107_31 - 1 Electrical Engineering Technology EET 107...

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Unformatted text preview: 1 Electrical Engineering Technology EET 107 Introduction to Circuit Analysis # 31 Professor Robert Herrick Purdue University © EET 107 - 31 Introduction to Circuit Analysis 2 Electrical Engineering Technology Zeroing Supplies And Superposition Purdue University © EET 107 - 31 Introduction to Circuit Analysis 3 Overview u u Zeroing Supplies Superposition Purdue University © EET 107 - 31 Introduction to Circuit Analysis 4 Zeroing a Voltage Supply 0V E + - R 0V Short ! Natural resistance ! Purdue University © EET 107 - 31 Introduction to Circuit Analysis 5 Zeroing a Voltage Supply 0V E + - R E R Replace E with a SHORT ! Purdue University © EET 107 - 31 Introduction to Circuit Analysis 6 Zeroing a Current Supply 0A R I 0A Open ! Natural resistance ! Purdue University © EET 107 - 31 Introduction to Circuit Analysis 7 Zeroing a Current Supply 0A R I R I Replace I with an OPEN ! Purdue University © EET 107 - 31 Introduction to Circuit Analysis 8 Voltage Supply – Internal Resistance Voltage supply • Ideal Internal Resistance = 0 Ω • Looks like a SHORT to other supplies SHORT E=0V e=0V Purdue University © EET 107 - 31 Introduction to Circuit Analysis 9 Current Supply – Internal Resistance Current supply • Ideal Internal Resistance = ∞ Ω • Looks like an OPEN to other supplies OPEN I=0A OPEN i=0A Purdue University © EET 107 - 31 Introduction to Circuit Analysis 10 Overview u u Zeroing Supplies Superposition Purdue University © EET 107 - 31 Introduction to Circuit Analysis 11 Superposition A technique to solve a multiple supply circuit. 1. For each supply, zero the other supplies and draw each its equivalent circuit 2. Find the desired I’s and V’s for each of these single supply circuits 3. Superimpose (add up) each desired I or V to find net result Purdue University © EET 107 - 31 Introduction to Circuit Analysis 12 Superposition - example 1 - find I 2A 9V I 3Ω + - + - 3V Previous method using laws only: ENET = 9 V – 3 V = 6 V clockwise clockwise RT = 3 Ω I = 6V/3Ω = 2A Purdue University © EET 107 - 31 Introduction to Circuit Analysis 13 Superposition - example 1 - find I 3Ω I 9V contribution 9V + - 2Α + - 3V contribution 3V I = I’ + I” = 3 A + (−1 A) = 2 A I’ 9V I” 3Ω + - 3Α Purdue University © 3Ω 1Α EET 107 - 31 + - 3V Introduction to Circuit Analysis 14 Superposition - example 2 - find I I 6mA 3kΩ + - 2 supplies Purdue University © 6kΩ 9V 2 sub-circuits EET 107 - 31 Introduction to Circuit Analysis 15 Superposition - example 2 - subcircuits I 6 mA contribution 3kΩ 6mA 6kΩ + - 9V 9V contribution I = I’ + I” I’ 6mA I” 6kΩ 3kΩ 3kΩ Purdue University © EET 107 - 31 6kΩ + - 9V Introduction to Circuit Analysis 16 Superposition - example 2 - 6mA contribution I 6mA contribution 6mA 6mA 3kΩ 3kΩ 6mA I’ 6kΩ 6kΩ 2mA Purdue University © + - 9V CDR I’ = 3 kΩ / 9 kΩ × 6 mA = 2 mA EET 107 - 31 Introduction to Circuit Analysis 17 Superposition - example 2 - 9V contribution I 6mA 3kΩ 6kΩ + - 9V 9V contribution I” Ohm’s Law I” = − 9V / 9kΩ = −1 mA Purdue University © EET 107 - 31 3kΩ 1mA 6kΩ + - 9V Introduction to Circuit Analysis 18 Superposition - example 2 - superimpose results I 6 mA contribution 3kΩ 6mA 6kΩ 1mA + - 9V 9V contribution I = 2 mA + (−1 mA) = 1 mA I’ 6mA 3kΩ I” 6kΩ 3kΩ 2mA Purdue University © EET 107 - 31 1mA 6kΩ + - 9V Introduction to Circuit Analysis 19 Superposition - example 2 - rest of circuit 6mA + 15V 6mA - 5mA 3kΩ 1mA + 15V - 6kΩ + 6V - + - 9V V6kΩ = 1 mA × 6 kΩ = 6 V I3kΩ = 6 mA − 1 mA = 5 mA V3kΩ = 5 mA × 3 kΩ = 15 V Vcurrent supply = V 3kΩ = 15 V Purdue University © EET 107 - 31 Introduction to Circuit Analysis 20 Superposition u Currents superimpose u Voltages superimpose u Power does not superimpose Purdue University © EET 107 - 31 Introduction to Circuit Analysis 21 Power Does NOT Superimpose NOT Example: power dissipated by 6 kΩ I’ power P ’ = (2 mA )2 × 6 kΩ = 24 mW I’’ power P’’ = (1 mA ) 2 × 6 kΩ = 6 mW