The physics behind an ignition system consist of points, capacitor and ignition coil​

The first step

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The problem is from a 12V DC system creates a spark with 25kVolt.

It is hard to step up a DC voltage to a higher level with simple components as points and capacitor.
An ignition coil has a ratio of typical 1 to 100 and a transformer only works with AC current and not DC, note that the ignition coil is a transformer.
There must be a way to generate 250V AC in to the ignition coil.

When breaking a current through a coil (inductor) it will generate high voltage, this can be seen when the current through a SU pump is switching on and off. To save the switch in the SU pump it was previous a capacitor across the switch, now it is a diode. The function of the capacitor was to reduce how fast the current was changing and that reduced the voltage.

The formulas for this is U=L*dI/dt .  U is in volt that will be generated, L is the inductance in Henry in the coil, I is the current in amps and t is the time in seconds.
The dI/dt is how fast the current was changed, I.e. dI will go from 4 amps down to 0 amps, then dI will be -4. The time dt is very short, mille seconds. That’s end up with a voltage that is quite high.
You cannot change L or dI (the current through the coil). The thing you can affect is the time with a capacitor.

Without any capacitor the current will drop to zero almost instant or say 1 thousands of a second.
This will give a voltage of let say 1000V. With a capacitor the time will be much lower so the switch will survive.

These figures are from a standard car relay. The relay consume 0.15A, that’s why the “lower” voltage (280V). Note the polarity, it is negative, the reason for this is that the current is going from 0.15A down to zero amps.

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