Supplementary Curse Ntes Adding and Subtracting AC Vltages and Currents As mentined previusly, when cmbining DC vltages r currents, we nly need t knw the plarity (vltage) and directin (current). In the case f alternating vltages and currents, the plarity and directin are peridically changing. There are mathematical and graphical ways t cmbine these, but these can be quite cumbersme and time cnsuming, especially when there are a number f cases t determine. Frtunately, Gerge Prteus Steinmetz develped a methd f cmbing AC signals which greatly simplifies this task; we knw it as Phasr Ntatin. It is based n the mre cmplicated general, mathematical trignmetric methd f cmbinatin, but is greatly simplified when applied t sinusidal vltages r currents f exactly the same frequency. Remembering the general parametric equatin fr a sinusidal signal, we ntice that if the frequency f all the signals cncerned is exactly the same, they will differ nly in magnitude and phase angle. It is these tw characteristics that are used t d a transfrmatin f the signal int a tw-dimensinal, cmplex valued number which can be mathematically cmbined quite simply by ding cmplex additin r subtractin t cmbine the signals. Once cmbined, they can be cnverted back in t a time functin representatin shuld that be required t determine time-related parameters. General transfrmatin f a sinusidal signal frm the time dmain t the phasr dmain: Time: v(t) = Asin( ωt +ϕ)v Table 1: Time t Phasr Cnversin Time Dmain t Phasr Dmain Cnversin Phasr: V = A V (plar frm), (rectangular frm) 2 ϕ = A 2 cs(ϕ ) + j A 2 sin(ϕ ) Nte: phasrs are cnventinally shwn as RS values, but the cncept wrks equally well fr maximum (i.e. peak) values. Reverse transfrmatin frm the phasr dmain back int the time dmain: Table 2: Phasr t Time Cnversin Phasr Dmain t Time Dmain Cnversin Phasr: V = B θv (plar frm) r V = (X +jy)v (rectangular frm) Time:, r v( t ) = 2 X 2 + Y 2 sin ωt + tan 1 Y v( t ) = 2Bsin(ωt + θ )V X V Of curse in the reverse case, there is knw way f determining ω withut additinal infrmatin! Denard Lynch Page 1 f 10
Supplementary Curse Ntes The Phasr Diagram: Phasrs are just vectrs in a 2-D cmplex plane. A Phasr Diagram can be used t illustrate their relatinship and additin. An example: Phasr Diagram j V 2 sinφ 2 V 1 V Tt V 1 sinφ 1 V 1 csφ 1 φ 1 φ 2 V 2 V 2 csφ 2 Figure 1: Phasr Diagram Final nte: Phasrs are transfrmatins f time-varying signals nly. Other AC circuit elements, as we will discuss next, will als be represented as cmplex-valued numbers, but are nt cnsidered phasrs. Denard Lynch Page 2 f 10
Supplementary Curse Ntes Oppsitin t Flw in AC Circuits In a steady-state DC circuit, resistrs, capacitrs and inductrs behave relatively simply: resistrs bey Ohm s Law, capacitr lk like pen circuits, and inductrs lk (almst) like shrt circuits. If the AC wrld, these elements can react a little differently. Resistrs still bey Ohm s Law, accrding t the instantaneus vltage r current at any given time. Thus, a resistr cnnected t an AC vltage surce will allw a current t flw accrding t Ω s Law at any instant, resulting in a sinusidal current f exactly the same frequency and phase as the vltage, and a magnitude predicted by Ω s Law at any instant. Fr inductrs that are cnnected t an AC surce: v L ( I sinωt) = ωli csωt = V csωt = V sin( ωt + ), wherev = ωli di d = L = L 90 dt dt Nte that the vltage leads the current by 90 0 (r, current lags vltage: eli). Define the term X L = ωl, as the Inductive Reactance, which is the Alternating Current wrld's ppsitin t flw fr an inductr, and is measured in hms, Ω. (L is the inductance in Henries, and ω is in Radians/secnd). Nte als that the reactance is directly prprtinal t bth the frequency and the Inductance. X L als = V I Similarly fr Capacitrs: which illustrates its equivalence t resistance, r ppsitin t flw. i C ( V sinωt) = ωcv csωt = I csωt = I sin( ωt + ), where I = ωcv dv d = C = C 90 dt dt Nte that here the current leads the vltage (vltage lags current, ice) by 90 0. The quantity, V I 1 ωc =, where C is in Farads (F) is called the Capacitive Reactance, X C and is the Alternating Current wrld's equivalent f Resistance, R, fr a capacitr and is measured in hms, Ω. Nte that in this case, the reactance is inversely prprtinal t frequency and capacitance. Nte that R, X L and X C are all scalar quantities representing ppsitin t flw, and will all bey Ω s Law at any instant in time. If we have cnverted ur vltages and currents t phasrs, we need t d a cnversin f R and X s that the ppsitin t flw will bey Ω s Law perfectly and simply in the Phasr Dmain. Denard Lynch Page 3 f 10
