sqrt{1.656\times10^{-13}} = 1.29\times10^{-7}\ \text{V} = 0.129\ \mu\text{V}$$
This fundamental noise sets the limit for sensitive measurements. It is unavoidable (a consequence of the fluctuation-dissipation theorem).
Result: $V_{rms} \approx 0.13\ \mu\text{V}$.
Answer
$$\boxed{= \sqrt{1.656\times10^{-13}} = 1.29\times10^{-7}\ \text{V} = 0.129\ \mu\text{V}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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