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Thévenin / Norton calculator

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Example Thévenin 5 V behind 1 kΩ ≡ Norton 5 mA across 1 kΩ. Open-circuit the terminals and you measure 5 V; short them and 5 mA flows.

Example Into 1 kΩ: 2.5 V across the load, 2.5 mA through it, 6.25 mW delivered while the source resistance wastes 6.25 mW (50 % efficient). You are sitting on the match point.

Any linear source, however many parts are inside it, behaves at one pair of terminals exactly like a voltage source behind a series resistance — or, just as truthfully, like a current source across a parallel resistance. The two swap in either direction with $I_n=\frac{V_{\mathrm{th}}}{R_{\mathrm{th}}}$ one way, $V_{\mathrm{th}}=I_n\times R_n$ the other, and $R_n=R_{\mathrm{th}}$ both ways: the resistance never changes, only how the source is written. Edit either the voltage or the current below and the other follows.

Hang a load on the terminals and the equivalent tells you what it gets: $V_L=V_{\mathrm{th}}\times \frac{R_L}{R_{\mathrm{th}}+R_L}$. The delivered power peaks when the load matches the source resistance — $P_{\max }=\frac{{V_{\mathrm{th}}}^2}{4R_{\mathrm{th}}}$ — and the classic mistake is reading that as "best". At the match point the source resistance burns exactly as much as the load, so it is 50 % efficiency: matching is what you want from an antenna or a sensor, never from a power supply.