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Obedience = Happiness

Obedience = Happiness. Mosiah 2:41

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Obedience = Happiness

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  1. 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. Discussion #14 – AC Circuit Analysis

  2. Lecture 14 – AC Circuit Analysis Phasors allow the use of familiar network analysis Discussion #14 – AC Circuit Analysis

  3. L R1 ix vs(t) R2 C + – ~ ia(t) ib(t) RLC Circuits Linear passive circuit elements: resistors (R), inductors (L), and capacitors (C) (a.k.a. RLC circuits) • Assume RLC circuit sources are sinusoidal ZL ZR1 Ix ZR2 ZC + – Vs(jω) Ia(jω) Ib(jω) Time domain Frequency (phasor) domain Discussion #14 – AC Circuit Analysis

  4. Zn ZN Z2 Z3 Z1 ∙ ∙ ∙ ∙ ∙ ∙ RLC Circuits - Series Impedances • Series Rule: two or more circuit elements are said to be in series if the current from one element exclusively flows into the next element. • Impedances in series add the same way resistors in series add ZEQ Discussion #14 – AC Circuit Analysis

  5. Z1 Z2 Z3 Zn ZN ZEQ RLC Circuits - Parallel Impedances • Parallel Rule: two or more circuit elements are said to be in parallel if the elements share the same terminals • Impedances in parallel add the same way resistors in parallel add Discussion #14 – AC Circuit Analysis

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

  7. R1 is(t) R2 C vs(t) + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF Discussion #14 – AC Circuit Analysis

  8. R1 is(t) R2 C vs(t) + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Note frequencies of AC sources Only one AC source - ω = 377 rad/s Discussion #14 – AC Circuit Analysis

  9. R1 Z1 = R1 is(t) Vs=10ej0 Is(jω) R2 C vs(t) Z2=R2 Z3=1/jωC + – + – ~ ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Note frequencies of AC sources • Convert to phasor domain Discussion #14 – AC Circuit Analysis

  10. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Note frequencies of AC sources • Convert to phasor domain • Represent each element by its impedance Discussion #14 – AC Circuit Analysis

  11. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Note frequencies of AC sources • Convert to phasor domain • Represent each element by its impedance • Solve using network analysis • Use node voltage and Ohm’s law Node a Discussion #14 – AC Circuit Analysis

  12. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Solve using network analysis • Use node voltage and Ohm’s law Node a Discussion #14 – AC Circuit Analysis

  13. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Solve using network analysis • Use node voltage and Ohm’s law Node a Discussion #14 – AC Circuit Analysis

  14. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Solve using network analysis • Use node voltage and Ohm’s law Node a Discussion #14 – AC Circuit Analysis

  15. Z1 = R1 Vs=10ej0 Is(jω) Z2=R2 Z3=1/jωC + – ~ RLC Circuits • Example 1: find is(t) • vs(t) = 10cos(ωt), ω = 377 rad/s R1 = 50Ω, R2 = 200Ω, C = 100uF • Solve using network analysis • Use node voltage and Ohm’s law • Convert to time domain Node a Discussion #14 – AC Circuit Analysis

  16. Rs R1 R2 is(t) i1(t) i2(t) L1 vs(t) L2 + – ~ RLC Circuits • Example 2: find i1 and i2 • vs(t)= 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH Discussion #14 – AC Circuit Analysis

  17. Rs R1 R2 is(t) i1(t) i2(t) L1 vs(t) L2 + – ~ RLC Circuits • Example 2: find i1 and i2 • vs(t)= 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Note frequencies of AC sources Only one AC source - ω = 377 rad/s Discussion #14 – AC Circuit Analysis

  18. Zs = Rs Rs R1 R2 is(t) Is(jω) Vs=155ej0 ZR2=R2 ZR1=R1 + – ~ I2(jω) i1(t) i2(t) L1 vs(t) I1(jω) L2 + – ~ ZL2=jωL2 ZL1=jωL1 RLC Circuits • Example 2: find i1 and i2 • vs(t)= 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Note frequencies of AC sources • Convert to phasor domain Discussion #14 – AC Circuit Analysis

  19. Zs = Rs Is(jω) Vs=155ej0 ZR2=R2 ZR1=R1 + – ~ I2(jω) I1(jω) ZL2=jωL2 ZL1=jωL1 RLC Circuits • Example 2: find i1 and i2 • vs(t)= 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Note frequencies of AC sources • Convert to phasor domain • Represent each element by its impedance Discussion #14 – AC Circuit Analysis

  20. Zs = Rs Is(jω) Vs=155ej0 + – I2(jω) ~ Z2=ZR2+ZL2 I1(jω) Z1=ZR1+ZL1 RLC Circuits • Example 2: find i1 and i2 • vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Solve using network analysis • Ohm’s law Discussion #14 – AC Circuit Analysis

  21. RLC Circuits • Example 2: find i1 and i2 • vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Solve using network analysis • KCL V(jω) Zs Is(jω) Vs=155ej0 + – Z2 Z1 ~ I1(jω) I2(jω) Discussion #14 – AC Circuit Analysis

