Choosing the Right Resistor to Limit Current to 0.5A for a 20V Power Adapter

Hello. I have a problem, actually two problems.
first, I don't know what section this question should be posted in, there is so much of it here.

I have a 20V and 5A power adapter, how do I make it charge 20V and 0.5A?? it's about charging the 18V battery. Please help.

The simplest way is a resistor, but then you have to give R a lot of power and thus unnecessary conversion of energy into heat. Search for current stabilizers.

resistor does not reduce the voltage? is it so that R connected in series reduce the voltage and parallel the current? I'm green in this matter. I have a few batteries to charge from 18V screwdrivers at the construction site and I need another charger .... and when I connected the one I have, I cooked one battery

without changing the voltage? You can't, limiting the current is just ... limiting the voltage, but don't worry about it. In this case, it is best to build a simple current source based on transistors.

it decreases but it all depends on its R and charging current. I don't know if I'm counting correctly, but it turned out that R = 1 ohm or even less.

Hello!
Ohm's law has been known for quite a long time, and it can also be used here: put a RESISTOR between the power supply and the battery. If the voltage difference is 2V (20V - 18V), and this difference is supposed to cause 0.5A flow, then Ohm's Law gives the resistance: 2V/0.5A= 4 ohms. In this resistor, the power will be 2Vx0.5A=1W, so its load capacity should be at least that.
However, if the battery voltage is 20V, and the rectifier voltage is also 20V, then the current stops flowing, regardless of the resistance.
However, charging the battery is not so simple: if it is a lead acid battery that has 9 2-volt cells, the final charging voltage is 21.6V, and a power supply that gives a lower voltage simply WILL NOT CHARGE this battery.
If it is an alkaline battery (e.g. NiMH) then there are 15 cells of 1.2V, but its final voltage will be 21.3V - and again, a rectifier that gives less - WILL NOT CHARGE that battery.
The voltage thing: if your rectifier gives 20V, is it a DC (stabilized) voltage or a ripple (rectified sine wave), and under what load?
And what happens if you put a capacitor on the output of the rectifier, e.g. 100uF/40V?

Hello :)

You can do the current limitation on the universal LM317 regulator.
The diagram of how to connect everything can be found on this page (CURRENT REGULATOR): http://diyaudioprojects.com/Technical/Voltage-Regulator/

There is a calculator under the diagram. You enter the resistance of the resistor R1 (instead of a comma, give a dot, otherwise it counts wrong) and you get the current limiting intensity. If you want it to be 0.5A, you must use a 2.5Ω resistor (at least 2W so as not to burn it).

Good luck

Rufus is absolutely right. The capacitor will show the max voltage at the output. And should be charged with a slightly higher voltage.

I do not know what kind of 18V battery it is, but similarly for a 12V car battery, we have a nominal voltage of 12V and the voltage to which we charge the battery to charge it is over 14V, so in your case the final charging voltage will be ? around 21.5 volts?

The voltage of 20V is the voltage of the power supply, but which one?
Is it stabilized voltage? is it rectified and filtered voltage?
So it may turn out that your 20V is too low voltage to charge the battery, if it is rectified and filtered, because it is the peak voltage.
Perhaps it is enough to disconnect the capacitors and the rest of the system and then measure the voltage behind the bridge and if it is over 15V, there is a chance that you will charge it (but slowly, for a long time and with a small current).
For example, at 16V on the bridge, the peak voltage is 22.6V, so the difference between the peak and the battery voltage allows it to be charged, but in the final phase of charging, with a small current, judging by your statement that the battery is also of low capacity (high internal resistance)