= Thermal noise worked example
For an original hypothetical calculation, stipulate $T=300\,\mathrm K$, $R=10\,\mathrm{k}\Omega$, and an ideal rectangular measurement <signal bandwidth> $B=20\,\mathrm{kHz}$. Using $k_{\rm B}=1.380649\times10^{-23}\,\mathrm{J/K}$ gives
$$
S_v^{(1)}=1.6567788\times10^{-16}\,\mathrm{V^2/Hz},\qquad
\sqrt{S_v^{(1)}}\approx12.872\,\mathrm{nV}/\sqrt{\mathrm{Hz}},
$$
$$
\langle v_{\rm oc}^2\rangle=3.3135576\times10^{-12}\,\mathrm{V^2},\qquad
v_{\rm oc,rms}\approx1.8203\,\mu\mathrm V.
$$
A matched $10\,\mathrm{k}\Omega$ <electrical load> receives $0.91016\,\mu\mathrm V$ <root mean square> from the source <resistor>, giving $P_{\rm av}=8.283894\times10^{-17}\,\mathrm W$. These are stipulated model predictions, not laboratory measurements. Increasing $R$ raises the <open-circuit voltage> <noise> but does not increase the matched available <power>. Restricting the measurement <signal bandwidth> reduces <root mean square> <noise> as its square root; it improves <signal-to-noise ratio> only if the desired <signal> is preserved.
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