Supplementary Curse Ntes Impedance, Z Define ne mre new term: Impedance, Z, which represents the ppsitin t flw in the phasr dmain. It is als measured in Ohms (Ω), and is defined as: phasr _ vltage, r phasr _ current Z = V I (Nte: since Z is just a rati, it can be calculated frm either RS r ax values.) Since the phasr vltage and phasr current in this definitin are cmplex valued, the impedance, Z, is als cmplex valued. It is als measure in Ohms (Ω) and can represent the ppsitin t flw f any element r cmbinatin f elements. Using ur previus determinatins f the reactance f the basic elements (R, L, C) and the phase relatinship f the vltage and current when they are cnnected t an AC surce, we can develp an expressin fr the impedance f each type. Fr resistrs, since vltage and current are in phase, this is straightfrward: Z R = V 0 I 0 = Z 0 Ω = X R 0 Ω = R 0 Ω Fr inductrs, since current is 90 0 behind the vltage (eli) Z L V 0 = = Z 90 Ω = X L 90 Ω = ω L 90 Ω = I 90 jωl Fr capacitrs, since current is 90 0 ahead f the vltage (ice) Z C V 0 = I + 90 = Z 90 Ω = X C 1 1 90 Ω = 90 Ω = j ωc ωc Fr any series cmbinatin f R-L-C in AC circuits, the impedance, Z = R +j(x L X C ), a linear cmbinatin f the individual elements impedance. Denard Lynch Page 4 f 10
Supplementary Curse Ntes Impedances in series r in parallel ad exactly like resitrs in series r in parallel, except that we must us the cmplex-valued numbers. Using this cmplex impedance, Z, alng with phasr vltage and current representatins, Ohm's Law can be applied in a straightfrward fashin t A.C. circuits made up f cmbinatins f resistive, inductive and capacitive cmpnents. Nte: that althugh Z is cmplex, it is nt a phasr as it is nt a sinusidally varying quantity. Nte: that reactance(x) and resistance(r) are scalars and nt cmplex valued. While impedance is a cmplex vectr which can be represented n a 2-D plane, Z des nt represent a time-varying quantity (like a phasr des). Impedance, Z, can als be represented and added n a 2-D plane called an Impedance Diagram (similar t, but nt a Phasr Diagram!): Impedance Diagram j Z L Z Tt Z R Z C Figure 2: Impedance Diagram Denard Lynch Page 5 f 10
Supplementary Curse Ntes Finally ur favurite laws. These are virtually identical t thse fr the DC wrld, except we use (cmplex-valued) phasrs: Ohms Law fr AC Circuits: V = IZ, and Z = V I Kirchhff s Laws fr AC circuits: KVL: The Σ phasr vltages arund a lp = 0 KCL: The Σ phasr currents int a nde = 0 Nrtn and Thévèn equivalents and current/vltage surce cnversin als wrk exactly the same as the DC cunterparts, except that phasr and cmplex impedance values are used. With phasr vltages and currents and [cmplex] impedances, we can apply almst all f the same rules and Laws we used in DC circuits in the AC wrld. Rectangular Crdinates Table 3: Cmplex-number ath Summary Plar Crdinates + (A + jb) + (C + jd) = (A+C) + j(b+d) Add like vectrs; usually easiest t cnvert t rectangular crdinates then add. If θ is the same fr bth, can add magnitudes (A + jb) - (C + jd) = (A-C) + j(b-d) (A + jb) (C + jd) = AB + j(bc+ad) + j 2 (BD) = (AC-BD) + j(bc+ad) (A + jb) (C + jd), need Cmplex Cnjugate Cmplex Cnjugate: (A + jb)* = (A - jb), then N = N* NN* = (cmplex prduct) (real number) and... (A + jb) (C + jd) = {(AC+BD) (C 2 +D 2 )} + j{(bc-ad) (C 2 +D 2 )} As abve, subtract like vectrs, but easiest t cnvert t rectangular crdinates first. If θ is the same fr bth, can subtract magnitudes (A θ 1 ) (Β θ 2 ) = (A B) (θ 1 +θ 2 ) (A θ 1 ) (Β θ 2 ) = (A B) (θ 1 θ 2 ) NOTE: quantities in the phasr dmain can be, and usually are, given in effective r RS terms. Generally assume that any Phasr quantities are RS unless therwise stated. Denard Lynch Page 6 f 10
