Rlc circuits det2033 chp2

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R-L-C Circuits DET 2033

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  1. 1. R-L-C Circuits DET 2033
  2. 2. How vector is represented?
  3. 3. A Purely Resistive AC Circuit Purely resistive ac circuit The figure shows an ac circuit consisting of a purely resistor to which an alternating voltage v = Vm sin t is applied.
  4. 4. Since the circuit contains only a pure resistor, the applied voltage has to overcome only the ohmic resistive drop due to the current flowing in the circuit. The instantaneous value of the current in the circuit is given by i = v / R = (Vm sin t) / R But Vm / R =Im (Maximum current) Therefore i = Im sin t
  5. 5. Since v = Vm sin t and i = Im sin t We can say that for a purely resistive circuit, both voltage and current are in same phase; i.e. they may have different peak values but they attain their zero and maximum values at the same time.
  6. 6. Phase Relationship And Vector Diagram
  7. 7. Power In A Purely Resistive Circuit The instantaneous value of the power drawn by this circuit is given by the product of instantaneous values of current and voltage. i.e. P = v * i P = (Vm sin t)(Im sin t) = Vm Im sin t = ((Vm Im)/2) (1-cos2 t) =[(Vm Im)/2] - [((Vm Im)/2) (cos2 t)]
  8. 8. Constant part=(Vm Im)/2 A fluctuating part=((Vm Im)/2) (cos2 t) The average value of the fluctuating part of the power [((Vm Im)/2) (cos2 t)] is zero over a complete cycle. So total power expended in producing heat in whole cycle is P = (Vm Im)/2 = Vrms Irms = V I watts Thus no part of the power cycle is negative at any instant i.e. power in a purely resistive circuit is never zero.
  9. 9. A Purely Inductive AC Circuit Pure inductor is said to be pure if it contains no resistance. Purely inductive ac circuit
  10. 10. Due to the self inductance of coil, there will be back emf produced which opposes change in current. v = Vm sin t = -(-L di/dt) di = (Vm/ L) sin t dt Integrating we get; i = -(Vm / L ) cos t = (Vm / L ) sin(t /2) Here i will be maximum when sin(t /2) = 1 Thus Im = Vm / L = Vm / XL Where XL = L, this is known as inductive reactance.
  11. 11. i = Im sin(t /2) v = Vm sin t Thus current flowing through purely inductive circuit lags by 90. Phase Relationship And Vector Diagram
  12. 12. Power In A Purely Inductive Circuit P = vi = (Vm sin t)(Im sin(t /2)) = (Vm sin t)(- Im cos t) = -(2Vm Im sin t cos t) /2 = -(Vm Im /2) sin2 t The average power consumed by a purely inductive circuit is zero.
  13. 13. A Purely Capacitive AC Circuit Pure capacitor has zero resistance. Thus when an alternating voltage is applied to plates of the capacitor, the capacitor is charged first in one direction then in another direction. Purely capacitive ac circuit
  14. 14. We know q = CV q = C Vm sin t The current is given by rate of change of charge i = dq/dt = d/dt (C Vm sin t ) = C Vm cos t = [Vm / (1/ C)] sin(t + /2) Thus in purely capacitive circuit current leads voltage by 90.
  15. 15. Here i will be maximum when sin(t + /2)=1 Xc = 1 / C is called capacitive resistance and it is measured in ohms. Im = Vm / (1/ C) i = Im sin(t + /2) = Im cos t Phase Relationship And Vector Diagram
  16. 16. Power In A Purely Capacitive Circuit P = vi = (Vm sin t)(Im cos t) =(2Vm Im sin t cos t) /2 =(Vm Im /2) sin2 t The average power consumed in a purely capacitive circuit is zero.
  17. 17. R-L Series Circuit Let V be the rms value of applied voltage. VR, the voltage across the resistance R. VL, the voltage across the inductance L. I, the rms value of the current flowing in the circuit. R-L series circuit
  18. 18. In series circuit current flowing through R & L will be the same but voltage is divided. Vector sum of VR & VL will be equal to V. We know for resistor, VR & I are in same phase whereas for pure inductor VL leads I by 90. LV I RV V Phasor diagram
  19. 19. .. V = (VR + VL) = (IR) + (IXL) = I (R) + (XL) = IZ Where Z=(R) + (XL) (Impedance) = (Resistance) + (Reactance)
  20. 20. From figure it can be said that voltage leads current by an angle such that, tan = VL / VR = IXL / IR = XL / R = tan(XL / R) Thus in R-L series circuit current lags voltage by an angle If V= Vm sin t I = Im sin(t- ) Where = tan(XL / R)
  21. 21. POWER IN R-L SERIES CIRCUIT P = VI In R-L series circuit, V = Vm sin t I = Im sin(t- ) P = [Vm sin t][Im sin(t- )] = [(Vm Im)/2](2 sin t sin(t- )) = [(Vm Im)/2] cos - [(Vm Im)/2] cos(2t- )
  22. 22. Constant part = [(Vm Im)/2] cos Variable part = [(Vm Im)/2] cos(2t- ) Average of variable power component over a complete cycle is zero. Thus average power over complete cycle is given by, Pav = Vm Im cos = Vrms Irms cos cos is known as Power Factor.
