[Mondial]

A resistor will dissipate the same amount of heat as the original ballast. No more and no less. The 49A ballast drops the line voltage by 49 V at a current of 0.3 A. The power dissipated is 49 x 0.3 = 14.7 Watts, so for safety you would want to use a 25 or 30 W resistor. By Ohms law, the resistance needed is 49 V / 0.3A = 163 Ohms. You could mount the resistor inside the original ballast housing.

A capacitor has a reactance which will allow the same voltage drop, but since the current and voltage are out of phase no power is dissipated. To drop the equivalent 49V at 60 Hz, you need a capacitor of about 10 uF. It must be non polarized as there is AC across the cap, so a film cap would be best, although a nonpolarized electrolytic might work if it had a tight enough tolerance.

The diode basically acts as a light dimmer, by cutting off half of the ac waveform, which reduces the effective voltage across the filaments. You may have to connect a resistor in series with the diode to provide the proper voltage. The one problem with using a diode alone is that there is no current limiting, so at initial turnon there is a current surge which can eventually blow out one of the filaments. The resistance of the filaments is very low when cold and increases as they heat up, so the turn on current can be very high and overstress the heater which lights up first. You may have noticed this surge in the heater of a 50L6 or 35Z5 when an AA5 AC-DC set is first turned on.