ECEN 301Discussion #14 – AC Circuit Analysis1 DateDayClass No. TitleChaptersHW Due date Lab Due date Exam 20 OctMon14AC Circuit Analysis4.5 NO LAB 21 OctTue NO LAB 22 OctWed15Transient Response 1 st Order Circuits OctThu 24 OctFri Recitation HW 6 25 OctSat 26 OctSun 27 OctMon16Transient Response 2 nd Order Circuits 5.5 LAB 5 28 OctTueExam 1 Schedule…

ECEN 301Discussion #14 – AC Circuit Analysis2 Obedience = Happiness Mosiah 2:41 41 And moreover, I would desire that ye should consider on the blessed and happy state of those that keep the commandments of God. For behold, they are blessed in all things, both temporal and spiritual; and if they hold out faithful to the end they are received into heaven, that thereby they may dwell with God in a state of never-ending happiness. O remember, remember that these things are true; for the Lord God hath spoken it.

ECEN 301Discussion #14 – AC Circuit Analysis3 Lecture 14 – AC Circuit Analysis Phasors allow the use of familiar network analysis

ECEN 301Discussion #14 – AC Circuit Analysis4 RLC Circuits Linear passive circuit elements: resistors (R), inductors (L), and capacitors (C) (a.k.a. RLC circuits) ÙAssume RLC circuit sources are sinusoidal Z R1 Z R2 V s (j ω) +–+– IxIx Z L ZCZC I a (jω) I b (jω) R 1 R2R2 ixix L C i a (t) i b (t) v s (t) +–+– ~ Time domainFrequency (phasor) domain

ECEN 301Discussion #14 – AC Circuit Analysis5 RLC Circuits - Series Impedances uSeries Rule: two or more circuit elements are said to be in series if the current from one element exclusively flows into the next element. ÙImpedances in series add the same way resistors in series add Z EQ Z1Z1 ∙ ∙ ∙ Z2Z2 Z3Z3 ZnZn ZNZN

ECEN 301Discussion #14 – AC Circuit Analysis6 RLC Circuits - Parallel Impedances uParallel Rule: two or more circuit elements are said to be in parallel if the elements share the same terminals ÙImpedances in parallel add the same way resistors in parallel add Z1Z1 Z2Z2 Z3Z3 ZnZn ZNZN Z EQ

ECEN 301Discussion #14 – AC Circuit Analysis7 RLC Circuits AC Circuit Analysis 1.Identify the AC sources and note the excitation frequency (ω) 2.Convert all sources to the phasor domain 3.Represent each circuit element by its impedance 4.Solve the resulting phasor circuit using network analysis methods 5.Convert from the phasor domain back to the time domain

ECEN 301Discussion #14 – AC Circuit Analysis8 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF R 1 R2R2 C i s (t) v s (t) +–+– ~

ECEN 301Discussion #14 – AC Circuit Analysis9 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF R 1 R2R2 C i s (t) v s (t) +–+– ~ 1.Note frequencies of AC sources Only one AC source - ω = 377 rad/s

ECEN 301Discussion #14 – AC Circuit Analysis10 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF R 1 R2R2 C i s (t) v s (t) +–+– ~ Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 1.Note frequencies of AC sources 2.Convert to phasor domain

ECEN 301Discussion #14 – AC Circuit Analysis11 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Represent each element by its impedance

ECEN 301Discussion #14 – AC Circuit Analysis12 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Represent each element by its impedance 4.Solve using network analysis Use node voltage and Ohm’s law Node a

ECEN 301Discussion #14 – AC Circuit Analysis13 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 4.Solve using network analysis Use node voltage and Ohm’s law Node a

ECEN 301Discussion #14 – AC Circuit Analysis14 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 4.Solve using network analysis Use node voltage and Ohm’s law Node a

ECEN 301Discussion #14 – AC Circuit Analysis15 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 4.Solve using network analysis Use node voltage and Ohm’s law Node a

