ECE 222 Electric Circuit Analysis II Chapter 13 Passive Elements In Frequency Domain Herbert G. Mayer, PSU Status 5/9/2016 For use at CCUT Spring 2016

Syllabus Phasor Transform in Circuit Analysis V-I Relationships Impedance vs. Reactance Exercises 1, 2 KVL in Frequency Domain KCL in Frequency Domain Exercises 3, 4 Delta Wye Transform Exercise 5 Bibliography

Phasor Transform in Circuit Analysis Phasor Transform in circuit analysis conducted in these 2 steps: 1. Specify relation between phasor voltage and phasor current at terminals of passive components 2. Derive phasor-domain version of Kirchhoff’s Laws Here we define relation between phasor current and phasor voltage at resistor, inductor, and capacitor Using Passive Sign Convention Recall Passive Sign Convention: When the reference direction of the current in an electric component is the same as the direction of the current through that component, use a positive sign in equations relating voltage to current; else use negative sign Used in Phasor Transform

Convention Note use of Greek symbol ϕ for phase angle Text book sometimes uses Greek letter θ Yet θ also used for angle, whose tan() is ωL/R We’ll use ϕ or θ for phase angle, as shown in context

V-I Relationship in Resistor Assuming a sinusoidally varying current i = i(t) through a resistor R With maximum amplitude Im, a phase angle ϕi, varying at rate ωt i = Im cos( ωt + ϕi ) Then the voltage v = v(t) at the terminals is v = v(t) = R Im cos( ωt + ϕi ) The phasor transform of this voltage equation is: V = R Im e jϕ V = R Im ϕi

V-I Relationship in Resistor But Im ϕi is nothing but the phasor representation of the sinusoidal current We can rewrite this equation, looking suspiciously like Ohm’s law, as: V = R I With V being phasor voltage and I the phasor current There is no phase shift between the two, i.e. when the voltage reaches its max, the current is at a max at the same moment Current and voltage are in phase Ohm’s Law for AC in phasor notation thus looks almost identical to Ohm’s Law in DC circuits in conventional notation

V-I Relationship in Resistor No phase shift caused by resistor R i(t) = 2 sin(x), v(t) = R i(t) v(t) t x i(t)

V-I Relationship in Inductor Assuming a sinusoidally varying current i = i(t) through an inductor L With maximum amplitude Im, a phase angle ϕi, varying at rate ωt i = Im cos( ωt + ϕi ) Given that voltage v is a function of inductivity L and the rate change of the current: v = L di / dt Then the voltage v = v(t) at the terminals is v = L di /dt = -ω L Im sin( ωt + ϕi ) Rewritten using cos(): v = -ω L Im cos( ωt + ϕi - 90o )

V-I Relationship in Inductor In phasor notation -ω L Im cos( ωt + ϕi – 90o ) becomes: V = -ω L Im e j( ϕ - 90 ) V = -ω L Im e jϕ e -j 90 Recall Euler’s identity: e -j 90 = cos( -90o ) + j sin( -90o ) Thus e -j 90 = -j; so we can rewrite voltage equation: V = j ω L Im e jϕ Phasor voltage and phasor current are related via: V = j ω L I Specifying the relation between phasor voltage and phasor current for an inductor L: Phasor voltage V at terminals of inductor L is j ω L times the phasor current I

V-I Relationship in Inductor Can also be expressed equivalently as: V = ω L 90o Im ϕi V = ω L Im 90+ϕi Indicating that current and voltage are out of phase in an inductor by exactly 90o In inductor L the current lags the voltage by exactly 90o Equivalently, the voltage in an inductor L is exactly 90o ahead of the current Expressed in seconds this phase shift is period T divided by 4 Equivalently, the phase shift is 1 / (4 f) with f being the frequency

V-I Relationship in Inductor Phase shift in inductor L: current lags voltage by 90o i(t) = K1 sin(x-π/2), v(t) = K2 sin(x) v(t) t x i(t)

V-I Relationship in Inductor In class engineering exercise: Students think up and explain a plausible explanation for . . . Why does the current through an inductor L lag behind the voltage across the terminals of L?

