Thevenin equivalents preserve the load behavior

A useful view from two terminals

A linear resistive network with independent sources can be represented at a chosen port by a Thevenin voltage $V_{TH}$ in series with resistance $R_{TH}$. The equivalent predicts the voltage and current of loads connected to that port. It does not preserve internal branch currents or every component's power. State the two port terminals before calculating.

$V_{TH}$ is the port's open-circuit voltage. To find $R_{TH}$ for this independent-source example, set ideal voltage sources to zero, replacing them with short circuits; set ideal current sources to zero, replacing them with open circuits. Dependent sources must remain active and require a test-source method, which is beyond this first example.

A loaded divider without starting again

A 5 V divider uses upper 10 kΩ and lower 10 kΩ resistors. Its open-circuit output is 2.5 V. With the 5 V source shorted, both resistors connect output to ground, so $R_{TH}=10\,\mathrm{k}\Omega\parallel10\,\mathrm{k}\Omega=5$ kΩ. Attach a 10 kΩ load. Output becomes $2.5\times10/(5+10)=1.666667$ V and load current is 0.166667 mA.

Original port: divider out, ground
Equivalent: 2.5 V source --- 5 kohm --- out --- load --- ground

The open-circuit voltage remained 2.5 V, but it was never a promise of 2.5 V under all loads. A source impedance of 5 kΩ explains the droop. Confirm using the original network: lower 10 kΩ in parallel with the load gives 5 kΩ, and 5 V times 5/(10+5) gives the same result.

What the equivalent cannot tell you

The Thevenin resistor's dissipation is not generally the total heating inside the original divider. Use the original circuit for component ratings. Likewise, a nonlinear diode network may only admit a local small-signal equivalent around one bias point. Use a source model whose scope matches the question, and retain the original diagram in your design notes so that an elegant simplification does not erase a physical limitation.