Calculating current a capacitor can deliver

Question

I made a simple coilgun circuit

• Capacitor 50 V 10000 uF
• DC power supply of 20 V - 5 A
• Coil of enamelled wire 18 AWG - 3.5 ohm

My calculations according to Ohm's law the current in the capacitor should be:

20 V (capacitor voltage after charged) / 3.5 ohms (resistance) = 5.7 A in the circuit

Why do I only between 1.5 to 2.2 A?

Answer 1

1. Your clamp meter can't capture the peak fast enough.
2. The current depends on the impedance of the coil, not just the resistance.
3. The current depends in the internal resistance of the capacitor as well as its impedance.

Your calculated current is probably incorrect, and your meter probably can't capture the real current because it changes too fast.

Answer 2

Shown below is current that flows through the coil after the "launch" switch is closed.

Coil inductance is not known, but coil resistance is very likely near 3.5 ohms - it is a relatively easy thing to measure. 3.5 ohms is assumed to be the dominant resistance in the capacitor's discharge path, and is held constant for all five plots. Capacitance is 10,000 microfarads and is assumed charged initially to 20V:

You can see that coil inductance greatly influences how current flows after the capacitor begins discharging. Note too, that maximum current occurs after the switch closes. One is most likely only interested in that first current peak, and the time when it occurs.
For the 0.03H coil, peak current occurs 16ms after switch closes (switch closes at the 0.1 second point). Current falls to near-zero about 0.1 second after switch closes.

Answer 3

This is how the capacitor current behaves if the inductance is 1 H. You can see that at 3.5s the current is essentially zero. The clamp meter cannot measure this.