Supplementary Curse Ntes Table 4: Phasr Representatin Summary Time Dmain Phasr Dmain (Plar) Phasr Dmain (Rectangular) Asin(ωt+θ) A +θ Acsθ + jasinθ Asin(ωt-θ) A θ Acsθ - jasinθ Acs(ωt+ θ )=Asinω(t+θ+90 0 ) Asinω(t+ θ )=Acsω(t+θ-90 0 ) -Asin(ωt+ θ )= Asinω(ωt+θ±180 0 ) -A ±θ = Α (±θ±180 0 ) If A +θ = (Acsθ + jasinθ) then A θ = (Acsθ - jasinθ) Denard Lynch Page 7 f 10
EE204 Basic Electrnics and Electric Pwer Curse Ntes Eg. 1: An R-L-C series AC circuit: Given the current magnitude f 2.48A, and the reactance values: V R = 116.6V V L = 93.5V V C = 65.7V Nte: these are scalar magnitudes! I=2.48 0 0 A + 47Ω X L = 37.7Ω X C =26.5Ω a) Determine the impedance f each element, and their ttal impedance as seem by the surce. b) Determine the phasr vltage acrss each element and the phasr vltage f the surce (Anther Nte: the default assumptin is that any values fr V r I are RS, and cmpnent values are fr reactance an in Ωs unless specified therwise.) We can nw write the impedance f each element (nte that the reactances, X, are already given): Z R = 47 0 0 Ω, Z L = 37.7 90 0 Ω, and Z C = 26.5-90 0 Ω And nw calculating the phasr vltages acrss each element: (Nte: we ll assume the values are nw in RS fr cnvenience.) V R = IZ R = (2.48 0 0 A)(47 0 0 Ω) = 116.6 0 0 V V L = (2.48 0 0 A)(37.7 90 0 Ω) = 93.5 90 0 V V C = (2.48 0 0 A)( 26.5-90 0 Ω) = 65.7-90 0 V And the phasr ttal is: E = V R + V L + V C = (116.6 +j0) + (0 +j93.5) + (0 j65.7) = 119.9 13.4 0 V The ttal impedance is: Z T = Z R + Z L + Z C = (47 +j0) + (0 +j37.7) + (0 j26.5) = (47 +j11.2)ω = 48.3 13.4 0 Ω We can d a check: Z T = V/I = 119.9 13.4 0 V/2.48 0 0 A = 48.3 13.4 0 Ω Denard Lynch 2012 Page 8 f 10 Sep 23, 2012
EE204 Basic Electrnics and Electric Pwer Curse Ntes Eg. 2: A series - parallel AC circuit I S V R Given: E S = 35 23 0 V = (32.2 +j13.68)v, f = 30kHz R = 10Ω, C =.22µF, L = 100µH E S I C (a) V C I L V L Find: I S, I C, I L, V R, V L, V C, Z T. First, calculate the impedances: Z C = j ωc = j 2π 30kHz ( ) = 24.11 900 Ω = 0 j24.11 ( ) = 21.34 62.05 0 Ω = 10 + j18.85 ( ).22 X10 6 F Z RL = Z R + Z L = 10 + jω L = 10 + j 2π 30kHz 100 X10 6 H Calculate the branch currents and vltages acrss the elements: I C = E s Z C = I RL = E s Z RL = 35 23 0 V 24.11 90 0 Ω = 1.45 1130 A 35 23 0 V 21.34 62.05 0 Ω = 1.64 39.050 A V C = I C Z C = 1.45 113 0 A ( )Ω ( )( 24.11 90 0 Ω) = 35 23 0 V = E S ( )( 10 0 0 Ω) = 16.4 39.05 0 V ( )( 18.85 90 0 Ω) = 30.91 50.95 0 V V R = I RL Z R = 1.64 39.05 0 A V L = I RL Z L = 1.64 39.05 0 A The ttal surce current is just the sum: I S = I C + I RL = 1.45 113 0 A + 1.64 39.05 0 A =.7687 23.25 0 A We can als check t make sure the vltages acrss R and L add t E S : E S = V R + V L = 16.4 39.05 0 V + 30.0 50.95 0 V = 35.0 23.0 0 V Finally, we can calculate the ttal impedance f the circuit: Z T = E S I S = 35.0 23.00 V.7687 23.25 0 A = 45.5.250 Ω Nte: These phasrs and impedances can be pltted in their respective 2-D cmplex planes. Hwever, nt all values can be graphically added. I C and I RL will add crrectly t I S, and V R and V L will add t E S, and Z R and Z L will ttal Z RL, ther cmbinatins may ( )Ω Denard Lynch 2012 Page 9 f 10 Sep 23, 2012
EE204 Basic Electrnics and Electric Pwer Curse Ntes require a separate apprach. T graphically add Z C and Z RL fr instance, we wuld have t cnvert them t admittances and add them n an Admittance Diagram. Eg. 3: Sme mre practice with AC circuits Find the unknwn vltages fr the fllwing circuits and express yu answer in plar ntatin. (Nte: there are tw pssible answers fr (b); prvide bth.) V C =80V V C =200V E S =150 0 0 V V R =? E S =150 0 0 V V L =? (b) a) Cmmn current, therefre V C must be 90 0 behind (lagging) V R. Their vectr sum must als equal E S. (Refer t the Phasr diagram) agnitude f V R = 150 2 80 2 ( ) = 126.9, and the angle fr V C, 80 θ = cs 1 150 = 57.8 0, s: V C =80-57.8 0 V V R = 126.9 (-57.8 0 + 90 0 ) = 126.9 32.1 0 V V R E S V C b) V C and V L are 180 0 apart (cmmn current), and the nly way their sum can be at 0 0 is if they are als bth n the hrizntal axis, with an angle f 0 0 r 180 0. Since V L + V C = E S, if V C is 200 0 0 V, V L must be 50-180 0 V. If V C = 200-180 0 V, then V L must be 350 0 0 V V L E S E S V L V C V C Denard Lynch 2012 Page 10 f 10 Sep 23, 2012