This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: through the 3k resistor. The problem must be done by treating the 3k resistor as a load and finding the Thevenin Equivalent Circuit of everything else where A & B are the open circuit points of it. ● Use a source transformation on the 18V source and 6k resistor to reduce the problem to a mesh problem with only current sources. ● Use a WyeDelta transformation on the three 6k resistors. (Draw new circuit diagrams at each step.) This is the result and also combined the parallel 4k resistors ● Calculate the open circuit voltage: With V T = 14 V, R T = 6 k Ω and a 3 k Ω load, the current will be I = 14/9 mA. A Using KCL, you can figure out the currents in the branches, then the voltage drops from A to B; V AB = 2x1 + 2x4 + 2x2 = 14 V 2mA 4mA 1mA 1mA 1mA With the sources removed, the resistance from A to B is clearly 6k Ω ....
View
Full Document
 Spring '10
 Shankar
 Volt, Resistor, 3k, 2 w, Thévenin's theorem, maximum power transfer, 4k

Click to edit the document details