  23. 23. R-C Series Circuit Let V be the rms value of applied voltage. VR, the voltage across the resistance R. VC, the voltage across the capacitance C. I, the rms value of the current flowing in the circuit. R-C series circuit
  24. 24. In series circuit current flowing through R & C will be the same but voltage is divided. Vector sum of VR & VC will be equal to V. We know for resistor, VR & I are in same phase whereas for pure capacitor Vc lags I by 90. CV I RV V Phasor diagram
  25. 25. .. V = (VR + VC) = (IR) + (IXC) = I (R) + (XC) = IZ Where Z=(R) + (XC) (Impedance) = (Resistance) + (Capacitance)
  26. 26. From figure it can be said that voltage lags current by an angle such that, tan = VC / VR = IXC / IR = XC / R = tan(XC / R) Thus in R-L series circuit current leads voltage by an angle If V= Vm sin t I = Im sin(t+ ) Where = tan(XC / R)
  27. 27. POWER IN R-C SERIES CIRCUIT P = VI In R-C series circuit, V = Vm sin t I = Im sin(t + ) P = [Vm sin t][Im sin(t + )] = [(Vm Im)/2](2 sin t sin(t + )) = [(Vm Im)/2] cos - [(Vm Im)/2] cos (2t + )
  28. 28. Constant part = [(Vm Im)/2] cos Variable part = [(Vm Im)/2] cos(2t + ) Average of variable power component over a complete cycle is zero. Thus average power over complete cycle is given by, Pav = Vm Im cos = Vrms Irms cos **cos is known as Power Factor.
  29. 29. R-L-C Series Circuit Consider a circuit consisting of R resistance, pure inductor of inductance L Henry and pure capacitor of C farads in series. I V R RV L LV CV C R-L-C series circuit
  30. 30. Let V be the rms value of applied voltage. VR, the voltage across the resistance R. VL, the voltage across the inductance L. VC, the voltage across the capacitance C. I, the rms value of the current flowing in the circuit.
  31. 31. Current is taken as reference VR is drawn in phase with it. VL is drawn leading by 90, Vc is drawn lagging by 90, Since VL and Vc are in opposition to each other there can be two cases: 1) VL > VC 2) VL < VC
  32. 32. Case-1:When VL > VC the phasor diagram would be; V would be the vector sum of VR and (VL - VC)
  33. 33. V = (VR) + (VL - VC) = (IR) + (IXL - IXC) = I [R+ (XL - XC)] V = I [R+ (XL - XC)] I = V / [R+ (XL - XC)] Z = [R+ (XL - XC)] tan = (VL VC) / VR = tan((VL VC) / VR)
  34. 34. When VL > VC current lags voltage by angle V = Vm sin t I = Im sin(t - ) Im = Vm / Z = V / [R+ (XL - XC)] Power consumed P = VI= [Vm sin t][Im sin(t- )] = [(Vm Im)/2](2 sin t sin(t- )) = [(Vm Im)/2] cos - [(Vm Im)/2] cos(2t- )
  35. 35. Average of variable power component over a complete cycle is zero. Thus average power over complete cycle is given by, Pav = Vm Im cos = Vrms Irms cos
  36. 36. Case-2:When VL < VC V = (VR) + (VC - VL) = (IR) + (IXC - IXL) = I [R+ (XC - XL)] V = I [R+ (XC - XL)] I = V / [R+ (XC - XL)] Z = [R+ (XC - XL)] tan = (VC VL) / VR = tan((VC VL) / VR)
  37. 37. When VL < VC current leads voltage by angle V = Vm sin t I = Im sin(t + ) Im = Vm / Z = V / [R+ (XC - XL)] Power consumed P = VI = [Vm sin t][Im sin(t + )] = [(Vm Im)/2](2 sin t sin(t + )) = [(Vm Im)/2] cos - [(Vm Im)/2] cos(2t + )
  38. 38. Average of variable power component over a complete cycle is zero. Thus average power over complete cycle is given by, Pav = Vm Im cos = Vrms Irms cos
  39. 39. END of CHAPTER 2