Voltage divider calculator
The output above assumes nothing is connected to it. The moment a load draws current, it sits in parallel with R2 and the output sags. The rule of thumb is that the divider current should be at least ten times the load current, so a divider feeding a 1 mA load wants at least 10 mA flowing through it, which for a 12 V supply means a total resistance of about 1.2 kΩ and 120 mW burnt continuously. For anything drawing real current, use a regulator.
A voltage divider outputs Vin × R2 ÷ (R1 + R2). Two equal 10 kΩ resistors on 12 V give 6 V at 0.6 mA. The output only holds if the load draws far less current than the divider itself: a factor of ten is the usual minimum.
How to design a voltage divider
Dividers are for sensing, not for supplying. Scaling a battery voltage down into an ADC input is the textbook use, and it works because an ADC input draws almost nothing. Choosing the values is a trade: high resistances waste less power but are more susceptible to noise and to the input impedance of whatever is reading them; low resistances are stiffer but burn current continuously, which matters enormously in a battery device. For an ADC divider, tens of kilohms is the usual compromise.
Questions
Vout = Vin × R2 ÷ (R1 + R2), where R2 is the resistor between the output and ground.
The load sits in parallel with R2, lowering the effective bottom resistance. Keep divider current at least ten times load current.
Only microamp loads. For anything real, use a regulator. A divider has no regulation and wastes power constantly.
For an ADC input, tens of kilohms balances noise against current. For a low-impedance reference, go lower and accept the current.
Yes. Two 5% resistors can put the output off by nearly 10%. Use 1% parts wherever the ratio matters.