AC steady state power analysis

AC steady state power analysis - 22/12/2008 ECSE ECSE 210:...

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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
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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
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Instantaneous Power
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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
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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
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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
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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
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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
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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
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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
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AC steady state power analysis - 22/12/2008 ECSE ECSE 210:...

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