# Heat transfer, heat from current

by Frostfire
Tags: current, heat, transfer
 Sci Advisor HW Helper PF Gold P: 2,532 Heat transfer, heat from current OK, that's going to be relatively straightforward: the efficiency is just $$\eta=\frac{I^2R_L}{I^2R_L+I^2R_S}=\frac{R_L}{R_L+R_S}=\frac{1}{1+R_SA/\rho L}$$ where $I$ is the current, $R_L=\rho L/A$ is the load resistance (the resistance of the heater), $R_S$ is the source resistance (the resistance of the power supply and wiring), $\rho$ is the resistivity of the heater material, and $L$ and $A$ are the length and cross-sectional area of the thermal heater. This is essentially the principle of power matching; you maximize power transfer when the load resistance matches the source resistance and the source resistance is minimized. Does this answer your question?