On Wednesday, April 15, 2020 at 5:57:07 AM UTC-7, wrote:

I measured the inrush current once again and found that the vertical of the scope was set for a 1X probe instead of the 10X actually being used. This meant that the peak current was 90A instead of 9A, which is a bit high. I added a 1.1 ohm resistor and the peak current dropped to 6A.

My understanding of that test and the numbers don't add up. I'm thinking you are charging a cap with a 12 V battery and measuring the current with and without and added 1.1 ohm series resistor?

If I combine the two measurements to find the open circuit voltage of the battery and overall wiring resistance I get 7.07 volts and 0.078 ohms.

Probably I don't understand what you are testing. We seem to have lots of time to kill. Could you send a picture of the circuits with the probe attached?

The circuit is very simply: a capacitor and a switch to a battery. In the second case, a 1.1 ohm resistor is added in series between the capacitor and the battery. You are mistaking a d.c. circuit and an a.c. circuit: when the switch closes current will flow from the battery to the capacitor. This current flow is modeled by a differential equation:

i = C dv/dt
or
dv/dt = i / C
Integrated wrt time gives you:
V = 1/C Int[i dt]

In other words, the voltage on the capacitor increases as current flows into it. If you look at it in the steady state, or d.c., you will miss this entirely.

I measured the current with a 0.0015 ohm current shunt and an oscilloscope.