2013+F+314+Ex+2+Pr_2Bsol

r1 2 r2 6 r3 18 r4 3 r5 24 r6 27

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rrent flows from terminal a to terminal b when they are connected with a wire. Thus the 6 Ω resistor is shorted out and carries no current. KVL, or mesh current equation: +(10 V ) + I ⋅ (4 Ω) = 0 thus I = −2.5 A € Note that the directions of the mesh current I and the Norton equivalent current are opposite, because the mesh current flows clockwise from terminal b to terminal a. Thus the Norton equivalent current equals +2.5 A. Answer: A © 2013 Alexander Ganago Page 6 of 30 Last printed 2013- 12- 08 6:33 PM File: 2013 F 314 Ex 2 Pr+sol+instr.docx 2013 Fall EECS 314 Exam 2 Problem 4 In the circuit shown on the diagram, Vs = 12 V Is = – 4 A (note the sign!) R1 = 2 Ω R2 = 6 Ω R3 = 18 Ω R4 = 3 Ω R5 = 24 Ω...
View Full Document

This note was uploaded on 02/06/2014 for the course EECS 314 taught by Professor Ganago during the Spring '07 term at University of Michigan.

Ask a homework question - tutors are online