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Inductive Reactance

Inductive Reactance. Electronics. Inductors in AC Circuits. Inductance. I nductance opposes a change in current. Inductors create a voltage that opposes the current. Counter EMF (voltage). v. L. Inductance. The Henry (symbol: H) is the SI unit of inductance .

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Inductive Reactance

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  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) v L

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

  5. Current Voltage Capacitor in DC

  6. ELI

  7. E L I Voltage Inductance Current Voltage leads Current in an Inductive Circuit

  8. Current Voltage Inductance in AC

  9. Voltage Current I = Ipsin(2πft) V = L di/dt Ip-Peak Current V = L d(Ipsin(2πft))/dt 2π-Cycle f-frequency(Hz) V = L Ip(2πf)cos(2πft) t-time(seconds) Vp = Ip(2πf)L

  10. Vp = Ip2πfL Vp/Ip= 2πfL R = 2πfL Inductive Reactance XL = 2πfL L is an Active Component

  11. 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. E=6V L=3mH f=60Hz XL = 2πfL I = E/XL XL = 2π(60Hz)(.003H) I = 6V/1.13Ω XL = 1.13Ω I = 5.3A

  12. Inductive/Resistive Circuit • 90° Phase Shift caused by Inductor • Impedance, Z, is calculated by adding XL and R vectorially. XL Z R

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

  14. Z XL R

  15. Current Voltage 4.6° Inductance in AC

  16. Inductance in AC Circuits

  17. Capacitor Values • 1mF = 1 X 10-3F • 1μF = 1 X 10-6F • 1nF = 1 X 10-9F • 1pF = 1 X 10-12F

  18. RLC Circuits

  19. RLC Circuits • 90° Phase Shift caused by Inductor • -90° Phase Shift caused by Capacitor Z XC XL R

  20. Xc = 1/(2πfC) XL = 2πfL Xc = 1/(2π(60)(1.5X10-6) XL = 2π(60)(0.65) Xc = 1768Ω XL = 245Ω Z XC XL 𝚹 = -81° ICE – Current leads Voltage by 81° R

  21. Resonance • The frequency where XL = XC • The Circuit becomes a purely resistive circuit Xc = 1/(2πfC) XL = 2πfL Xc = XL 1/(2πfC) = 2πfL 1/(4π2CL) = f2 = f

  22. = f 161 Hz= f

  23. Worksheet Lab 2-6 Capacitors In AC Circuits Problems

  24. 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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