This preview shows pages 1–2. Sign up to view the full content.
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 Wye-Delta 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 and a 3 k 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 ....
View Full Document
This note was uploaded on 07/17/2011 for the course E E 240 taught by Professor Shankar during the Spring '10 term at University of Alberta.
- Spring '10