  22. RLC Circuits • Example 2: find i1 and i2 • vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Solve using network analysis • Ohm’s Law V(jω) Zs Is(jω) Vs=155ej0 + – Z2 Z1 ~ I1(jω) I2(jω) Discussion #14 – AC Circuit Analysis

  23. RLC Circuits • Example 2: find i1 and i2 • vs(t) = 155cos(ωt)V, ω = 377 rads/s, Rs = 0.5Ω, R1 = 2Ω, R2 = 0.2Ω, L1 = 0.1H, L2 = 20mH • Convert to Time domain V(jω) Zs Is(jω) Vs=155ej0 + – Z2 Z1 ~ I1(jω) I2(jω) Discussion #14 – AC Circuit Analysis

  24. R1 L vs(t) R2 C + – ~ ia(t) ib(t) RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF Discussion #14 – AC Circuit Analysis

  25. R1 L vs(t) R2 C + – ~ ia(t) ib(t) RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources Only one AC source - ω = 1500 rad/s Discussion #14 – AC Circuit Analysis

  26. R1 L ZR1 ZL vs(t) R2 C + – ZC ~ Vs(jω) ZR2 + – ia(t) ib(t) ~ Ia(jω) Ib(jω) RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources • Convert to phasor domain Discussion #14 – AC Circuit Analysis

  27. ZR1 ZL ZC Vs(jω) ZR2 + – ~ Ia(jω) Ib(jω) RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF • Note frequencies of AC sources • Convert to phasor domain • Represent each element by its impedance Discussion #14 – AC Circuit Analysis

  28. Vs(jω) + – ~ RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Solve using network analysis • Mesh current + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #14 – AC Circuit Analysis

  29. Vs(jω) + – ~ RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Solve using network analysis • Mesh current + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #14 – AC Circuit Analysis

  30. Vs(jω) + – ~ RLC Circuits • Example 3: find ia(t) and ib(t) • vs(t) = 15cos(1500t)V, R1 = 100Ω, R2 = 75Ω, L = 0.5H, C = 1uF +ZR1– +ZL– • Convert to Time domain + ZC – + ZR2 – Ia(jω) Ib(jω) Discussion #14 – AC Circuit Analysis

  31. I + V – Load Source Load Load I I VT(jω) IN(jω) + V – + V – ZT + – ZN AC Equivalent Circuits Thévenin and Norton equivalent circuits apply in AC analysis • Equivalent voltage/current is complex and frequency dependent Thévenin Equivalent Norton Equivalent Discussion #14 – AC Circuit Analysis

  32. Z3 Z1 Z3 Z1 a a + – Z2 Z2 ZT Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computation of Thévenin and Norton Impedances: • Remove the load (open circuit at load terminal) • Zero all independent sources • Voltage sources short circuit (v = 0) • Current sources open circuit (i = 0) • Compute equivalent impedance across load terminals (with load removed) ZL NB: same procedure as equivalent resistance Discussion #14 – AC Circuit Analysis

  33. Z3 Z1 a Z3 Z1 a + – Z2 + VT – Vs(jω) + – Z2 Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computing Thévenin voltage: • Remove the load (open circuit at load terminals) • Define the open-circuit voltage (Voc) across the load terminals • Chose a network analysis method to find Voc • node, mesh, superposition, etc. • Thévenin voltage VT = Voc NB: same procedure as equivalent resistance Discussion #14 – AC Circuit Analysis

  34. Z3 Z1 a Z3 Z1 a + – Z2 IN Vs(jω) + – Z2 Vs(jω) Z4 Z4 b b AC Equivalent Circuits Computing Norton current: • Replace the load with a short circuit • Define the short-circuit current (Isc) across the load terminals • Chose a network analysis method to find Isc • node, mesh, superposition, etc. • Norton current IN = Isc NB: same procedure as equivalent resistance Discussion #14 – AC Circuit Analysis

  35. L Rs vs(t) + vL – + – C ~ RL AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF Discussion #14 – AC Circuit Analysis

  36. L Rs vs(t) + vL – + – C ~ RL AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources Only one AC source - ω = 103 rad/s Discussion #14 – AC Circuit Analysis

  37. L ZL Rs Zs vs(t) + vL – + – C ~ Vs(jω) RL ZC + – ~ ZLD AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources • Convert to phasor domain Discussion #14 – AC Circuit Analysis

  38. AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF • Note frequencies of AC sources • Convert to phasor domain • Find ZT • Remove load & zero sources ZL Zs ZC Discussion #14 – AC Circuit Analysis

  39. AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF ZL • Note frequencies of AC sources • Convert to phasor domain • Find ZT • Remove load & zero sources • Find VT(jω) • Remove load Zs Vs(jω) ZC + VT(jω) – + – ~ NB: Since no current flows in the circuit once the load is removed: Discussion #14 – AC Circuit Analysis

  40. ZT VT(jω) + – ~ ZLD AC Equivalent Circuits • Example 4: find the Thévenin equivalent • ω = 103 rads/s, Rs = 50Ω, RL = 50Ω, L = 10mH, C = 0.1uF ZL Zs Vs(jω) ZC + – ~ ZLD Discussion #14 – AC Circuit Analysis

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