Norton's Theorem - Example of A Norton Equivalent Circuit

Example of A Norton Equivalent Circuit

In the example, the total current Itotal is given by:

I_\mathrm{total} = {15 \mathrm{V} \over 2\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \| (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega)} = 5.625 \mathrm{mA}.

The current through the load is then, using the current divider rule:

I = {1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega \over (2\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega)} \cdot I_\mathrm{total}
= 2/3 \cdot 5.625 \mathrm{mA} = 3.75 \mathrm{mA}.

And the equivalent resistance looking back into the circuit is:

R_\mathrm{eq} = 1\,\mathrm{k}\Omega + 2\,\mathrm{k}\Omega \| (1\,\mathrm{k}\Omega + 1\,\mathrm{k}\Omega) = 2\,\mathrm{k}\Omega.

So the equivalent circuit is a 3.75 mA current source in parallel with a 2 kΩ resistor.

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