Electronic – Where do the power leads attach to this simple sound circuit

Power dissipation is not preserved in Thévenin and Norton equivalent circuits.

For a simple example, consider a 1 volt voltage source with 1 ohm across it. The power dissipation is 1 watt.

But the Thévenin equivalent circuit is just the voltage source by itself. 0 watts.

Remember that, in circuit analysis, "wires" are simply a tool we use to show connected components. You should try to see circuits as a combination of nets; a net simply describes how component pins are connected to each other. I know there can only be two different voltage readings in your circuit because your circuit only has two nets: ground, and 10V. Think about it. Every pin of every component in this circuit (in this case, a voltage source and resistor) is hooked up to only those two nets. No other nets exist.

Remember, voltage is always measured between nets, so there's really only one voltage to measure in your circuit, since you only have two nets -- ground, and 10V. It makes no sense to talk about the voltage at different points along a wire; just imagine the wire didn't exist, and the component(s) were hooked directly to each other. For example, if you hooked the little ground symbol straight up to your resistor (without the wire labeled "B"), that wouldn't affect the operation of the circuit at all.

Explanation: voltage doesn't change in an ideal wire. In other words, if you were to build that circuit, and measure the voltage (relative to ground) at every point along the circuit, you'd measure 10V all along the wire that hooked your positive voltage source to the resistor. Then, on the "bottom side" of the resistor, you'd measure 0V, all along the wire that hooked the resistor to ground.

Voltage only gets dropped across loads. Resistors, incandescent bulbs, motors, LEDs (and other semiconductor devices), etc.

So, if you added a second 1k resistor in series below the one you already have, you'd (again) measure 10V at the top of the first resistor, then you'd measure 5V in between the two resistors, and then you'd measure 0V at the bottom of the second resistor.

NB: We're talking about ideal wires here. In reality, every wire is a combination inductor/capacitor/resistor. It drops voltage. It stops sudden changes in current. It stops sudden changes in voltage. Wires are nasty things!

Also, you may have gotten confused by the "edge cases" that violate circuit theory; For example, if you replaced the resistor in your circuit with a wire, it would appear that the wire would have to drop 10V. But that circuit is invalid, and you can't apply analysis techniques to it.