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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 isIn a rectangular signal bandwidth $B$, the power delivered by that source contribution to the load isbecause $(R-R_L{)}^2\ge 0$. Equality occurs at $R_L=R$: this is matched available noise power. Matching halves the voltage, quarters its power spectral density, and gives $P_{\mathrm{av}}=k_{\mathrm{B}}TB$. 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 $2k_{\mathrm{B}}TR$, equivalent to $4k_{\mathrm{B}}T(R/2)$. One contribution alone is $k_{\mathrm{B}}TR$. 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.

Model an ideal resistor of constant electrical resistance $R>0$ at uniform absolute temperature $T$ by its Thévenin equivalent: a zero-mean random voltage source in series with a noiseless $R$. In the classical range its one-sided, open-circuit voltage power spectral density, per hertz, iswhere $k_{\mathrm{B}}$ is the Boltzmann constant. Thus a positive-frequency interval of width $df$ contributes $4k_{\mathrm{B}}TRdf$ to the mean square voltage. This is approximately white noise only where the electrical resistance is effectively constant and $hf\ll k_{\mathrm{B}}T$, with $h$ the Planck constant. The ideal constant spectrum cannot describe arbitrarily high frequencies.^[ref]

For a real voltage, an equivalent two-sided power spectral density is $S_v^{(2)}(f)=2k_{\mathrm{B}}TR$ over signed frequencies. With a real linear time-invariant system of frequency response $H(f)$, the same output mean square isAngle brackets denote an expected value; for this zero-mean noise the mean square is its variance. Do not integrate $4k_{\mathrm{B}}TR$ over both signs of frequency: that counts the same fluctuations twice. All integrals here use frequency $f$ in hertz, not angular frequency $\omega =2\pi f$. The amplitude spectral density is $\sqrt{4k_{\mathrm{B}}TR}$ in $\mathrm{V}/\sqrt{\mathrm{Hz}}$, whereas the root mean square voltage requires a specified measurement signal bandwidth.