Lect07-Ch5-SinusoidalAnalPt1

Lect07-Ch5-SinusoidalAnalPt1 - Steady-State Sinusoidal...

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1. Sinusoidal Voltages and Currents 2. Phasors 3. Complex Impedance 4. Circuit Analysis with Phasors and Complex Impedance Steady-State Sinusoidal Analysis* *Hambley Chapter 5 – Sinusoidal Analysis
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Sinusoidal Voltage (Hertz) frequency 2 1 T 2 period (degrees) angle phase ) (rad./sec. frequency angular ge peak volta = = = = = = = = π ω θ T f T Vm ) cos( ) ( + = t Vm t v WHERE (amplitude)
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° = = = ° = ° ° + = ° = ° + = 60 200 10 , ) 60 200 cos( 10 90 30 200 cos 10 ), 90 - cos(z sin(z) : using 30 200 sin 10 - , , Vm thus t (t) v ) t ( (t) v ) t ( (t) v x x x Sinusoidal Voltage Sinusoids: reference is cosine EXAMPLE: ω θ
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Average Power in Sinusoidal System R ) ( 1 ) ( 1 ) ( 1 ) ( E : period one in delivered Energy ) ( ) ( : R resistance to by v(t) delivered Power 2 0 2 0 2 0 0 T 2 = = = = = = T T T T T dt t v T dt R t v T Pavg dt t p T T E Pavg dt t p R t v t p
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RMS: Equates Average AC & DC Power Dissipation in Resistor • DC (Direct Current): I DC , Power P DC = I DC R Average DC power dissipated: • AC (Alternating Current): I AC =I o cos( ω t + θ ) Average AC power dissipated: Equate: P AC,avg = P DC,avg RMS: I RMS ± I DC,equivalent = I o / 2 0.707 I o 2 P DC , avg = 1 T I DC 2 R T dt = I DC 2 R = 2 π T Circular frequency P AC , avg = 1 T I o 2 cos 2 ( t + ) R 0 T dt = I o 2 R T 1 2 0 T 1 + cos2( 2 t T + ) dt = I o 2 R 2 I DC 2 R = I o 2 R/2
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Root Mean Square (RMS) and Average Power I rms = 1 T i 2 ( t ) dt 0 T P avg = I rms ( ) 2 R V rms = 1 T v 2 ( t ) dt 0 T P avg = V rms ( ) 2 R Voltage Current Definition: Equivalent average AC/DC heating of resistor, over single period P AC , avg = I o 2 R 2 = I o 2 R = I rms 2 R
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Root Mean Square (RMS) of a Sinusoid m m rms T m rms m V . V V dt t V T V t V t v 707 0 2 : to reduces ) ( cos 1 ) cos( ) ( 0 2 2 = = + = + = θ ω See Text Page 203 for Development
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Example 5.1 : Power Delivered to Resistor v ( t ) = 100cos(100 π t ) and R = 50 Ω V rms = V m = 100 = 70.7 P avg = V rms ( ) 2 R = 70.7 ( ) 2 50 = 100 W p(t ) = v 2 ( t ) R = 100cos(100 t ) ( ) 2 50 = 200cos 2 (100 t ) also, f = 100 2 = 50 T = 1 50 = 20 ms T See Book pp. 205-206 for RMS of Triangular Waveform Ã
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