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Assuming a 2.6V forward drop V_F of LED1 and V_BATT=18V, the dissipated power of R2 is around 15.4V*20mA = 308mW which is too much for a 1206 resistor (~250mW).
Assuming a 2.6V forward drop V_F of LED1 and V_BATT=18V, the dissipated power of R2 is around 15.4V*20mA = 308mW which is too much for a 1206 resistor (~250mW).