I am building a voltage divider from the B+ supply to run a LED indicator that the Standby switch is closed/HT fuse is not blown/amp should be active/It also acts as a bleeder to bleed the large filter caps when the amp is powered off. Some people may call it an idiot light.
The question I have is about the current devloped across the LED resistor and the how that current interacts with the divider.
So the plan is: Vsource is 350Vdc divided by a 1 megOhm and a 22kOhm providing about 7.5Vdc for the LED circuit, which is just a 470 Ohm resistor feeding the 10mm Blue superbright LED. The LED will have 10 mA of current. So how does the current not interfere with the divider? Or if it is where is the current being supplied from...the lower 22kR, the 1 MegR or all three resistors?