Electronic – Checking for a short in a circuit with a large capacitor

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.

The quick answer is no. A capacitor won't help, AND you don't need it. If you have a 400 - 800 AHr battery, it will provide you with all the juice you need for a small appliance such as you've specified. Additionally, without going into details, practical capacitors simply don't have the energy and power density you want.

If you're worried about transient load currents (as you should be), you need to pay attention instead to your inverter specifications. Inverters don't generally have much excess margin on their load current specs, so you need to get one which seems oversized compared to your average load. The exact amount of this overage will depend on exactly what appliances you're using.

Recharging via exercise bike won't hurt, but it won't help a whole lot, either. The general rule of thumb for humans doing work (like driving a generator) is 100 watts. This is (roughly) the solar power incident at noon on a 1 square foot solar cell. Figure 20% efficiency for the cell, and a 1-hour workout on the bike is more or less the equivalent of an extra square foot of solar cell over the course of a bright day. Like I say, it won't hurt, but its major impact will be to make you feel better about exercising and not wasting energy.

By the same token, running appliances for short periods of time need to be taken into account, but in general they are not dominant in your energy budget. Let's say a blender draws 10 amps at 120 volts. For a really good inverter with a 12-volt input, that's an inverter draw of 100 amps. But, it only lasts for 1 minute, so the total battery drain is 100 amps x 1/60 hours, or about 1.5 amp-hours.