MET 487 Instrumentation and Automatic Control. Lecture 3

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1 MET 487 Instrumentation and Automatic Control Lecture 3 Electrical Circuits and Components Lecture 2 - By P. Lin 1 Electrical Circuits and Components Basic Electrical Elements Resistor, Capacitor, Inductor Kirchhoff s s Laws KVL, KCL Series Circuits, Parallel Circuits Circuit Analysis Voltage and Current Sources and Meters Input and Output Impedance Alternating Current Circuit AC Circuit Analysis Impedance Matching Grounding and Electrical Interference Electrical Safety August 30, 2005 Lecture 2 - By P. Lin 2

2 Basic Electrical Elements Resistor Ohm s s Law: V = I*R, I = V/R, R = V/I Wire Resistance L R = ρ A Example: find the resistance of a copper wire: 1.0 mm in diameter, 10 m long Solution: ρ = 1.7x10-8 Ωm, D = m, r = D/2, L = 10m A = π r 2 = π (D 2 /4) = 7.8 x 10-7 m 2 R = ρ L/A = 0.22 Ω August 30, 2005 Lecture 2 - By P. Lin 3 Color Coded Resistor Lecture 2 - By P. Lin 4

3 Color Coded Resistor - Examples August 30, 2005 Lecture 2 - By P. Lin 5 Series Resistors Rt = R1 + R2 sum of resistance (ohms) Rt = = 220 Ω Two resistors plus wire resistance in series Rt = R1 + R2 + R3 Rt = = Ω Lecture 2 - By P. Lin 6

4 I Parallel Resistors A Vs I1 + I2 + - R1 R2 - I = Vs/R1 + Vs/R2 = Vs/(1/R1 + 1/R2) = Vs/ [(R1 *R2)/(R1 + R2)] = Vs/Rt Rt = (Product OVER Sum) If Vs = 12V, R1 = 10kΩ,, R2 = 10kΩ, Rt = 10k*10k/(10k+10k) = 5k I = 12/5k = 2.4 ma,, I1 = I2 = 1.2 ma V1 = V2 = Vs = 12 v Lecture 2 - By P. Lin 7 MATLAB Examples MATLAB Sum Calculator: enter the following lines at the MATLAB command window: >> ans = 220 >> Rt = Rt = 220 >> R1 = 100; >> R2 = 120; >> R1 + R ans = August 30, 2005 Lecture 2 - By P. Lin 8

5 Voltage Divider (Resistors in Series) IF R1 = 10K, VR2 = 12*10K/(10K+10K) = 6V IF R1 = 5K, VR2 = 12*10K/(5K+10K) = 12*2/3 = 8 V IF R1 = 2K, VR2 = 12*10K/(2K+10K) = 12*10/12 = 10 V Lecture 2 - By P. Lin 9 Electric Power Example: Electric Power Calculation, for R = 15 ohms, voltage = 120 volts: P = V^2 /R (watts). MATLAB Solution: >>R = 15.0; >>V = 120; >>P = V^2 / R P = 960 August 30, 2005 Lecture 2 - By P. Lin 10

6 Prefix and Power Some Prefixes for SI Units (International Standard) Power Prefix yocto zepto atto femto pico 10-9 nano 10-6 micro Abbrevi ation y z a f p n μ August 30, 2005 Lecture 2 - By P. Lin 11 Prefix and Power Some Prefixes for SI Units (International Standard) Power Prefix 10-3 milli 10-2 centi 10-1 deci 10 1 deka 10 3 kilo 10 6 mega 10 9 giga Abbrevi ation m c d da k M G Source: August 30, 2005 Lecture 2 - By P. Lin 12

7 Prefix and Power Some Prefixes for SI Units (International Standard) Power Prefix 10 tera 10 peta 10 exa 10 Zetta 10 Yotta Abbreviation T P E Z Y August 30, 2005 Lecture 2 - By P. Lin 13 Capacitor Capacitor q( t) 1 V ( t) = = I d C C ( τ ) τ I( t) = C dv dt Capacitors in Series Ceq = C1*C2/(C1 + C2) t 0 Capacitor in Parallel Ceq = C1 + C2 Ceq C1 C2 August 30, 2005 Lecture 2 - By P. Lin 14

8 Inductor Inductor dλ dφ V ( t) = L = L dt dt λ = LI, φ = LI di V ( t) = L dt 1 I( t) = L t 0 V ( τ ) dτ Inductors in Series Leq = L1 + L2 August 30, 2005 Lecture 2 - By P. Lin 15 Inductor Inductors in Parallel Leq = (L1* L2)/(L1 + L2) August 30, 2005 Lecture 2 - By P. Lin 16

