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उदाहरण R = 10 mΩ → nearest E24 10 mΩ, which gives 100 mV at 10 A — that is the burden voltage your load no longer gets (0.83 % of a 12 V rail) — and 1 W of heat

उदाहरण A 2512 chip at ~40 °C/W runs 40 °C above ambient (65 °C), which is 100 % of the 1 W it is still rated for at 25 °C — past the 60 % anyone should design to, so go one package larger or accept a smaller sense voltage. At 100 ppm/°C that rise alone shifts the reading by 0.4 %, so sense the two terminals directly (Kelvin) and keep the copper around it cool

$R_s=\frac{V_{\mathrm{sense}}}{I_{\mathrm{fs}}}$ — but the sense voltage is the real decision, because it alone fixes the heat: $P={I_{\mathrm{fs}}}^2{R}_s=V_{\mathrm{sense}}\times I_{\mathrm{fs}}$, with R cancelling out. 100 mV at 10 A is a full watt no matter which resistor you buy.

The mistake is choosing a generous sense voltage for a clean signal and then putting it in a package that cannot get rid of the power: $\mathrm{\Delta T}=P\times R_{\mathrm{th}}$, and a shunt that runs hot changes value while you are reading it. A package's watts are quoted at 70 °C and fall to nothing at its maximum operating temperature, so the ambient you give below decides how much of that rating is actually left. Driving an amplifier instead? $V_{\mathrm{sense}}=\frac{V_{\mathrm{adc}}}{G}$ is the sense voltage its gain implies at the top of the ADC's range — a ceiling, not a target, so the shunt is rounded DOWN to a stocked value in that mode.