= Resistor-capacitor equilibrium noise
{title2=$\langle v_C^2\rangle=k_{\rm B}T/C$}
Connect one physical <resistor> $R$ across an ideal lossless <capacitor> $C$, with no applied source and an ideal readout that draws no <electric current>. Represent the <resistor> by its noisy <Thévenin equivalent>. The resulting <RC low-pass frequency response> gives the stationary <capacitor> voltage <power spectral density>
$$
S_C^{(1)}(f)=\frac{4k_{\rm B}TR}{1+(2\pi fRC)^2}.
$$
Its mean square follows by the <effective noise bandwidth> calculation:
$$
\langle v_C^2\rangle=4k_{\rm B}TR\int_0^\infty\frac{df}{1+(2\pi fRC)^2}
=4k_{\rm B}TR\frac1{4RC}=\frac{k_{\rm B}T}{C}.
$$
This is also the <equipartition theorem>: the mean <capacitor energy> is $\tfrac12 C\langle v_C^2\rangle=\tfrac12 k_{\rm B}T$. Raising $R$ makes the low-<frequency> spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final <variance>.
The integral to infinity belongs to the ideal classical model. For a physical <electrical circuit>, assume <thermal equilibrium>, linear components with negligible extra dissipation, and a corner far below the <frequencies> where parasitic effects, frequency-dependent <electrical resistance>, or $hf\sim k_{\rm B}T$ invalidate the model; the omitted high-<frequency> tail must be negligible. A finite-bandwidth instrument measures only part of this <variance>. Its input <capacitance> contributes to the total $C$, and its finite input <electrical resistance> introduces both loading and its own <thermal electrical noise>.
For $T=300\,\mathrm K$ and $C=1\,\mathrm{nF}$, the predicted equilibrium <variance> is $4.141947\times10^{-12}\,\mathrm{V^2}$ and the <root mean square> voltage is $2.0352\,\mu\mathrm V$. At $R=10\,\mathrm{k}\Omega$, $RC=10\,\mu\mathrm s$ and $B_{\rm n}=25\,\mathrm{kHz}$. At $R=100\,\mathrm{k}\Omega$, $RC=100\,\mu\mathrm s$ and $B_{\rm n}=2.5\,\mathrm{kHz}$; the equilibrium <root mean square> voltage is identical.
\Image[/thermal-electrical-noise-rc-equilibrium.png]
{title=Changing <electrical resistance> changes noise timescales but preserves $k_{\rm B}T/C$}
{description=Original calculated comparison at $T=300\,\mathrm K$ and $C=1\,\mathrm{nF}$. Left: the physical parallel <resistor> and <capacitor> and their noisy <Thévenin equivalent>. Middle: equilibrium one-sided voltage spectra for $10$ and $100\,\mathrm{k}\Omega$ have the same full integral. Right: mean square voltage grows toward the same equilibrium value from an exactly discharged initial <capacitor>. Curves describe ensemble <expected values>, not measured or simulated voltage traces.}
To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the <resistor> at $t=0$ while maintaining its <temperature>. The <RC low-pass filter>'s causal <impulse response> is $h(t)=e^{-t/(RC)}/(RC)$ for $t\geq0$. Classical ideal <white noise> with two-sided <power spectral density> $2k_{\rm B}TR$ gives
$$
\langle v_C(t)^2\rangle=2k_{\rm B}TR\int_0^t h(u)^2\,du
=\frac{k_{\rm B}T}{C}\left(1-e^{-2t/(RC)}\right).
$$
The factor $2k_{\rm B}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\rm B}TR$. They give the same long-time limit. The <variance> approaches equilibrium with <time constant> $RC/2$, although a deterministic initial voltage decays with <time constant> $RC$. If the <capacitor> already starts in <thermal equilibrium>, its initial <variance> supplies the missing exponentially decaying term and the total <variance> stays constant.
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