Superimposed power P’’ = 24 mW – 6 mW = 18 mW Actual power Purdue University © P = (1 mA) 2 × 6 kΩ = 6 mW EET 107 - 31 Introduction to Circuit Analysis 22 + 24Ω 9A 8Ω 3A V - 4 supplies - + Example 3: Superposition - find V 9V 1Ω 2Ω + - 9V 4 sub-circuits ? Not necessarily! Reduce first? Purdue University © EET 107 - 31 Introduction to Circuit Analysis 23 - + Superposition - find V + 24Ω 9A 8Ω 3A 9V V 1Ω 2Ω - + - 9V • Combine parallel current supplies • Combine parallel resistors • Combine series voltage supplies • Combine series resistors Purdue University © EET 107 - 31 Introduction to Circuit Analysis 24 + 24Ω 9A 8Ω V 3A - • I=9A–3A=6A - + Superposition - find V – reduce circuit 1Ω 9V 2Ω + - 9V into top node • parallel R = 24 Ω // 8 Ω = 6 Ω • E = 9 V + 9 V = 18 V • series R = 1 Ω + 2 Ω = 3 Ω Purdue University © EET 107 - 31 Introduction to Circuit Analysis 25 Superposition - find V – simplified circuit 6A 6Ω 3Ω + V + - 18V - • Current supply I=6A • Parallel resistor R=6Ω • Voltage supply E = 18 V • Series resistor R=3Ω Purdue University © EET 107 - 31 Same V Introduction to Circuit Analysis 26 Superposition - example 3- 2 subcircuits 6A contribution + 6A V + - - + 6Ω 6Ω 6A 3Ω V’ + 3Ω 6Ω - Purdue University © 18 V 18V contribution V” 3Ω + - 18V - EET 107 - 31 Introduction to Circuit Analysis 27 Superposition - example 3- 2 subcircuits 6A contribution + 6Ω 6A V 3Ω + - 18V - V’ = 6 A × (6 Ω // 3 Ω) = 6 A × 2 Ω = 12 V Ω) + 6A 6Ω V’ 3Ω - Purdue University © EET 107 - 31 Introduction to Circuit Analysis 28 Superposition – example 3 - 2 subcircuits + 6A 6Ω V 3Ω + - - 18 V 18V contribution V” = (6 Ω / 9 Ω) 18 V = 12 V + 6Ω V” 3Ω + - 18V - Purdue University © EET 107 - 31 Introduction to Circuit Analysis 29 Superposition - example 3 - superimpose 6A contribution + 6Ω 6A V 3Ω + - - 18 V 18V contribution V = V’+ V” = 12 V + 12 V = 24 V + 6A 6Ω V’ + 3Ω 6Ω - Purdue University © V” 3Ω + - 18V - EET 107 - 31 Introduction to Circuit Analysis 30 Superimpose u DC signals: I superimpose V superimpose u AC signals: i superimpose v superimpose Purdue University © EET 107 - 31 Introduction to Circuit Analysis 31 Circuit Analysis with AC Supplies - example 3kΩ 18V 1kHz iL + 3mA 1kHz es is 6kΩ Assume rms Individually solve then superimpose (add) es is è i’L è i”L Purdue University © EET 107 - 31 Introduction to Circuit Analysis 32 Circuit Analysis with AC Supplies - example iL 2mA 3kΩ 18V 1kHz + es 3mA 1kHz open 6kΩ es contribution I’L = 18 V / 9 kΩ = 2 mArms Purdue University © EET 107 - 31 Introduction to Circuit Analysis 33 Circuit Analysis with AC Supplies - example short 18V 1kHz 3kΩ + iL 3mA 1kHz is 1mA 6kΩ is contribution I”L = 3 kΩ / 9 kΩ × 3 mA = 1 mArms Purdue University © EET 107 - 31 CDR Introduction to Circuit Analysis 34 Circuit Analysis with AC Supplies - example iL 3mA 3kΩ 18V 1kHz + 3mA 1kHz iL = i’L + i”L 6kΩ Superposition IL rms = 2 mA + 1 mA = 3 mArms Purdue University © EET 107 - 31 Introduction to Circuit Analysis 35 Circuit Analysis with AC Supplies - example 3kΩ 18V 1kHz + 3mA 1kHz iL 3mA + 6kΩ vL − + 18 Vrms − VL rms = 3 mA × 6 kΩ = 18 Vrms VL p = √2 × 18 Vrms = 25.5 Vp Purdue University © EET 107 - 31 Introduction to Circuit Analysis 36 Circuit Analysis with AC Supplies - example Sketch VL p = 25.5 V p sine wave 1kHz T = 1/f = 1/1kHz = 1 ms vL(t) +25.5V 1ms 0V t -25.5V Purdue University © EET 107 - 31 Introduction to Circuit Analysis 37 Circuit Analysis with AC Supplies - example vL(t) +25.5V 1ms 0V -25.5V t 51.0Vpp VL pp = 2 × 25.5 Vp = 51.0 Vpp Purdue University © EET 107 - 31 Introduction to Circuit Analysis 38 Contribution Powers Superimpose? example example 3kΩ 18V 1kHz iL + 3mA 1kHz 6kΩ es P'average= (2 mA) 6kΩ = 24mW is P''average= (1mA) 6 kΩ = 6mW Assume rms 2 2 Paverage total signal = (3 mA )2 6 kΩ = 54 mW Purdue University © EET 107 - 31 ΝΟ Introduction to Circuit Analysis 39 Summary Zeroing Supplies And Superposition Purdue University © EET 107 - 31 Introduction to Circuit Analysis ...
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