Externet,
I hope I have my head screwed on correctly about this. First of all, let's talk about no interruptions and no bypass resistor. The capacitor in series with the resistance will change the current and voltage phase. If the reactance and the resistance are equal, then the current will lead the voltage by 45°. If the reactance is much larger than the resistance, then the phase difference will approach 90°. Also, if the reactance and resistance are equal, then the voltage will be 0.707 of the source, not 0.500 like it would be if the reactance were replaced by a resistor of equal ohms. Let's assume you can live with the above conditions.
You threw out some values of capacitance (0.5uf), hertz (60), voltage (120 volts rms), and interruption time (12 cycles), so let's use those. You can calculate new values if they are different. The worse thing that can happen is that the interruption occurs when the cap is fully energized and starts up again around 12 cycles later before the cap is fully deenergized. In 3 time constants, the cap will deenergize down to 5% of it max value. Its max value is 1.414 of 120 = 170 volts. 5% of 170 is 8.5 volts. One third of 12 cycles is 4 cycles, so the time constant is 4/60 = .067 seconds. Therefore the determination equation is (Rp||Rl)*0.5E-6 = 0.067 secs , where Rp is the resistor in parallel with C and Rl is the load resistor. At worse, the recovery voltage will be 170+8.5 volts. After you figure out Rp, you can determine if you can live with its power dissipation. Maybe the load resistor will be enough to deenergize the capacitor sufficiently.
Ratch