Solution 4 - ECE 205 Section A, B, and C. Homework #4...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE 205 Section A, B, and C. Homework #4 Solutions 2.22 At the node, KCL requires that 0 0 v 2 10 4 v + + = 0 v 0 = –4.444V The current through the controlled source is i = 2V 0 = -8.888A and the voltage across it is v = (6 + 4) i 0 (where i 0 = v 0 /4) = 10 111 . 11 4 v 0 = Hence, p 2 v i = (-8.888)(-11.111) = 98.75 W 2.26, If i 16 = i o = 2A, then v = 16x2 = 32 V 8 4 A 8 v i == , 42 8 A, i 16 vv i 2481 6 16 8 4 2 30 A x iiiii =+++ = +++= 22222 16 2 8 4 4 8 2 16 960 W Pi R x x x x = = +++ = or 30 32 960 W x v x = 6 Ω 2v 0 + v 0 - 10A
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2.30. We start by combining the 6-ohm resistor with the 2-ohm one. We then end up with an 8-ohm
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/03/2010 for the course ECE 201 taught by Professor C during the Spring '10 term at Miami University.

Page1 / 2

Solution 4 - ECE 205 Section A, B, and C. Homework #4...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online