Class 12 Physics Alternating Current Phasor diagram for Series LCR circuit

Phasor diagram for Series LCR circuit

  • Resistor, inductor and capacitor all are in series.
  • Source voltage V = Vm sinωt and current I = Im sin (ωt + f).
  • VL = voltage across inductor, VC =voltage across capacitor, VR = voltage across resistor and V = source voltage.
  • Peak values: - VL =  Im XL , VC = ImXC , VR = ImR  and  V =Vm .
  • Resistor VR and I are in phase .Inductor I lag behind the VL.
  • In Capacitor VC lag behind the IL.
  • From the phasor diagram we can see that the VL  and  VC  are exactly in opposite direction with each other and are in same line.
  • The length of the phasors represents the peak values (Im XL, ImXC  and ImR)  .
  • From the circuit , VR + VC + VL = V.
  • Refer  figure(2)  from the phasors  a  right triangle is obtained whose hypotenuse =V .
  • Using  Pythagoras  theorem, V2R + (VC - VL)2 =  V2m
  • V2m = (ImR)2  +  [ImXC   -  Im XL]2
  • = I2mR2  + I2m(XC - XL)2
  • V2m  = I2m  [R2 + (XC - XL)2]
  • Vm  =  Im √( R2 + (XC - XL)2)
  • Comparing the above equation with V=IR , then R = √( R2 + (XC - XL)2).
  • Therefore  Z = √( R2 + (XC - XL)2)

 (a) Relation between the phasors VL, VR, VC, and I,

(b) Relation between the phasors VL, VR, and (VL + VC) for the circuit.

 

(a) Phasor diagram of V and I.

(b) Graphs of v and i versus w t for a series LCR circuit where XC > XL.

Problem:- A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 μF.

Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.  

Answer:-

(a) To find the impedance of the circuit, we first calculate XL and XC.

XL = 2 π ν L

= 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W

XC = (1/2 π ν C)

= (1/ 2 × 3.14 × 50 × 796 x 10-6)

= 4Ω

Therefore,

Z = √R2 + (XL − XC )2 =√(3)2 + (8 − 4)2

= 5 Ω

(b) Phase difference, f = tan–1 ϕ (XC – XL)/(R)

=tan-1 ((4-8)/(3))

=-53.10

Since ϕ is negative, the current in the circuit lags the voltage across the source.

(c) The power dissipated in the circuit is

 P2 = I R

Now, I = (IM)/(√2)

=(1/√2 )(283/5)

=40 A

Therefore, P = (40A )2× 3W = 4800

(d) Power factor = cos ϕ = cos53.1° = 0.6

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