ECEN 301Discussion #14 – AC Circuit Analysis16 RLC Circuits uExample1: find i s (t) Ùv s (t) = 10cos(ωt), ω = 377 rad/s R 1 = 50Ω, R 2 = 200Ω, C = 100uF Z 1 = R 1 Z 2 =R 2 Z 3 =1/jωC I s (jω) V s =10e j0 +–+– ~ 4.Solve using network analysis Use node voltage and Ohm’s law 5.Convert to time domain Node a

ECEN 301Discussion #14 – AC Circuit Analysis17 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH R s R1R1 i s (t) v s (t) +–+– ~ L2L2 R2R2 L1L1 i 1 (t) i 2 (t)

ECEN 301Discussion #14 – AC Circuit Analysis18 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH R s R1R1 i s (t) v s (t) +–+– ~ L2L2 R2R2 L1L1 i 1 (t) i 2 (t) 1.Note frequencies of AC sources Only one AC source - ω = 377 rad/s

ECEN 301Discussion #14 – AC Circuit Analysis19 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH R s R1R1 i s (t) v s (t) +–+– ~ L2L2 R2R2 L1L1 i 1 (t) i 2 (t) Z s = R s Z R1 =R 1 Z R2 =R 2 I s (jω) V s =155e j0 +–+– ~ Z L1 =jωL 1 Z L2 =jωL 2 I 1 (jω) I 2 (jω) 1.Note frequencies of AC sources 2.Convert to phasor domain

ECEN 301Discussion #14 – AC Circuit Analysis20 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH Z s = R s Z R1 =R 1 Z R2 =R 2 I s (jω) V s =155e j0 +–+– ~ Z L1 =jωL 1 Z L2 =jωL 2 I 1 (jω) I 2 (jω) 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Represent each element by its impedance

ECEN 301Discussion #14 – AC Circuit Analysis21 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH Z s = R s Z 1 =Z R1 +Z L1 Z 2 =Z R2 +Z L2 I s (jω) V s =155e j0 +–+– ~ I 1 (jω) I 2 (jω) 4.Solve using network analysis Ohm’s law

ECEN 301Discussion #14 – AC Circuit Analysis22 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH Z s Z1Z1 Z2Z2 I s (jω) V s =155e j0 +–+– ~ I 1 (jω) I 2 (jω) 4.Solve using network analysis KCL V(jω)

ECEN 301Discussion #14 – AC Circuit Analysis23 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH Z s Z1Z1 Z2Z2 I s (jω) V s =155e j0 +–+– ~ I 1 (jω) I 2 (jω) 4.Solve using network analysis Ohm’s Law V(jω)

ECEN 301Discussion #14 – AC Circuit Analysis24 RLC Circuits uExample2: find i 1 and i 2 Ùv s (t) = 155cos(ωt)V, ω = 377 rads/s, R s = 0.5Ω, R 1 = 2Ω, R 2 = 0.2Ω, L 1 = 0.1H, L 2 = 20mH Z s Z1Z1 Z2Z2 I s (jω) V s =155e j0 +–+– ~ I 1 (jω) I 2 (jω) 5.Convert to Time domain V(jω)

ECEN 301Discussion #14 – AC Circuit Analysis25 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t)

ECEN 301Discussion #14 – AC Circuit Analysis26 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t) 1.Note frequencies of AC sources Only one AC source - ω = 1500 rad/s

ECEN 301Discussion #14 – AC Circuit Analysis27 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF R 1 v s (t) +–+– ~ R2R2 L C i a (t) i b (t) 1.Note frequencies of AC sources 2.Convert to phasor domain Z R1 V s (jω) +–+– ~ Z R2 I a (jω) Z L Z C I b (jω)

ECEN 301Discussion #14 – AC Circuit Analysis28 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF Z R1 V s (jω) +–+– ~ Z R2 I a (jω) Z L Z C I b (jω) 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Represent each element by its impedance

