EE203-SUNYBuffalo-11-Chapter10-02

EE203-SUNYBuffalo-11-Chapter10-02 - SMALL for Big Things...

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Unformatted text preview: SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Recall: rms value (from Chapter 9.1) EE 203 Circuit Analysis 2 Lecture 11 Chapter 10.3 rms Value, Power Calculations rms (root-mean-square) value of a periodic function = square root of the mean value of the squared function = √(Vm2 ½) Kwang W. Oh, Ph.D., Assistant Professor SMALL (Nanobio Sensors & MicroActuators Learning Lab) Department of Electrical Engineering University at Buffalo, The State University of New York 215E Bonner Hall, SUNY-Buffalo, Buffalo, NY 14260-1920 Tel: (716) 645-3115 Ext. 1149, Fax: (716) 645-3656 kwangoh@buffalo.edu, http://www.SMALL.Buffalo.edu EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 1/10 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 2/10 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York rms value rms Given Given an equivalent resistive load, R, and an equivalent time period, T, the rms value of a sinusoidal source delivers the same energy to R as does a dc source of the same value. A dc source of 100 V delivers the same energy in T seconds that a sinusoidal source of 100 Vrms delivers, assuming equivalent load resistance resistance. Energywise, the effect of the two sources is identical. Effective value being used interchangeably with rms value. rms value: effective value EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 3/10 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 4/10 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Effective Value Example 10.3 Vrms ⎛ 625 ⎞ ⎟ 2 Vrms ⎜ 2⎠ P= =⎝ 50 R A ramp with 120 V, 100 W a resistance of 1202/100 (=144 Ω) Irms = Ieff = 120/144 = 0.833A Ipeak = 0.8332 = 1.19A EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo V = m = 625 2 2 Lecture 11 | Chapter 10 | 2/5 | 5/10 Vrms = I rms R → I rms 2 V = rms = R 625 2 50 2 2 ⎛ 625 ⎞ ⎛ 625 ⎞ ⎛ 625 ⎞ ⎜ ⎜ ⎟ ⎟ ⎜ 2⎠ 2⎠ 2 ⎟ 50 = ⎝ 2 P = I rms R = ⎜ 50 = ⎝ ⎟ 50 2 50 ⎜ 50 ⎟ ⎝ ⎠ EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo 2 Lecture 11 | Chapter 10 | 2/5 | 6/10 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Complex Power EE 203 Circuit Analysis 2 Lecture 11 Chapter 10.4 Complex Power Complex Power Kwang W. Oh, Ph.D., Assistant Professor SMALL (Nanobio Sensors & MicroActuators Learning Lab) Department of Electrical Engineering University at Buffalo, The State University of New York 215E Bonner Hall, SUNY-Buffalo, Buffalo, NY 14260-1920 Tel: (716) 645-3115 Ext. 1149, Fax: (716) 645-3656 kwangoh@buffalo.edu, http://www.SMALL.Buffalo.edu (VA) Apparent Power EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 7/10 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 8/10 SMALL for Big Things University at Buffalo SMALL for Big Things University at Buffalo nanobioSensors & MicroActuators Learning Lab The State University of New York nanobioSensors & MicroActuators Learning Lab The State University of New York Example 10.4 Example 10.4 P = Veff I eff cos(θ v − θ i ) = 240 I eff 0.8 8000 = 41.67 A 240 ⋅ 0.8 cos θ = 0.8 ∴ I eff = Vrms = Veff = 240 V, P = 8 kW, cos θ = 0.8 kW S = P + jQ = S cos θ + j S sin θ = 8 + j10 ⋅ 0.6 = 8 + j 6 kVA cos 2 θ + sin 2 θ = 1 ∴ sin 2 θ = 0.36 → sin θ = 0.6 S= P 8 = = 10 kVA cos θ 0.8 ∴θ = cos −1 (0.8) = 36.87° Z= Veff I eff = 240 = 5.76 41.67 ∴ Z = Z ∠θ = Z cos θ + j Z sin θ = 5.76∠36.87° = 5.76 ⋅ 0.8 + j 5.76 ⋅ 0.6 = 4.608 + j 3.456 Ω EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 9/10 EE 203 Circuit Analysis 2 | Spring 2008 | Prof. Kwang W. Oh | EE@SUNY-Buffalo Lecture 11 | Chapter 10 | 2/5 | 10/10 ...
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This note was uploaded on 10/19/2011 for the course EE 203 taught by Professor Staff during the Spring '08 term at SUNY Buffalo.

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