The smaller capacitors will not pass as much current at 60Hz, so the bulb will not light (or glow as brightly)
We can calculate the expected wattage:
For 1uF:
1 / (2 * pi * 60 * 1e-6) = 2652 Ohms.
If we add this to the bulb resistance (say 50 ohms as it will be somewhere in between cold and hot) we get 2700 ohms (we can be rough here as the bulb resistance makes little difference)
For a 115V line, that's 115 / 2700 = ~43mA.
So the apparent wattage will be around 115V * 43mA = ~5W
The real wattage dissipated by the bulb will be much less, around 100mW.
We calculate this by I^2 * R. So with a 50 ohm filament resistance we get 0.043^2 * 50 = 92mW.
Not enough to light the bulb up. For the 2.2uF the result would be around 900mW (may glow a bit?)
If we do the same for the 22uF capacitor, we get:
1 / (2 * pi * 60 * 22e-6) = 120 Ohms.
sqrt(120^2 + 285^2) = 309 Ohms.
115V / 309 = ~372mA
115V * 372mA = ~43W (~39W dissipated by filament)
So you see the larger cap makes a big difference.
You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric, and is ideally infinity.
The formula you use is correct, but the value you come out with is NOT the ESR, is the leakage resistance. Once the capacitor is charged, if you leave it it slowly discharges trough the leakage resistor with a time constant \$R_{leak}\cdot C\$, so \$R_{leak}\$ is what you calculated, approximately \$50M\Omega\$, that is plausible.
To calculate the ESR you need to measure how long does it take the capacitor to discharge through a much smaller resistor, let's call it \$R_{dis}\$. When you discharghe the capacitor through \$R_{dis}\$ the total resistance through which it discharges is actually \$R_{dis}+R_{ESR}\$, so using the very same formula you used for the leakage resistance you can calculate the ESR.
But is it really that easy? Of course not.
The ESR is hopefully quite small, tenths of milliohms if you have a very good capacitor up to a few ohms. Since in the formula you have \$R_{dis}+R_{ESR}\$ you don't want an eccessive \$R_{dis}\$ to mask \$R_{ESR}\$. Ok then! Why don't we choose \$R_{dis}=0\Omega\$? Easy question:
- \$0\Omega\$ resistance does not exist. But i can make it small!
- Time. You need to be capable to measure how long does it take to the capacitor to discharge.
If you charge the capacitor to a certain voltage it will take \$\tau\ln{2}\approx0.7\cdot\tau\$ where \$\tau=RC\$. If \$R=R_{ESR}+R_{dis}=1\Omega+1\Omega=2\Omega\$ and \$C=680\mu F\$ that's less than 1ms. Without proper equipment, that is a properly set oscilloscope, you can't easily measure the ESR.
Last but not least, keep in mind that electrolytic capacitors values have a tolerance of \$\pm10\%\$, that leads to:
$$
R_{ESR}=\frac{t_{dis}}{\left( C\pm C/10\right)\ln{2}} - R_{dis}
$$
with the above numbers, t=1ms, C=\$680\mu F\$, \$R_{dis}=1\Omega\$, this translates to:
$$
R_{ESR}\in\left[0.91,1.33\right]\Omega
$$
That's 10% down and over 30% up.
Best Answer
You want a rapidly rising magnetic field at ns timescales, and you worry about cap esR?? ESL is more likely to be the gotcha, but even then....
You may want to start by thinking about the inductance of your magnet, and then how much voltage you will need to make the current change quickly in that inductor (Inductors really do not do step changes in current).
By the time you are playing in ns speeds you really want to be thinking EM not just E or H as often your circuit geometry means you have to think fields and waves, not lumped element.