V-I Relationship in Capacitor Assuming a sinusoidally varying voltage v = v(t) through an capacitor C With maximum amplitude Vm, a phase angle ϕv, varying at rate ωt v = Vm cos( ωt + ϕv ) Given that current i is a function of capacity C and the rate change of the voltage: i = C dv / dt Then current i = i(t) at the terminals is i = C dv /dt = -ω C Vm sin( ωt + ϕv ) Rewritten using cos(): i = -ω C Vm cos( ωt + ϕv - 90o )

V-I Relationship in Capacitor In phasor notation -ω C Vm cos( ωt + ϕv - 90o ) becomes: I = -ω C Vm e j( ϕ - 90 ) I = -ω C Vm e jϕ e -j 90 Euler’s identity: e -j 90 = -j Rewrite current equation: I = j ω C Vm e jϕ Phasor current and phasor voltage are related via: I = j ω C V Expressing the phasor voltage as a function of the phasor current: V = I / ( j ω C )

V-I Relationship in Capacitor Voltage across the terminals of a capacitor C lags the current by 90o V = 1 / (ωC) -90o Im ϕio V = Im / (ωC) ( ϕi - 90 )o Or we say: The current leads the voltage in a capacitor, by 90o To conclude: all 3 equations expressing the V-I relation exhibit the consistent form of V = Z I Where Z is the impedance, in all 3 cases the ratio of the voltage phasor by the current phasor, a beautiful consistency Impedance of resistor is R, of an inductor is jωL, and of a capacitor is 1/jωC = j / (-ωC)

V-I Relationship in Capacitor Phase shift in capacitor C: voltage lags current by 90o i(t) = K3 sin(x+π/2), v(t) = K4 sin(x) v(t) i(t) x

V-I Relationship in Capacitor In class engineering exercise: Students think up and explain a plausible explanation for . . . Why does the current through a capacitor C run ahead of the voltage across the terminals of C?

Impedance vs. Reactance Resistance is force against the motion of electrons, a force observable in all conductor, most notably in resistors. Resistance often symbolized by letter R, measured in Ohm (Ω) Reactance is inertia against the motion of electrons caused by electric or magnetic fields, exhibited by inductors and capacitors. AC current experiencing reactance creates voltage drop 90o out of phase with the current. Often symbolized by letter X, also measured in Ohm (Ω) Impedance is comprehensive opposition to electron flow, including both resistance and reactance; may include imaginary part. Often symbolized by the letter Z, also measured in Ohm (Ω), possibly imaginary. Inverse is admittance Key difference between reactance and impedance is imaginary vector part j shown for impedance

Impedance vs. Reactance Admittance is the general measure, how easily a circuit allows a current to flow. The SI unit of admittance is Siemens [S]. The inverse of admittance is impedance. General is meant to convey that the current may have real and imaginary parts Susceptance is simply the imaginary part of admittance. Hence the SI unit of susceptance is Siemens [S] Conductance is simply the real part of admittance. Hence the SI unit of conductance is Ohm [Ω]

Impedance vs. Reactance Impedance measured in Ohm Impedance of resistor is R Impedance of inductor is jωL Impedance of mutual inductance is jωM Impedance of capacitor is 1 / (jωC) = j (-1/jωC) Recall that 1/j = -j = j (-1)

Impedance vs. Reactance Resistance R, Reactance X, and Impedance Z applied to a resistor, an inductor, and a capacitor; note phase shift!

