{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ACReview - ECE225 REVIEW PROBLEMS Problem 1(Nodal Analysis...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
ECE225 REVIEW PROBLEMS SPRING 2010 Problem 1 (Nodal Analysis) Problem 2 (Mesh Analysis) Problem 3 (Superposition) Problem 4 (Superposition) Problem 5 (Thevenin Equivalent Circuit) 1
Background image of page 1

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

View Full Document Right Arrow Icon
Problem 1 (Nodal Analysis) 4 0 V o - + o ^ V j4 -j5 - + 2 0 = ˜ V o - 4 6 0 o 2 + ˜ V o - j 5 + ˜ V o j 4 4 6 0 o 2 = 1 2 + 1 - j 5 + 1 j 4 ˜ V o 2 = (0 . 5 - j 0 . 05) ˜ V o ˜ V o = 2 0 . 5 - j 0 . 05 = 40 10 - j 1 = 40 101 6 - 5 . 71 o = 3 . 98 6 5 . 71 o V 2
Background image of page 2
Problem 2 (Mesh Analysis) o 6sin 2t 4 - + - + 2 H 4 0.25 F o i - j4 o 10 0 V ^ ^ ^ 6 -90 V V o I 2 I 1 I -j2 + 10cos 2t V - + ( 10 = 4 ˆ I 1 - j 2( ˆ I 1 + ˆ I 2 ) - j 6 = j 4 ˆ I 2 - j 2( ˆ I 1 + ˆ I 2 ) ( (4 - j 2) ˆ I 1 - j 2 ˆ I 2 = 10 - ˆ I 1 + ˆ I 2 = - 3 ˆ I 1 = fl fl fl fl fl 10 - j 2 - 3 1 fl fl fl fl fl fl fl fl fl fl (4 - j 2) - j 2 - 1 1 fl fl fl fl fl = 10 - j 6 4 - j 4 ˆ I 2 = fl fl fl fl fl (4 - j 2) 10 - 1 - 3 fl fl fl fl fl fl fl fl fl fl (4 - j 2) - j 2 - 1 1 fl fl fl fl fl = - 2 + j 6 4 - j 4 ˆ I o = ˆ I 1 + ˆ I 2 = 10 - j 6 4 - j 4 + - 2 + j 6 4 - j 4 = 8 4 - j 4 = 2 1 - j 1 = 2 6 0 o 2 6 - 45 o = 2 6 45 o = 1 . 414 6 45 o i o ( t ) = 2 cos(2 t + 45 o ) = 1 . 414 cos(2 t + 45 o ) A 3
Background image of page 3

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

View Full Document Right Arrow Icon
Problem 3 (Superposition) DC CIRCUIT - + 4 i (t) o 1 H
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}