ECEN 301Discussion #14 – AC Circuit Analysis29 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 4.Solve using network analysis Mesh current

ECEN 301Discussion #14 – AC Circuit Analysis30 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 4.Solve using network analysis Mesh current

ECEN 301Discussion #14 – AC Circuit Analysis31 RLC Circuits uExample3: find i a (t) and i b (t) Ùv s (t) = 15cos(1500t)V, R 1 = 100Ω, R 2 = 75Ω, L = 0.5H, C = 1uF +Z R1 – V s (jω) +–+– ~ + Z R2 – I a (jω) +Z L – + Z C – I b (jω) 5.Convert to Time domain

ECEN 301Discussion #14 – AC Circuit Analysis32 AC Equivalent Circuits Thévenin and Norton equivalent circuits apply in AC analysis ÙEquivalent voltage/current will be complex and frequency dependent Load +V–+V– I Source V T (jω) +–+– ZTZT Load +V–+V– I I N (jω) ZNZN Load +V–+V– I Norton Equivalent Thévenin Equivalent

ECEN 301Discussion #14 – AC Circuit Analysis33 AC Equivalent Circuits Computation of Thévenin and Norton Impedances: 1.Remove the load (open circuit at load terminal) 2.Zero all independent sources ÙVoltage sources short circuit (v = 0) ÙCurrent sources open circuit (i = 0) 3.Compute equivalent impedance across load terminals (with load removed) NB: same procedure as equivalent resistance Z L Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 Z 3 Z2Z2 Z 4 a b ZTZT

ECEN 301Discussion #14 – AC Circuit Analysis34 AC Equivalent Circuits Computing Thévenin voltage: 1.Remove the load (open circuit at load terminals) 2.Define the open-circuit voltage (V oc ) across the load terminals 3.Chose a network analysis method to find V oc Ùnode, mesh, superposition, etc. 4.Thévenin voltage V T = V oc Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b +VT–+VT– NB: same procedure as equivalent resistance

ECEN 301Discussion #14 – AC Circuit Analysis35 AC Equivalent Circuits Computing Norton current: 1.Replace the load with a short circuit 2.Define the short-circuit current (I sc ) across the load terminals 3.Chose a network analysis method to find I sc Ùnode, mesh, superposition, etc. 4.Norton current I N = I sc Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b Z 1 V s (j ω) +–+– Z 3 Z2Z2 Z 4 a b ININ NB: same procedure as equivalent resistance

ECEN 301Discussion #14 – AC Circuit Analysis36 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL–

ECEN 301Discussion #14 – AC Circuit Analysis37 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL– 1.Note frequencies of AC sources Only one AC source - ω = 10 3 rad/s

ECEN 301Discussion #14 – AC Circuit Analysis38 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF R s v s (t) +–+– ~ RLRL L C +vL–+vL– 1.Note frequencies of AC sources 2.Convert to phasor domain Z s Z LD +–+– ~ ZLZL ZCZC V s (jω)

ECEN 301Discussion #14 – AC Circuit Analysis39 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Find Z T Remove load & zero sources Z s ZLZL ZCZC

ECEN 301Discussion #14 – AC Circuit Analysis40 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF 1.Note frequencies of AC sources 2.Convert to phasor domain 3.Find Z T Remove load & zero sources 4.Find V T (jω) Remove load Z s +–+– ~ ZLZL ZCZC V s (jω) + V T (jω) – NB: Since no current flows in the circuit once the load is removed:

ECEN 301Discussion #14 – AC Circuit Analysis41 AC Equivalent Circuits uExample4: find the Thévenin equivalent Ùω = 10 3 rads/s, R s = 50Ω, R L = 50Ω, L = 10mH, C = 0.1uF Z s Z LD +–+– ~ ZLZL ZCZC V s (jω) Z T +–+– ~ V T (jω) Z LD