To expand on Jeff's comments -
Power = Volts * Current = VI
V = Current * Resistance (I*R) OR by rearrangement
I = V/R
Hence, by substitution
Power = (Current*Current)*Resistance = I^2*R OR
Power = V^2/R (that is power goes as voltage squared)
Hence, if you half the voltage, you reduce the power draw by a factor of 4. So for a resistance heater, a small reduction in voltage reduces the power draw considerably. However, transformers are not lossless either (they use a little energy due to resistance of the coils). As Jeff and Joe said, it's easier to just by a lower wattage heater. On another note, the fan contributes very little to the power draw so I doubt the startup current is much more than the operating current.
Power = Volts * Current = VI
V = Current * Resistance (I*R) OR by rearrangement
I = V/R
Hence, by substitution
Power = (Current*Current)*Resistance = I^2*R OR
Power = V^2/R (that is power goes as voltage squared)
Hence, if you half the voltage, you reduce the power draw by a factor of 4. So for a resistance heater, a small reduction in voltage reduces the power draw considerably. However, transformers are not lossless either (they use a little energy due to resistance of the coils). As Jeff and Joe said, it's easier to just by a lower wattage heater. On another note, the fan contributes very little to the power draw so I doubt the startup current is much more than the operating current.