Electrical substitution heat balance ($P_{\mathrm{abs}}=P_{h,d}-P_{h,l}$)

An explicit lumped model makes the cancellation understandable. Treat the receiver as having uniform temperature $T$, heat capacity $C$, heater power $P_h$ deposited in the receiver, net parasitic input $P_{\mathrm{par}}$, and outward heat transfer $Q_{\mathrm{out}}(T,T_b)$ to a reference at temperature $T_b$. With incident radiant flux $P_{\mathrm{in}}$ and radiometric absorptance $a$, conservation of energy givesThe dark state has no admitted beam. Compare settled dark and light states, denoted $d$ and $l$; each has $dT/dt=0$. Subtracting their balances giveswhere $\Delta Q_{\mathrm{out}}=Q_{\mathrm{out},l}-Q_{\mathrm{out},d}$ and $\Delta P_{\mathrm{par}}=P_{\mathrm{par},l}-P_{\mathrm{par},d}$. At identical $T$ and $T_b$, with unchanged surroundings and parasitic heating, both correction differences vanish. Thus $P_{\mathrm{abs}}=\Delta P_h$, and $P_{\mathrm{in}}=\Delta P_h/a$. Notice the signs: extra stray thermal radiation admitted with the beam also reduces heater power, so it must be subtracted from that reduction before attributing the result to the intended beam.

Measure the heater's own terminal voltage $V_h$ and electric current $I_h$: for steady electrical heating, $P_h=V_h{I}_h=I_h^2{R}_h$, where $R_h$ is its electrical resistance. A supply voltage that includes wire voltage drops counts Joule heating in the leads too; only the portion deposited in the receiver belongs in this balance. Changes in that deposited lead heat belong in $\Delta P_{\mathrm{par}}$. Optical-electrical heating non-equivalence adds a further instrument-specific correction when the uniform-temperature model is insufficient.^[ref]