Exercise 1 Given an inductor of 20 mH with an alternating current i(t) = 10 cos( 104 t + 30o ), calculate a.) the inductive reactance, and b.) impedance of the inductor a.) Reactance named x here is: x1 = ωL ω = 104, L = 20 * 10-3 [H] Thus x1 = 104 * 20 * 10-3 = 200 [Ω] b.) Impedance y1 is simply j * reactance x1, hence: y1 = j 200 [Ω]

Exercise 2 Given a capacitor of 5 μF, with an alternating voltage at its terminals of v(t) = 30 cos( 4*103 t + 325o ), calculate a.) the capacitive reactance, b.) impedance of the capacitor a.) Capacitive reactance named x2 here is: x2 = -1/ωC ω = 4*103, C = 5 * 10-6 [F] Thus x2 = -1/ (4*103 * 5 * 10-6 ) x2 = -50 [Ω] b.) Capacitive impedance y2 is simply j * reactance: y2 = -j 50 [Ω]

Kirchhoff’s Laws in Frequency Domain

Kirchhoff Voltage Law For any closed path in an electric circuit we know that KVL holds: The sum of all voltages equals 0 [V] v1 + v1 + v3 . . . + vn = 0 In circuit with sinusoidal sources, capacitances, and inductances this becomes complex Vm1 cos( ωt + θ1 ) + Vm2 cos( ωt + θ2 ) + . . . = 0 Where Vmk is maximum voltage of k-th sinusoidal voltage, with phase shift θk, part of some closed path Using Euler’s identity, we rewrite for all n parts: Re{ Vm1ejωt ejθ1 ) } + Re{ Vm2ejωt ejθ2 ) } + Re{ . . . = 0

Re{ ( Vm1 ejθ1 + Vm2 ejθ2 + . . . ) ejωt } = 0 Kirchhoff Voltage Law With the term ejωt factored out this becomes: Re{ ( Vm1 ejθ1 + Vm2 ejθ2 + . . . ) ejωt } = 0 And since all partial voltages have the same frequency, simplify rewriting: Re{ ( V1 + V2 + . . . + Vn ) ejωt } = 0 Since ejωt is not = 0, we can divide by ejωt yielding: V1 + V2 + . . . + Vn = 0 Which is Kirchhoff’s Voltage Law applied to phasor voltages KVL Σ vn = 0 for n voltages vn applies to sinusoidal voltages in the time domain, while Σ Vn = 0 is the equivalent expression for n phasor voltages in the frequency domain

Kirchhoff Current Law For any node in an electric circuit we know that KCL holds: The sum of all currents converging at that node equals 0 [A] i1 + i1 + i3 . . . + in = 0 Using the same logic as for the voltages it holds: I1 + I2 + . . . + In = 0

Series Impedances Thus we can easily compute the equivalent impedance Z from a series of n known impedances Zi By adding up all i, i = 1..n impedances Zi Z = Z1 + Z2 + . . . Zn Proof: Since the identical phasor current I runs through all impedances Zi, we know: V = Z1 I + Z2 I + . . . Zn I V = I ( Z1 + Z2 + . . . Zn ) Z = V / I

Exercise 3 Sinusoidal voltage source vs = 750 cos( 5000 t + 30o) drives a series load in steady state With R = 90 Ω, L = 32 mH, and C = 5 μF a.) Derive a frequency domain equivalent circuit b.) Use phasor method to compute steady state current

Exercise 3 Students: how large is ω ? Students: Compute the inductor’s impedance ZL Students: Compute the capacitor’s impedance ZC

a.) Exercise 3 Equivalent Circuit ω = 5000 rad/s -- by problem definition With L = 32 mH, compute impedance ZL ZL = j ω L ZL = j 5000 * 32 * 10-3 ZL = j 160 Ω With C = 5 μF, compute impedance ZC ZC = -j / ( ω C ) ZC = - j / ( 5000 * 5 * 10-6 ) ZC = - j 40 Ω

a.) Exercise 3 Equivalent Circuit The phasor transform for the source voltage is: Vs = 750 30o See equivalent circuit below:

b.) Exercise 3 Steady State Current Students, compute Z at terminals of voltage source! Think of Ohm’s law

b.) Exercise 3 Steady State Current Like regular Ohm’s law, the phasor current is voltage divided by impedance Z at VS terminals is: Z = 90 + j 160 – j 40 Z = 90 + j 120 Z = ( 902 + 1202 )½ arctan( 120 / 90 ) Z = 150 53.13o

b.) Exercise 3 Steady State Current We have the voltage VS and impedance Z, so can compute the current I = VS / Z I = 750 30o / 150 53.13o I = 5 -23.13o Thus the steady state current i(t) is: i(t) = 5 cos( 5000 t - 23.13o ) [A]

