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Wired power for battery tools

Likely, I'm at the bottom of ranked beginners here. I do try reading electronics textbooks, but the actual education yield is small. The abstract information needs experience to become knowledge and experience raises questions that can't be readily answered by an inanimate information source. So, thank goodness for forums.
I was given a collection of 18v battery tools, without the batteries, of course. I'd like to power them from a wired source. An external dc power supply that can supply about 20 amps. I rewound the secondary on a microwave transformer (13A/120V) with 12 ga wire to get 19VAC output. Without a load, the adapted transformer primary draws 3A. I connected a 100A full wave rectifier to it and got an 18VDC output. Since I had a junk drawer large electrolytic capacitor, could I use it to smooth the dc somewhat? I've read that to run a dc motor the dc current doesn't have to be smooth, so this is academic. When I connected the capacitor (63V/8000uf) in parallel to the output of the rectifier, the output voltage became 28VDC. But why? I know this is the dumbest of all questions here.
120V electricity is a sinewave that alternates and goes from 0V to a peak voltage then to 0V again and to the alternate peak polarity.

The 18V is an average voltage of half the sinewave that goes from 0V to the peak voltage of 1.414 times the average voltage, over and over.
18V x 1.414= 25.5V peak that the capacitor charges to. You measured higher at 28V maybe because the voltage was not loaded with a high current.

A DC motor expects to be powered from smooth DC, not pulses from a rectifier. The pulses might damage the motor and the fluctuating speed might be bad. It will also produce a hum sound.

Without a load, your modified transformer should barely draw any 120V current, yours is wasting 120V x 3A= 360W!