I've been trying to build a lead-acid charger for quite a while. Burned a few transformers, tried bulb in series with the battery: not enough current. Today I pulled a capacitor from an induction motor. I wired that in series with the battery and get quite a bit of current and the battery voltage stays within it's limits so it's not a direct short.
What I don't understand is how the extra voltage is dropped.
Say I feed in 120 volts and the current is 5 amps. The output is near 24 volts (2x 12v batteries in series). Would the current be multiplied and the battery taking in 20+ amps or would the battery see only 5 amps? The capacitors became warm but not hot. I don't have a functioning amp meter so I can't figure this one out on my own.
Side note: the AC is rectified by a full wave bridge rectifier, so the battery gets charged with DC current. My current setup with the capacitor seems to charge the battery very quickly, but not dangerously fast.
What I don't understand is how the extra voltage is dropped.
Say I feed in 120 volts and the current is 5 amps. The output is near 24 volts (2x 12v batteries in series). Would the current be multiplied and the battery taking in 20+ amps or would the battery see only 5 amps? The capacitors became warm but not hot. I don't have a functioning amp meter so I can't figure this one out on my own.
Side note: the AC is rectified by a full wave bridge rectifier, so the battery gets charged with DC current. My current setup with the capacitor seems to charge the battery very quickly, but not dangerously fast.