this is an easy question. it deals with ratios. And known voltages to calculate a rough but close answer. Let's investigate use answers giving in a past comment. One was 16000kv. Another is 5 amps. Let's say the primary voltage is 12 volts at 5 amps. And it produces a 16000 volt spark. What's the miliamperage of the secondary voltage. So we take the two voltages 12÷16000 is a ratio of .00075. Now take that ratio and multply that by the 5 amps from the primary coil. So .00075×5 is .00375 miliamps. Now I'll admit I'm not formally educated. But I do study and I'm an extra class ham radio operator and a former auto mechanic building race engines for friends that have won races with the engine. One time holding the pole at road Atlanta ga. In front of all those V8 mustangs. So if that builds confidence in what I've propose to you. then the better for you.

12 volt car coils take from 3 amps to 5 amps depending on performence. Point driven car coils get full 12 volts during starting at designed full current and after started and rpms increas during speed the switch after starting returns to run position where resistance is switched into the circuit to keep high current from burning up coil.high rpms cause heat so resistance andthe coil has oil inside to help dissapate the internal heat.the primary is wound on thr outside of the many turns of inner fine wire and that coil takes milliamps. The # 19 heavy wire on the outside is 150 feet long. I reuse this wire as it unwinds very eazy and use it to design 12 volts to 36 volt power supplies with a mosfet circuit driven from a 12 volt battery

At my work, we repair small engines such as chainsaws, string trimmers, cutoff saws, etc etc. 9 out of 10 times. the ignition coils outpuit 16kV as a "free voltage" and 8 kV as a "working voltage". IE 16 000 volt with out it trying to spark under the environment of compressino in the cylinder and 8000 volts under the regular working conditions of a sparkplug.

active| newest | oldestthis is an easy question. it deals with ratios. And known voltages to calculate a rough but close answer. Let's investigate use answers giving in a past comment. One was 16000kv. Another is 5 amps. Let's say the primary voltage is 12 volts at 5 amps. And it produces a 16000 volt spark. What's the miliamperage of the secondary voltage. So we take the two voltages 12÷16000 is a ratio of .00075. Now take that ratio and multply that by the 5 amps from the primary coil. So .00075×5 is .00375 miliamps. Now I'll admit I'm not formally educated. But I do study and I'm an extra class ham radio operator and a former auto mechanic building race engines for friends that have won races with the engine. One time holding the pole at road Atlanta ga. In front of all those V8 mustangs. So if that builds confidence in what I've propose to you. then the better for you.

Input or output?

L

To be honest, I don't know. But that should help someone else.

L

If you mean the input current, for a short duration it's high I and relatively low V

Steve