MET 487 Instrumentation and Automatic Control Lecture 3 Electrical Circuits and Components http://www.etcs.ipfw.edu/~lin 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
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 = 0.0010m, 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
Color Coded Resistor - Examples August 30, 2005 Lecture 2 - By P. Lin 5 Series Resistors Rt = R1 + R2 sum of resistance (ohms) Rt = 100 + 120 = 220 Ω Two resistors plus wire resistance in series Rt = R1 + R2 + R3 Rt = 100 + 120 + 0.01 = 220.01 Ω Lecture 2 - By P. Lin 6
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: >> 100 + 120 ans = 220 >> Rt = 100 + 120 Rt = 220 >> R1 = 100; >> R2 = 120; >> R1 + R2 + 0.01 ans = 220.01 August 30, 2005 Lecture 2 - By P. Lin 8
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
Prefix and Power Some Prefixes for SI Units (International Standard) Power Prefix 10-24 yocto 10-21 zepto 10-18 atto 10-15 femto 10-12 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: http://www.frontierusa.com/ August 30, 2005 Lecture 2 - By P. Lin 12
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
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
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
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 = 377.7 t V rms = 120 V m = V rms 1.414 = 169.7v, Φ = 45º = π/4 = 0.7854 radian V(t) ) = V m sin(2πft + Φ) ) = 169.7 sin(377.7 t + 0.7854) Volt August 30, 2005 Lecture 2 - By P. Lin 19 Alternating Current Click -> Debug -> Run August 30, 2005 Lecture 2 - By P. Lin 20
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 = 377.7 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 = 120.0 + 120.0i August 30, 2005 Lecture 2 - By P. Lin 21 Alternating Current Rectangular form V = Vm*cos cos(φ) ) + j Vm sin(φ) = 169.7 * cos(45º) ) + j 169.7 * sin (45º) = 169.7 * cos(π/4) + j 1169.7*sin(π/4) MATLAB Example >> V = 169.71 69.7*cos(pi/4)+j*169.7*sin(pi/4) V = 120.0 +120.0i August 30, 2005 Lecture 2 - By P. Lin 22
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
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) % 0.4636 pi % 180 pi % ----- = ---- % theta_degree theta theta_degree = (180*theta)/pi % 26.5615 degree = 0.4636 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 = 1.000 0.500i I_theta_deg = -26.6 I_mag = 1.118 VR = 8.0000-4.0000i VL = 3.0000 + 6.0000i VC = -1.0000-2.0000i KVL = 0 Lecture 2 - By P. Lin 26
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