Exercise 4 Sinusoidal current source is = 8 cos( 200,000 t + 0o) drives a complex load in the steady state: Detail shown below, with resistors, cap, and inductor a.) Derive a frequency domain equivalent circuit b.) Derive steady state expressions for v, i1, i2, and i3

Exercise 4 Students, how large is the phase shift for the current source? (trick question  ) Students, define ω Students, compute the inductor’s impedance ZL in the frequency domain Students, compute the capacitor’s impedance ZC

a.) Exercise 4 No phase shift for the current source; that as the trick! Problem specifies ω = 200,000 [s-1] ZL = j ω L ZL = j 2 * 105 * 40 * 10-6 ZL = j 8 [Ω] ZC = -j / ω C ZC = -j / ( 2 * 105 * 10-6 ) ZC = -j 5 [Ω]

a.) Exercise 4 Redraw the circuit with the new component units!

b.) Exercise 4 To compute the voltage V across the current source, compute the total impedance, then use Ohm’s Law To compute the total impedance, since we have parallel circuits, we compute the conductance, in complex circuit named admittance, units Siemens [S] Here we have 3 partial admittances Y1 to Y3 Adding up 3 admittances Y1 to Y3 yields total admittance Y, and the inverse is the impedance, which is needed, Z = 1 / Y Once voltage V is known, currents I1 to I3 can be found

b.) Exercise 4 Y1 = 1 / 10 = 0.1 [S] Y2 = 1 / ( 6 + j8 ) = ( 6 - j8 ) / ( 6 + j8 ) ( 6 - j8 ) Y2 = ( 6 - j8 ) / ( 36 + 64 – j48 + j48 ) Y2 = ( 6 - j8 ) / 100 Y2 = 0.06 – j 0.08 [S] Y3 = 1 / -j 5 Y3 = j 0.2 [S] Y = Y1 + Y2 + Y3 = 0.16 + j 0.12 Y = 0.2 36.86o [S] Z = 1 / Y = 5 -36.86o [Ω]

b.) Exercise 4 Voltage is current times impedance: V = Z I V = 8 0o * 5 -36.86o V = 40 -36.86o

b.) Exercise 4 Now we can compute the currents; students compute right now in class I1

b.) Exercise 4 Now we can compute the currents I = V / Z I1 = V / 10 = 40 -36.86o / 10 I1 = 4 -36.86o = 3.199 - j2.4 [A] I2 = 40 -36.86o / ( 6 + j8 ) = 40 -36.86o / 10 53.13o I2 = 4 -90o = - j4 [A] I3 = 8 53.13o = 4.8 + j6.4 [A] I = I1 + I2 + I3 = 3.199 - j2.4 –j4 + 4.8 + j6.4 = 8 [A]

Delta-to-Wye Transformation

Delta-to-Wye for Resistors We know the delta-to-wye transformation for resistive circuits from ECE 221 The identical methodology is used in circuits with any and all passive elements, i.e. inductors and capacitors For resistors repeated here; then shown for general circuits

Delta-to-Wye for Resistors: Proof Define resistance R1-2 between points N1 to N2 2 ways: R1-2 = Rc || ( Ra + Rb ) = R1 + R2 R1-2 = Rc * ( Ra + Rb ) / ( Ra + Rb + Rc ) = R1 + R2 Similarly: R2-3 = Ra * ( Rb + Rc ) / ( Ra + Rb + Rc ) = R2 + R3 R3-1 = Rb * ( Ra + Rc ) / ( Ra + Rb + Rc ) = R1 + R3 Now we have 3 equations with 3 unknowns R1 to R3, easy to solve algebraically!

