= Thermal noise loading
An external <electrical load> changes the terminal <noise>. Connecting a purely resistive load $R_L>0$ to the noisy <resistor> gives the <voltage divider> ratio $R_L/(R+R_L)$. The source <resistor>'s contribution at the load is
$$
S_{v_L}^{(1)}(f)=4k_{\rm B}TR\left(\frac{R_L}{R+R_L}\right)^2.
$$
In a rectangular <signal bandwidth> $B$, the <power> delivered by that source contribution to the load is
$$
P_{R\to L}=4k_{\rm B}TB\frac{RR_L}{(R+R_L)^2}\leq k_{\rm B}TB,
$$
because $(R-R_L)^2\geq0$. Equality occurs at $R_L=R$: this is matched available noise <power>. Matching halves the voltage, quarters its <power spectral density>, and gives $P_{\rm av}=k_{\rm B}TB$. Dividing the squared <open-circuit voltage> by $R$ would instead give four times this available <power>.
These expressions isolate the source <resistor>'s contribution. A real load <resistor> also generates <thermal electrical noise>. At the same <temperature> its independent contribution produces the equal reverse <power> flow, so the net thermal <power> transfer is zero at <thermal equilibrium>. For example, two equal <resistors> at the same <temperature> have total terminal <power spectral density> $2k_{\rm B}TR$, equivalent to $4k_{\rm B}T(R/2)$. One contribution alone is $k_{\rm B}TR$. The fictitious noiseless series $R$ in the <Thévenin equivalent> is not an additional physical source: adding its <noise> again would double count the original <resistor>.
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