You have taken the first step along the path used in many contemporary (with us) fan-cooling systems, namely, reducing fan noise and cooling effect by dropping the supply-voltage to the fan. The voltage-drop may be arbitrary, as in your case, or mediated by a circuit responding to a thermistor incorporated into the fan. Your way, if you take account of the necessary heat dissipation from the dropping resistor—dropping 2V across a fixed 33Ω resistor to supply 10V to the fan needs that the resistor safely dissipate 0.15W, which must be added to the (reduced) heat generated by the fan—you may need a 1-W or larger (wattage) resistor to cope, but it is certainly not an impossible ask. (I have assumed a 12-V 80-mA 40-mm square fan for the example.) Use of a larger (say, 75Ω resistance) resistor to drop the voltage to the fan further to 8V will increase the wattage to be dissipated in the resistor with the square of the current passed by the resistor and with the first power of the resistor's ohmic value to 0.64W. You can cope with that by juggling the resistor's wattage rating, but powering a fan built for 12-V supply with 8V may cause it to fail to shift much air. In fewer words, it is a trade-off.
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