Delta-to-Wye for Resistors Goal of delta to wye transformation in resistive circuit is replacement of Ra, Rb, and Rc in delta circuit at terminals N1, N2, and N3 by other components R1, R2, and R3 with identical behavior, but combined via wye Electrical attributes at N1, N2, and N3 have to stay fixed

Wye-to-Delta for Resistors The inverse operation for resistors is shown below Again: identical method works for general circuits with capacitors and inductors: assuming Ra, Rb, and Rc in delta circuit, and R1, R2, and R3 in wye circuit Electrical attributes at N1, N2, and N3 stay constant

Delta-to-Wye for Impedances Goal of delta to wye transformation is replacing Za, Zb, and Zc in delta circuit at terminals Na, Nb, and Nc by other components Z1, Z2, and Z3 yielding a wye, but with identical electric behavior Using abbreviation: Σ = Za + Zb + Zc, then: Z1 = ( Zb + Zc ) / Σ Z2 = ( Za + Zc ) / Σ Z3 = ( Za + Zb ) / Σ

Wye-to-Delta for Impedances Goal of wye-to-delta transformation is replacing Z1, Z2, and Z3 in wye circuit at terminals Na, Nb, and Nc by other components Za, Zb, and Zc yielding a delta with identical electric behavior Za = ( Z1Z2 + Z2Z3 + Z3Z1 ) / Z1 Zb = ( Z1Z2 + Z2Z3 + Z3Z1 ) / Z2 Zc = ( Z1Z2 + Z2Z3 + Z3Z1 ) / Z3

Exercise 5: Delta-to-Wye Transform We analyze circuit to the right, better resolution next slide: 2 Δ Decide, which to convert to Wye Known NoVoMo and MeCuMo methods do not work here Conversion uses series of all delta components 3 times in denominator; we pick simpler one Upper Δ: Σ = 73.2 - j1.6 Ω Lower Δ: Σ = 30 + j40 Ω Select lower, seems simpler

Exercise 5 Delta Circuit Circuit below actually contains two delta circuits, one at points a) b) c), and the other at points b) c) d)

Exercise 5 Using formula Z1 = ( Zb + Zc ) / Σ three times, with suitable index substitutions, and Σ = 30 + j40: Z1 = ( Zb + Zc ) / Σ Z1 = ( 20 + j60 ) ( 10 ) / ( 30 + j40 ) Z1 = ( 200 + j600 ) ( 30 - j40 ) / ( ( 30 + j40 ) ( 30 – j40 ) ) Z1 = ( 6k –j8k + j18k + 24 k ) / ( 900 + 1600 ) Z1 = 12 + j4 [Ω] Z2 = 10 ( - j20 ) / ( 30 + j40 ) Z2 = -3.2 - j2.4 [Ω] -- negative resistance! Z3 = ( 20 + j60 )(-j20) / ( 30 + j40 ) Z3 = 8 – j24 [Ω]

Exercise 5 – Substitution Circuit Here the equivalent circuit, with delta replaced by wye The top left Y branch Z1 consists of 12+j4 Ω, connecting to the new node n) The top right Z2 branch of Y consists of -3.2-j2.4 Ω, note the negative resistance The bottom neck of the Y consists of the series 8 - j24 Ω See better resolution on the next page

Exercise 5 – Substitution Circuit

Exercise 5 – Substitution Circuit Cleverly  designed, in the left branch of the parallel circuit the components –j4 and j4 cancel out to 0 Ω In the right branch –j2.4 and j2.4 amount to 0 Ω In the right branch 63.2 and -3.2 Ω add up to 60 Ω Which are parallel to 12 Ω Resulting in a single ohmian resistance of 10 Ω This is in series with 8 Ω and –j24 Ω So the total circuit is the series of 18 Ω in series with a capacitor of –j24 Ω

Exercise 5 – Simplified Circuit Now the computation of all units in this circuit is quite simple

Bibliography Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13: 9780-13-376003-3 Radian: https://www.khanacademy.org/math/geometry/cc-geometry-circles/intro-to-radians/v/introduction-to-radians Phasor: https://en.wikipedia.org/wiki/Phasor Euler’s Formula: https://en.wikipedia.org/wiki/Euler%27s_formula https://en.wikipedia.org/wiki/Admittance https://en.wikipedia.org/wiki/Susceptance https://en.wikipedia.org/wiki/Negative_resistance