Norton's Theorem - Example of A Norton Equivalent Circuit

Example of A Norton Equivalent Circuit

In the example, the total current I_total 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.

Read more about this topic: Norton's Theorem

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