PHYS345 Electricity and Electronics

Detailed Norton Generator Example

Most of the numerical work has already been done in the Detailed Thévenin Generator Example.

Redrawn to emphasize Norton approach

Norton Generator

Norton generator

Determination of Open-Circuit Voltage

Please refer to the Thévenin Example for details on finding Voc = 3.71 V.

Determination of Short-Circuit Current

Please refer to the Thévenin Example for details on finding Isc = 1.05 mA.

Determination of Norton Resistance

RTh = Voc / Isc
RTh = 3.53 k

Norton Generator Ready for Use!

Norton generator ready for use

Now the current can be found by finding the equivalent resistance of the parallel combination of the load resistance and the Norton resistance; this resistance is 2.22k. Application of Ohm's law using the equivalent resistance and the Norton current then yields the voltage across the pair of resistors: (2.22k)(1.05 A) = 2.33 V. Thus the current through the load resistor is I = (2.33 V) / 6k = 0.39 mA.


"http://www.physics.udel.edu/~watson/phys345/examples/norton-example.html"
Last updated Sept. 15, 1998.
Copyright George Watson, Univ. of Delaware, 1998.