# HW 3s - Hw 3 Solutions 1.20 Since 2 p = 0-30><6 6x12...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Hw 3 Solutions 1.20 Since 2 p = 0 -30><6 + 6x12 + 3V0 + 28 + 28x2 - 3x10 = 0 72 + 84 + 3V0 = 210 or 3V0 = 54 W=l§¥ 2.7 La)6branches,4nodes , Cb) ’7 branC‘hesj s—r’odes 2.8 IZmA 8'“ ﬂ 9 mA =i1+12, i1=—4mA 9=8+i2, iz=1mA 12+i3=9, i3=—3mA 2.12 Forloop 1= -20 -25 T10 +v1 = 0 —> v1 =35v Forloop 2= —10 +15 -V2 = 0 —> v» =5v For loop 3, 47; +172 +V3 = 0 —> v; = 30v 2.21 Applying KVL, -15 + (1+5+2)| + 2 V)( = 0 But V)( = SI, —15 +8| + 10| =0, I = 5/6 Vx = 5| = 25/6 = 4.167 V 2.22 At the no de: KCL requires that 33+10+2v0=0——> v0: 4.444v The current through the controlled source is i_= 2V0 = -8.888A and the voltage across it is v=(6+4) io (\vhereio =vox'4)= 10::— = —ll.lll Hence= mm: (-8.888)(—11.111) = 93.75 w ...
View Full 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

HW 3s - Hw 3 Solutions 1.20 Since 2 p = 0-30><6 6x12...

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

View Full Document
Ask a homework question - tutors are online