Pay attention to the arithmetic.

If your load takes 0.5A @ 9V

R = 9V / 0.5A = 18Ω

Your load represents a resistor of 18Ω.
Your voltage divider becomes 6Ω + [18Ω] where the [18Ω] is already in the load and is not an additional resistor.
This is a far cry from your 10KΩ + 30KΩ.
They both present a 1:3 ratio but the current through the 10KΩ resistor is too low to power your project or even do any damage.
Hence nothing was damaged.

The recommended solution is to get a 7809 regulator or a buck DC-DC converter.
If you are driving a string of LEDs that take 0.5A then what you need is a constant 0.5A supply, not a constant voltage supply.

Tell us what you are trying to do.

 

Out of interest. If Circuit A is 9V 0.5A and Circuit B is 12V 6A. I divide the 12V into 12V and 9V. When the 12V circuit pulls 6A, I guess the 9V circuit will also see 6A?

Would this explain why my circuit went pop?

Kevin

 

Out of interest. If Circuit A is 9V 0.5A and Circuit B is 12V 6A. I divide the 12V into 12V and 9V. When the 12V circuit pulls 6A, I guess the 9V circuit will also see 6A?

Would this explain why my circuit went pop?

Kevin

NO!

Your house probably has an electric panel that can supply something like 240 V at 50 A of current (I'm guessing what a typical household service panel is like in the U.K., but the specifics don't really matter). Does that mean that when everything is turned off in your house that you are drawing 50 A? Does that mean that if you are using a hair dryer that turning on the stove makes your hair dryer draw more current? No. The current rating on a voltage supply is telling you the maximum current that it can supply.

You have a 12 V 6 A supply. This means that it will supply 12 V and can deliver as much as 6 A. Don't try to get it to deliver more than 6 A. But if you only power something that needs 10 mA, it will supply just that 10 mA.

When you connect a voltage divider using 10 kΩ and 30 kΩ resistor you will get about 9 V at the junction of the two resistors and you will be drawing about 0.3 mA from the 12 V supply. If you try to power a 9 V device that needs more than about 0.03 mA of current, then this won't work.

In order to use a resistive voltage divider to provide 0.5 A at 9 V to your circuit, you need to be drawing about 5 A of current through the divider when no load is connected to it. Even if your 12 V supply can do that, you don't want to do that. You'll be wasting about 60 W have power and have to deal with getting rid of the heat.

Instead do as others have suggested. Either use a 9 V regulator IC that can deliver the needed 500 mA or us a DC-DC converter. Which is better depends on what you are powering with it. If you are powering lights and such, then the noise associated with the DC-DC converter will be negligible and it will be a better solution.

As to why your circuit popped, we can't tell unless you provide a schematic of what you actually connected to what (as opposed to what you had intended to connected to what) and a better description of how it failed.

 

Jeez @WBahn - consider this noob schooled.