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Inductive Reactance Electronics. Inductors in AC Circuits.

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Presentation on theme: "Inductive Reactance Electronics. Inductors in AC Circuits."— Presentation transcript:

1 Inductive Reactance Electronics

2 Inductors in AC Circuits

3 Inductance Inductance opposes a change in current. Inductors create a voltage that opposes the current. Counter EMF (voltage) L v

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5 Inductance The Henry (symbol: H) is the SI unit of inductance. It is named after the American physicist Joseph Henry..

6 Current Voltage Capacitor in DC

7 ELI

8 Voltage InductanceCurrent Voltage leads Current in an Inductive Circuit

9 Current Voltage Inductance in AC

10 Current Voltage I = I p sin(2πft) I p -Peak Current 2π-Cycle f-frequency(Hz) t-time(seconds) V = L di/dt V = L I p (2πf)cos(2πft) V p = I p (2πf)L V = L d(I p sin(2πft))/dt

11 V p = I p 2πfL R = 2πfL Inductive Reactance X L = 2πfL L is an Active Component V p /I p = 2πfL

12 Calculate the maximum current in a coil which has an inductance of 3 mH. The frequency is 60Hz. The maximum voltage across the coil is 6 V. X L = 2πfL L=3mH E=6V f=60Hz X L = 2π(60Hz)(.003H) X L = 1.13Ω I = E/X L I = 6V/1.13Ω I = 5.3A

13 Inductive/Resistive Circuit 90° Phase Shift caused by Inductor Impedance, Z, is calculated by adding X L and R vectorially. XLXL R Z

14 What is the impedance of a 100mH choke in series with a 470 Ω resistor with a 12V, 60Hz applied across them? What is the phase angle between voltage and current? X L = 2πfL L=100mH E=12V f=60Hz X L = 2π(60Hz)(.1H) X L = 37.7Ω R=470Ω

15 XLXL R Z

16 Current Voltage Inductance in AC 4.6°

17 Inductance in AC Circuits

18  1mF = 1 X 10 -3 F  1μF = 1 X 10 -6 F  1nF = 1 X 10 -9 F  1pF = 1 X 10 -12 F Capacitor Values

19 RLC Circuits

20 90° Phase Shift caused by Inductor XLXL R Z XCXC -90° Phase Shift caused by Capacitor

21 X c = 1/(2πfC) X c = 1/(2π(60)(1.5X10 -6 ) X c = 1768Ω X L = 2πfL X L = 2π(60)(0.65) X L = 245Ω XLXL R Z XCXC = -81° ICE – Current leads Voltage by 81°

22 Resonance The frequency where X L = X C The Circuit becomes a purely resistive circuit X c = 1/(2πfC) X L = 2πfL 1/(2πfC) = 2πfL X c = X L 1/(4π 2 CL) = f 2 = f

23 161 Hz= f

24 Worksheet Lab 2-6 Capacitors In AC Circuits Problems

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28 Reactance Test Classwork Lab 7, Book 2 – Capacitive Reactance Worksheets Lab 2-5 Lab 2-6 Lab 2-8 Lab 2-9 Lab 2-11 Lab 2-12 Lab 2-13 } Due the day of the test!!


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