Review of DC Electric Circuit. DC Electric Circuits Examples (source:

Review of DC Electric Circuit DC Electric Circuits Examples (source: http://hyperphysics.phyastr.gsu.edu/hbase/electric/dcex.html) 1

Review - DC Electric Circuit Multisim Circuit Simulation DC Circuit Construction, PhET Interactive Simulations, http://phet.colorado.edu/en/simulation/circuit-construction-kit-dc 2

Electric Power Claculation DC Electric Power = Voltage * Current P = V * I = (joules/coulomb)*(coulomb/sec) = joules/sec = watts Ohm s Law: V = I * R P = V * I = (I*R)*I = I 2 *R P = V * I = V * (V/R) = V 2 /R Other References Electric Power HyperPhysics, http://hyperphysics.phyastr.gsu.edu/hbase/electric/elepow.html 3

DC Relay Circuit with R, L and C. DC Circuit DC Power Supply V (volts), Resistor R (Ohms, mω, kω, MΩ, etc), Inductor L (Henry, mh, μh), Capacitor C (Farad, μf, nf, pf) Circuit Configuration Relay coil: L and R --- in series 12V DC coil: Calculated current 64.9 ma Measured current 64.x 59.x ma Contact 120V, 1/6 HP 240V, 1/3 HP Normally Open (NO) contact Normally Closed (NC) contact C in parallel with the relay coil E 12 V L R Relay 158.9mH 185Ω C 330µF 25V DC 12V, AC 120V 1/6 HP, 240V 1/3 HP 4

Single Phase AC Motor On/Off Control Solid State Relay (SSR) Magnetic Switch (or relay) AC 120V, ¼ HP Single Phase Motors, 4-pole See the complete demo from Feb 21, 2013 5

Example 2-1 FIGURE 2-2 Waveforms for the voltage and current represented by equation 2.3. Volt/Ampere 200 150 100 50 0-50 -100-150 % sinewave2.m Vm = 170; Vrms = Vm/sqrt(2); f = 60; T = 1/f; dt = T/100; t = 0: dt: 2*T; w = 2*pi*f; vt = Vm*sin(w*t); Im = 30; Irms = Im/sqrt(2); theta = pi/6; it = Im * sin(w*t - theta); plot(t, vt, t, it), grid on xlabel('time sec'), ylabel('volt/ampere'); -200 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 Time sec 6

Example 2-1 Phasor Diagrams (continue) FIGURE 2-3 reference. Phasor diagrams for the voltage and current of Example 2-1; (a) voltage as reference, (b) current as reference, (c) cos( t) as v t = 170 sin ωt = 120 2 sin(ωt) i t = 30 sin ωt π 6 = 21.21 2 sin(ωt π 6 ) cos (30 ) = cos (30 /180 *pi) = 0.866 P = VI cosθ = 120 * 21.21 * 0.866 = 2.2 kw 7

Table 2-1 Time Domain and Phasor Domain Conversion. Time Domain Phasor Domain 100 sin ωt + 2π/9 (0.707)100/_40 =70.7/_40 30 sinωt (0.707)(50)/_0 = 35.35/_0 75cosωt = 75sin(ωt +π/2) 100cos(ωt π/6) = 100sin(ωt+π/3) 2 120 sinωt = 170sinωt 120/_0 2 165 sin(ωt 7π/18) = 233sin (ωt 7π/18) 120/_0 2 10 sinωt = 14.14 sin(ωt + 0π) 10/_0 2 2400 sin(ωt + π/4) = 3394sin (ωt + π/4) 2400/_0 (0.707)(75)/_90 = 53/_90 (0.707)(100)/_60 = 70.7/_60 v t = V max sin ωt = 170 sinωt = Vr ms 2 sin ωt = 120 2 sin(ωt) 8

FIGURE 2-4 Illustration of rotating phasor; (a) at time t = 0, (b) at time t = /3, at time t =, and (d) at time t =. 9

FIGURE 2-5 Illustration of relationship between polar and rectangular quantities for a phasor or complex number. 10

Table 2-1 Symbol Notation for Several Complex Quantities. Quantity Phasor or Comple x numbe r symbol Magnit ude Symbol Polar form Voltage V V or V V/_α or V /_α Current I I or I I/_β or I /_β Rectangular form V r + j V j I r + j I j Complex power S S or S S/_θ or S /_θ P + j Q v t = V max sin ωt = 170 sinωt = Vr ms 2 sin ωt = 120 Z /_θ 2 sin(ωt) Impedance Z Z or Z Z/_θ or R + jx L R - jx C 11

