# AC steady state power analysis - ECSE ECSE 210 Electric...

This preview shows pages 1–11. Sign up to view the full content.

ECSE 210: Electric Circuits ECSE 210: Electric Circuits 2 22/12/2008 Chapter Chapter 11 11 Steady Steady-State Power Analysis State Power Analysis 1

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

View Full Document
Next Topic: Power 12/22/08 Recall: Power as a function of time + ) ( t i Passive Sign Convention: Power supplied to Electric Circuit/ Circuit Element - ) ( t v circuit/circuit element: ) ( ) ( ) ( t v t i t p ower as a function of time when the 2 Power as a function of time when the currents and voltages are sinusoids
Instantaneous Power

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

View Full Document
Instantaneous Power 22/12/2008 + ) ( t i Passive Sign Convention: ower upplied Electric Circuit/ Circuit Element - ) ( t v Power supplied to circuit/circuit element: In the case of sinusoidal xcitation in the steady state: ) ( ) ( ) ( t v t i t p ) cos( ) ( V m t V t v  excitation in the steady state: ) cos( ) ( I m t I t i ) cos( ) cos( ) ( I V m m t t V I t p s( s( 1 s( s( Recall:   ) cos( ) cos( 2 ) cos( ) cos( 4
Instantaneous Power 22/12/2008 ) cos( ) cos( ) ( I V m m t t V I t p ) 2 cos( 2 ) cos( 2 ) ( I V m m I V m m t V I V I t p Constant Periodic in time 1. For circuits with time varying inputs (e.g., sinusoidal put) the stantaneous ower is a function of time input) the instantaneous power is a function of time denoted as p(t). 2. For circuits with sinusoidal inputs in the steady state , (t) as a constant component and a periodic p(t) has a constant component and a periodic component. 5

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

View Full Document
Example 1 22/12/2008 ) ( t i 50 ) 10 cos( 10 ) ( 4 t t v V Given v(t), need to find i(t ) + - + ) ( t v 12mH ) 10 cos( 10 4 t V V 10 0 o V 20 0 0 - 120 50 50 j L j Z o 4 . 67 130 V 10 0 o 6 9 7 4 o ma I Z 130 67.4 o 76.9   67.4 7 0 s( 6 4 o mA ) 4 . 67 10 cos( 9 . 76 ) ( t t i p ( t ) 76.9 cos(10 4 t 67.4 o ) 10 cos(10 4 t ) mW ) 4 . 67 10 2 cos( 2 769 ) 4 . 67 cos( 2 769 4 o o t 6
Example 1 22/12/2008 0 10 t) (V) 0 0.2 0.4 0.6 0.8 1 1.2 x 10 -3 -10 v( t 100 00 0 i(t) (mA) 0 0.2 0.4 0.6 0.8 1 1.2 -3 -100 0.5 1 (W) 0 0.2 0.4 0.6 0.8 1 1.2 10 -3 -0.5 0 p(t) Time (s) Watts ) 4 . 67 10 2 cos( 38 . 0 15 . 0 ) ( 4 o t t p x 10 7

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

View Full Document
Average Power 22/12/2008 In the case of periodic excitation in the steady-state, the instantaneous power p(t) is a periodic function of time. The average value of p(t) over one period (the verage power is a useful indicator of how average power ) is a useful indicator of how much power is absorbed/supplied. The average power is NOT a function of time for a periodic excitation! For a general periodic function x(t), such that x(t+T)=x(t) the average value X is given by g gy T t ve o d x X ) ( 1 t ave o T 8
Average Power 22/12/2008 In the case of Example 1: ) 4 . 67 10 2 cos( 38 . 0 15 . 0 ) ( 4 o t t p P avg 1 T p ( ) d 0 T 1 T 0.15 d 0 T 1 T 0.38cos(2 10 4  67.4 o ) d 0 T 15 . 0 zero P avg 0.15 W 9

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

View Full Document
Average Power 22/12/2008 General case ) 2 cos( 2 ) cos( 2 ) ( I V m m I V m m t V I V I t p constant P avg 1 T I m V m 2 cos( V   I ) d 0 1 T I m V m 2 cos(2  V   I ) d 0 zero verage Power P avg I m V m 2 cos( V   I ) 10 Average Measure of ac power
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/29/2011 for the course ECSE 291 taught by Professor N/a during the Spring '11 term at McGill.

### Page1 / 65

AC steady state power analysis - ECSE ECSE 210 Electric...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online