PHYS345 Electricity and Electronics

Answer for P4.16

Open-Circuit Condition:
Circuit for Thevenin voltage
 

The current source pushes 3 A of current through the lower 2 ohm resistor, thus there is a 6 V difference across it with polarity as shown,

In the top portion, the 8 V source across the 2+j2 ohm impedance would create a current of 2-j2 A through the top 2 ohm resistor; thus it has a voltage of 4-j4 V across it.

The sum of the two, 10-j4 V is the Thevenin voltage.

 
Killing the Sources:
Circuit for Thevenin impedance
 

Shorting the voltage source and opening the current source yields this combination of passive elements. The j2 ohm and 2 ohm elements at the top are in parallel and have the equivalent impedance of 1+j ohm. The Thevenin impedance is thus 3+j ohm.

Using the Generator:
Thevenin generator
The total impedance of the equivalent Thevenin circuit is 3-j ohm, so the current through the resistor is
I = (10-j4)/(3-j) = (10.8\/-21.8) / (3.16\/-18.4) = 3.41\/-3.4a


"http://www.physics.udel.edu/~watson/phys345/protected/exercises/answers/P4-16.html"
Last updated Oct. 1, 1998.
Copyright George Watson, Univ. of Delaware, 1998.