Complex Number Manipulation A = A α = A r + ja j = Acos α + jasin α B = B β = B r + j B j = Bcos β + jb sinβ Addition: C = A + B = (A r + B r ) + j(a j + B j ) = C r + C j = C γ Subtraction: C = A - B = (A r - B r ) + j(a j - B j ) = C r + C j = C γ Multiplication: C = A * B = A α * B β = A*B (α+β) Division: C = A / B = A α / B β = A/B (α-β) 12

FIGURE 2-6 Addition of phasors. j = 90 degree operator A = A α = A r + ja j = Acos α + jasin α B = B β = B r + j B j = Bcos β + jb sinβ C = A + B = (A r + B r ) + j(a j + B j ) = C r + C j = C γ 13

Example 2-2 Given two phasors: A = 50 40 and B = 30-30 Find: a) C = A + B b) C = A * B c) C = A B Solution (MATLAB function): >> A = 50*exp(j*(40/180)*pi) A = 38.3022 +32.1394i >> B = 30*exp(j*(-30/180)*pi) B = 25.9808-15.0000i a) C = A + B >> C = A + B C = 64.2830 +17.1394i b) C = A * B = 50*30 /_40+(-30) = 1500 /_10 c) C = A B = 50/30 /_40-(-30) = 1.67 /_70 14

FIGURE 2-7 A phasor and its conjugate. A = A α = A r + ja j A* = A -α = A r - ja j (The complex conjugate of A) 15

Impedance in the AC Circuits FIGURE 2-8 AC circuit. Inductive Reactance X L = jωl = j2πf L Capacitive Reactance X C = -j1/(ωc) = -j1/(2πf C) Impedance Z = R +j(x L X C ) j 90 degree operator j*j = -1, 1/j = -j 16

Example 2-3 FIGURE 2-8 AC circuit. Given f = 60 Hz, Vrms = 120V, R = 40 ohms, L = 0.1592 Henry, C = 33.16 μf. Find: a) X L, and X C b) Current in each leg: I R, I L, I C c) Total current I = I R + I L + I C d) Draw a phasor diagram showing the source voltage and all the currents e) Total impedance Z 17

Example 2-3 FIGURE 2-8 AC circuit. Given f = 60 Hz, Vrms = 120V, R = 40 ohms, L = 0.1592 Henry, C = 33.16 μf. Find: a) X L, and X C b) Current in each leg: I R, I L, I C c) Total current I = I R + I L + I C d) Draw a phasor diagram showing the source voltage and all the currents e) Total impedance Z 18

FIGURE 2-9 Phasor diagram for Example 2-3(d). 19

FIGURE 2-10 triangle. (a) Power triangle for inductive load, (b) power triangle for capacitive load, and (c) several equations for the power 20

FIGURE 2-11 Resistive load. 21

FIGURE 2-12 Resistive-inductive (R-L) load. 22

FIGURE 2-13 Purely inductive load. 23

FIGURE 2-14 Parallel R-L circuit for use with Example 2-5. 24

FIGURE 2-15 Series R-L circuit for use with Example 2-6. 25

FIGURE 2-16 Summary of relationships for an inductive (RL) circuit: (a) the circuit, (b) the impedance triangle, (c) phasor diagram showing voltage and langging current, and (d) the power triangle. 26

FIGURE 2-17 Summary of relationships for a capacitive (RC) circuit: (a) the circuit, (b) the impedance triangle, (c) phasor diagram showing voltage and leading current, and (d) the power triangle. 27

FIGURE 2-18 Wye-connected coils of a three-phase generator. 28

FIGURE 2-19 Subscript notation for three-phase wye and delta. 29

FIGURE 2-20 Angular relationships for phase and line currents in a delta system. 30

FIGURE 2-21 Angular relationships for phase and line voltages in a wye system. 31

FIGURE 2-22 Line-to-neutral voltages in the time domain. 32

FIGURE 2-23 Phasor diagrams for Example 2-8. 33

FIGURE 2-24 Pole-mounted capacitor bank. 34

FIGURE 2-25 400 kvar power factor correction capacitors. 35

FIGURE 2-26 Real and reactive power vs. power factor. 36

FIGURE 2-27 Effect of poor power factor. 37

FIGURE 2-28 Power factor correction by addition of power triangles. 38

FIGURE 2-29 Figure for Problem P2-3. 39