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electromagnetic and magnetic circuit principles

electromagnetic and magnetic circuit principles. voltage and current waveforms. Example. When an instantaneous voltage of 500 Sin (314t + π/4) is applied to a series circuit of R and L, the current is found to be 10Sin (314t - π/6). Calculate:

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electromagnetic and magnetic circuit principles

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  1. electromagnetic and magnetic circuit principles

  2. voltage and current waveforms

  3. Example When an instantaneous voltage of 500 Sin (314t + π/4) is applied to a series circuit of R and L, the current is found to be 10Sin (314t - π/6). Calculate: i) Peak voltage ii) Frequency iii) Phase angle iv) Impedance v) Resistance vi) Inductance

  4. Example • Peak voltage = 500v • Frequency f = 314/2π = 50 Hz. • Phase angle π/4 + π/6 = 45o + 30o = 75o • Impedance Z = V/I = 500/10 = 50 Ω

  5. Example • Resistance = Z Cos φ = 50 x 0.2588 = 12.94 Ω • Inductive reactance - XL= Z Sin φ = 50 x 0.9659 = 48.30Ω • Inductance L = XL/ 2πf = 48.30/2π x 50 • = 0.1537 H

  6. Circuit possessing resistance only instantaneous value of voltage and current v = VmsinӨ and i = Vm/R sinӨ i = ImsinӨ = Imsinπ2Өft

  7. phasor diagram for the resistive circuit VR VR

  8. phasor diagram for the resistive circuit

  9. Circuit possessing inductance only: instantaneous value of induced e.m.f.: e = -L.di/dt = 2πfLIm instantaneous value of applied voltage v = 2πfLIm cos 2πft = 2πfLImsin(2πft+π/2)

  10. applied voltage, induced E.M.F., and current waveforms

  11. phasor diagram for the inductive circuit VL leads by 90o VL VR

  12. Inductive reactance Vrms/Irms= 0.707Vm/0.707Im = 2πfL = inductive reactance [XL] I = V/2πfL = V/XL[ohms]

  13. Example • An inductor of 0.6H and negligible resistance is connected across a 120 V a.c. supply. • Calculate the current when the frequency is: • i) 30 Hz • ii) 200 Hz

  14. Example • i) XL= 2πfL = 2π x 30 x 0.6 = 113 Ω • IL= V / XL= 120 / 113 = 1.06 A • ii) XL = 2πfL = 2π x 200 x 0.6 = 753 Ω • IL= V / XL= 120 / 753 = 0.159 A

  15. Circuit possessing capacitance only v = Vmsin θ = Vmsin 2πft i = C dv/dt i = 2πfCVm cos 2πft = 2πfCVmsin(2πft+π/2)

  16. Waveforms for capacitive circuit

  17. phasor diagram for the capacitive circuit VR VC lags by 90o VC

  18. Capacitive reactance Vrms/Irms= 0.707Vm/0.707Im = 1/(2πfC) = capacitive reactance [XL]

  19. Example • A capacitor of 0.6 μF is connected across a 120 V ac supply. Calculate the current when the frequency is: • i)30Hz • ii) 200 Hz

  20. i) XC= 1 / 2πfC = 1 / 2π x 30 x 0.6 x 10-6 = 8842 Ω • IC= V / XC= 120 / 8842 = 13.6 mA • i) XC= 1 / 2πfC = 1 / 2π x 200 x 0.6 x 10-6 = 1326 Ω • IC= V / XC = 120 / 1326 = 0.09 A

  21. Series Resonance • The resonance of a series RLC circuit occurs when the inductive and capacitive reactances are equal in magnitude but cancel each other because they are 180 degrees apart in phase. The sharp minimum in impedance which occurs is useful in tuning applications. The sharpness of the minimum depends on the value of R and is characterized by the "Q" of the circuit.

  22. Series Circuits (R,L,C) impedance [Z] =√{R2 +(XL -XC)2} φ = phase angle = tan-1(XL-XC)/R), Cos-1 R/Z, Sin-1 = (XL - XC) / Z phasor diagram

  23. Example • A 10 Ω resistor and 150μF capacitor are connected in series across a 200 Hz, 200 V ac supply. Calculate: • i) Circuit impedance • ii) Current • iii) Phase angle

  24. Example i) Circuit impedance. XC= 1 / 2πfC = 1 / 2π x 200 x 150 x 10-6 = 5.305 Ω Z = √ R2 + XC2 = √ 102 + 5.3052 = √128.143 = 11.32 Ω ii) Current = I = V/Z = 200 / 11.32 = 17.67 A iii) Phase angle = tan-1 XC / R = 27.95 degrees leading

  25. Series Resonance (R,L,C) XL =1/XC f = 1/{2π√(LC)} phasor diagram definition: acceptor circuit graph of current and impedance plotted against Z

  26. Q factor (at resonance) Q = XL/R = 1/R √(L/C) bandwidth - (f2-f1) - definition of half-power points Q = fr/(f2-f1)

  27. Parallel Circuits (R,L,C) supply current = √V/R + V/XL+ XCV) φ = phase angle = phase difference VSand IS φ = tan-1 (IL - IC)/IR phasor diagram

  28. Parallel Resonance (R,L,C) f = 1/(2πL) √(L/C - R2) phasor diagram definition: rejector circuit dynamic impedance RD= L/CR Q factor (at resonance) = XL/R

  29. Terms • Resistance is the opposition to current flow by a resistor • Reactance, is similar, it is the interference of a capacitor or an Inductor to current flow • XL is inductive reactance and XC is capacitive reactance • Impedance (Z) is actually the overall opposition to current presented by the circuit

  30. Conductance, Susceptance, and Admittance are the opposites to Resistance, reactance and impedance

  31. Impedance triangle Impedance Z Reactance X Resistance R

  32. Admittance triangle Conductance G Susceptance B Admittance Y

  33. Conductance [G] = R/Z2. Is 1/R when X is = 0 • admittance [Y] = 1/Z = R/Z2 • susceptance [B] = X/Z2. Is 1/X when X is = 0 • Y = G+ jB and tanφ = B/G

  34. R and L in series Z = R +jXL= Z < φ admittance = Y = 1/Z = (R/Z2 - jXL/Z2) = G – jBL = Y < -φ

  35. R and C in series • Z = R -jXC= Z<-φ • admittance = Y = 1/Z = (R/Z2 + jXC/Z2) = G + jBC = Y<φ

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