enee204Lectures_16_17_Gomez

# enee204Lectures_16_17_Gomez - Welcome to ENEE 204 Basic...

This preview shows pages 1–4. Sign up to view the full content.

1 Lecture 16 1 Welcome to ENEE 204 Basic Circuit Theory Lecture 16 Node and Mesh Analysis Midterm II: Nov. 3, 2005 Chapters 4-6 (up to Mesh Analysis) 2 Determining the elements of the admittance matrix and source vector Diagonal elements = sum of admittances connected to the specific numbered node Off-diagonal elements = - (sum of admittances between the pair of specific nodes) Current source vector entries = sum of current sources entering and exiting the specific node THIS IS ALL YOU NEED TO UNDERSTAND AND REMEMBER. YY Y Y Y = I I I 11 12 1n 21 22 2n n1 n2 nn 1 2 n s (1) s (2) s (n) v Lecture 16 3 Nodal analysis is straightforward for circuits with current sources in parallel with admittances ANYTHING HARDER??? Current source is in series with an admittance Voltage source is in series with an admittance Voltage source is directly between two nodes Lecture 16 4 Nodal analysis is straightforward for circuits with current sources in parallel with admittances Current source is in series with an admittance Voltage source is in series with an admittance Voltage source is directly between two nodes Y 3 I s1 I s2 Y 1 Y 2 + _ + _ + _ a I s b a Y b V s Y b V s a

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

View Full Document
2 Lecture 16 5 Case 1: Current source in series with an admittance I s V x + _ Y v a b This component has no effect on the rest of the circuit. TRICK: Eliminate Y (replace by a short) from the circuit for the purpose of nodal analysis. Once v a and v b are known, find V x . Lecture 16 6 After the problem is solved ( a , b , etc are calculated) we can readily find V x . I s V x + _ Y a b V Y + Y can change the power produced by the current source. It does not effect the rest of the circuit. Y I V s y = y x b a V V v v = Y I v v V s b a x + = Lecture 16 7 Example 2 3A 10 9A 3 = 0 1 5 = 3 9 10 2 10 1 10 1 10 2 2 1 v v No effect: replace with short = 50 70 2 1 v v Lecture 16 8 Continued … 2 3A 10 10 9A 3 = 0 1 5 = 50 70 2 1 v v + V x2 V x1 = 1 = 70 Volts + V x1 V x2 = 2 + 15 V = 65 15 + 2 2 v Y I V s x = 2 2 15 v V V x =
3 Lecture 16 9 CASE 2: A voltage source in series with an admittance Y v a b V s I s = V s . Y Y a b Trick: convert voltage source into a non-ideal current source. Lecture 16 10 Example 12 Ω 96 V 10 Ω 72 A 20 Ω 60 Ω i 12 Ω 8 A Norton equivalent 10 Ω 72 A 20 Ω 60 Ω 12 Ω 8 A Lecture 16 11 Continued: Solving the problem using node potential 10 Ω 72 A 20 Ω 60 Ω 12 Ω 8 A 2 3 = 0 1 = 8 72 - 2 . 0 1 . 0

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.

## This note was uploaded on 04/07/2008 for the course ENEE 204 taught by Professor Gomez during the Fall '04 term at Maryland.

### Page1 / 16

enee204Lectures_16_17_Gomez - Welcome to ENEE 204 Basic...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online