I tried the both methodes and the two supplies method worked better for me beacose with the one resistor from the voltage devider was overheatig and also i didn't see difference between using the both methodes.

With no power wasted in any resistors the 2-supply arrangement had to be better. And was my guess correct, that those were two laptop computer power supplies?

 

With no power wasted in any resistors the 2-supply arrangement had to be better. And was my guess correct, that those were two laptop computer power supplies?

No they are 12v adapters for monitors but i modified them to give 20 volts

 

No they are 12v adapters for monitors but i modified them to give 20 volts

THANKS! But my guess was close.

 

THANKS! But my guess was close.

Yeah they are the same the difference is 1-2 volts
Laptops use 18 or 19 vdc

 

What were the values of the resistors where one of them overheated?
Each resistor in the input voltage divider has 20V across it. Then for a 1/4W resistor to overheat its value must be less than 1.6k.
The Samplifier IC anyo circuit I posted shows 33k and the datasheet for the similar LM1875 shows 22k.

 

What were the values of the resistors where one of them overheated?
Each resistor in the input voltage divider has 20V across it. Then for a 1/4W resistor to overheat its value must be less than 1.6k.
The Samplifier IC anyo circuit I posted shows 33k and the datasheet for the similar LM1875 shows 22k.

I used 33k and 4.7k and the 4.7k overheated,not instantly but after 10 minutes it was smoking

 

I used 33k and 4.7k and the 4.7k overheated,not instantly but after 10 minutes it was smoking

The voltage divider resistors should both have the same value so that they produce "half the supply voltage".
33k in series with 4.7k as a divider have a current of 40V/(33k + 4.7k)= 1.06mA. The heating in the 4.7k resistor is 1.06mA squared x 4.7k= 5.3mW which is almost nothing.
A 4.7k 1/4W resistor will overheat when it has a voltage of 34V.