Unit 15 - Capacitance and RC Circuits F07

Unit 15 - Capacitance and RC Circuits F07 - 17 7 29-Oct Mon...

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7 17 29-Oct Mon AC Motors Primer Ch13 AC/DC Generators/Oscope SA 13-1, 13-2 18 31-Oct Wed Ch 9-13 Summary Ch9-13 (SA 10-3, 11-3, 11-5, 12-2) Summary Handout 19 2-Nov Fri Exam #3 - 8 20 5-Nov Mon Inductance & RL Circuits Ch14 RLC Circuits SA 14-1 21 7-Nov Wed Inductance & RL Circuits Ch14 (SA 14-2, 15-2) 22 9-Nov Fri Capacitance and RC Circuits Ch15 SA 15-1 9 12-Nov Mon Veteran's Day: No School RLC Circuits 23 14-Nov Wed Tuned RLC Circuits Ch16 (SA 16-2, 16-3) SA 16-1 24 16-Nov Fri Diodes & Power Supplies Ch17 SA 17-1 10 25 19-Nov Mon Diodes & Power Supplies Ch17 NO LAB 21-Nov Wed Thanksgiving Holiday 23-Nov Fri Thanksgiving Holida y 11 26 26-Nov Mon Ch 14-17 Summary Ch14-17 Diodes and Power Supplies Summary Handout 27 28-Nov Wed Exam #4 - (SA 17-3, 17-4, 17-6, 17-7) 28 30-Nov Fri Final Exam Review 5-Dec Wed Final Exam 0710-1000)
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Ch14: Inductance in AC Circuits Ch14: Inductance in AC Circuits Series RL Circuits: A series circuit contains an 250 mH choke and a 100 resistor connected to a 120 V 60 Hz AC source. Find the time constant, impedance, current, voltage drops, phase angle, apparent power, true power, and power factor. Draw the phasor diagram. Solution: τ = L/R = .25/100 = 2.5 ms X L = 2 π fL = (2)(3.14)(60)(0.25) = 94.2 Z = (R 2 + X L 2 ) = 100 2 + 94.2 2 ) = 137.38 I T = E S /Z = 120/137.38 = 0.873 A E R = I R x R = (0.873)(100) = 87.3 V E L = I L x X L = (0.873)(94.2) = 82.24 V θ = cos-1(R/Z) = cos-1(100/137.38) = cos-1(.7279) = 43.29° P a = E S x I T = (120)(0.873) = 104.76 VA P t = E S x I T x cos θ = (120)(0.873)(.7279) = 76.25 W PF = P t /P a = 76.25/104.76 = .7279 = cos θ
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Draw the phasor diagram (X L /R) 94.2 Z = 137.38 X L θ = 43.29° R 100 Ch14: Inductance in AC Circuits Ch14: Inductance in AC Circuits
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Parallel RL Circuits: Given a circuit that contains an 250 mH choke in parallel with a 100 resistor connected to a 120 V 60 Hz AC source. Find the time constant, branch currents, total current, impedance, phase angle, apparent power, and true power. Draw the phasor diagram Solution: τ = L/R = .25/100 = 2.5 ms X L = 2 π fL = (2)( π )(60)(.25) = 94.2 I R = E S /R = 120/100 = 1.2 A I L = E S / X L = 120/94.2 = 1.27 A I T = (I R 2 + I L 2 ) = (1.2 2 + 1.27 2 ) = 1.75 A θ = tan-1(I L /I R ) = tan-1(1.27/1.2) = tan-1(1.0583) = 46.62° Z = E S /I T = 120/1.75 = 68.57 P a = E x I = (120)(1.75) = 210 VA P t = E x I x cos θ = (120)(1.75)(0.6868) = 144.23 W PF = P t /P a (= cos θ ) = 144.22/210.0 = .6868 Ch14: Inductance in AC Circuits Ch14: Inductance in AC Circuits
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Draw the phasor diagram (I L /I R ) I C ` I R IT = 1.75 A Ch14: Inductance in AC Circuits Ch14: Inductance in AC Circuits
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Electronic Systems Electronic Systems Chapter 15: Chapter 15: Capacitance and RC Circuits
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Ch15: Capacitance Ch15: Capacitance Today’s Stuff . . . - Capacitors - Capacitance - Types of capacitors - Capacitor transient response - Capacitance in DC circuits - Capacitance in AC circuits - LC circuits
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Ch15: Capacitance Ch15: Capacitance Why an IT graduate should know this . . .
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Unit 15 - Capacitance and RC Circuits F07 - 17 7 29-Oct Mon...

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