9 Kirchhoff Voltage Law I2 + R2 - - V2 + I3 I1 R1 V1 + + I V3 R3 - E = 12V V4 - R4 + I4 August 30, 2005 Lecture 2 - By P. Lin 17 Kirchhoff Current Law I1 I3 I2 R1 V1 R2 V2 R3 V3 + + E = 12V E = 5V August 30, 2005 Lecture 2 - By P. Lin 18

10 Alternating Current AC Signal (voltage) V(t) ) = V m sin(ωt t + Φ) ) = V m sin(2πft + Φ) V(t) ) = Vdc + Vm sin(ωt t + Φ) --- with DC offset V m -- Amplitude (volt) V rms -- Root-Mean Square, or Effective value f -- Frequency (Hz) ω= = 2 π f t -- Radian frequency (rad( rad/sec) Φ = ωδt -- Phase Angle, leading or lagging T = 1/f -- Period (second) Example f = 60 Hz, T = 1/f = 16.7 ms, ω = 2πft 2 = t V rms = 120 V m = V rms = 169.7v, Φ = 45º = π/4 = radian V(t) ) = V m sin(2πft + Φ) ) = sin(377.7 t ) Volt August 30, 2005 Lecture 2 - By P. Lin 19 Alternating Current Click -> Debug -> Run August 30, 2005 Lecture 2 - By P. Lin 20

11 Alternating Current AC Signal (voltage) - Time domain equation V(t) ) = V m sin(ωt t + Φ) ) = Vm sin(2πft + Φ) Euler s s Formula e j(ωt+ t+φ) = cos(ωt t + Φ) ) +j sin(ωt t +Φ) + j= -1, also called 90 degree operator Polar Form Vrms = 120 v, Φ = 45º,, f = 60 Hz, ω = rad/sec V = V m e j(ωt+ t+φ) => V m /Φ = 120 /45/ 45º MATLAB Example: >> V = 169.7*exp(j*pi/4) V = i August 30, 2005 Lecture 2 - By P. Lin 21 Alternating Current Rectangular form V = Vm*cos cos(φ) ) + j Vm sin(φ) = * cos(45º) ) + j * sin (45º) = * cos(π/4) + j *sin(π/4) MATLAB Example >> V = *cos(pi/4)+j*169.7*sin(pi/4) V = i August 30, 2005 Lecture 2 - By P. Lin 22

12 AC Circuit Analysis A RLC circuit is shown on this slide, find a) Total impedance Z b) Voltage and current across each components Lecture 2 - By P. Lin 23 AC Circuit Analysis(continue) Analysis: Domain knowledge XL = 2πfL, where L is the inductance in Henry, f is the frequency of ac source XC = 1/(2 πfc), where C is the capacitance in Farard Z = R + j(xl XC) -- total impedance, where j shows the imaginary component of a complex number I = E/Z, total current VR = I*R, voltage drop across resistor VL = I*XL, voltage drop across the inductor VC = I*XC voltage drop across the capacitor Lecture 2 - By P. Lin 24

13 AC Circuit Analysis (continue) MATLAB Program %RLC_1.m f = 60; R = 8; % Peak value of the sine wave e = 10; XL = j*6; XC = -j*2; Z = R + (XL+XC) theta = angle(z) % pi % 180 pi % = ---- % theta_degree theta theta_degree = (180*theta)/pi % degree = pi mag_z = abs(z) Phasor Representation of Impedance Z = 8 + j*4 Lecture 2 - By P. Lin 25 AC Circuit Analysis (continue) MATLAB Program (cont.) %RLC_1.m % I = e/z I_thea_degree = angle(i) * (180)/pi I_mag = abs(i) VR = I*R VL = I*XL VC = I*XC KVL = e (VR + (VL + VC)) I = i I_theta_deg = I_mag = VR = i VL = i VC = i KVL = 0 Lecture 2 - By P. Lin 26

14 Power in Electrical Circuits Power: P = W/T = dw/dt Instantaneous Power in Resistive Circuits P = VI = I 2 R = V 2 /R Average Power Pavg = 0.5*(V m *I m )*cos cos(θ) V m = 2 2 * V rms ; I m = 2* 2*I rms Pavg = V rms *I rms *cos(θ) Average Power Consumed by a Resistor Pavg = V rms *I rms = RI 2 rms = V 2 rms /R August 30, 2005 Lecture 2 - By P. Lin 27 Power in Electrical Circuits Average Power Consumed by an AC Network Pavg = V rms *I rms *cos(θ) = I 2 rms Z * cos(θ) = (V 2 rms / Z ) * cos(θ) Power Factor (PF) PF = cos(θ): 0.75, 0.8, 0.85, 0.9, Lecture 23 - By P. Lin 28

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