According to ohm's law, why would a light bulb burn brighter the instant prior to it burning out? answer: ohm's law is R=V/I the ratio of potential difference to current is a constant for a given conductor. any resistance that does not change with temperature, voltage or the direction of charges obey ohm's law. at the instant the flow of the potential drop be higher but then due to the energy transferring to thermal the light bulb started to lose its brightens. the brighten of the bulb depends on the current that pass through, so I will assume that at the begining the current flow is larger until its reach other resistor causing the voltage to split into difference value depending on the resistance. Is it right?