Estimating the capacitance of the capacitor.
The two-stroke engine fires each rev, and revs at 40/sec. Each shot of fuel lasts, say, 6 degrees (maybe less) of crank turn. ie 6/360 X 1/40 secs. [
4/10^4]........u
Our current is 4 mAmp (we've been told), [
4/10^3].......w
That current flows for the time stated, so it requires a
charge of (amp X sec)
=uw or [
1.6/10^6] Coulombs.
Charge (known) is proportional to volts (10^2) and capacitance (unknown).
Capacitance (C) in microFarads means multiplying by
10^6.
So,
C = 10^6 times 1.6/10^6 divided by 10^2,
ie about
0.02 muF, which is convenient if true!
Conclusion, we use a 0.02 muF capacitor.
Malc
