Phasor Representation
Phasor Diagram is a graphical representation of the relation
between two or more alternating quantities in terms of
magnitude and direction.
1
Series RL circuit
v = Applied RMS voltage
i = Resultant RMS current
VR= i R= Voltage drop across R ( in phase with i)
VL= i XL= Voltage drop across L (right angle to i )
XL = Inductive Reactance
Phasor Diagram Impedance Triangle 2
Series RC circuit
v = RMS of applied voltage
i = RMS of resultant current
VR= i R= Voltage drop across R
( in phase with i)
XC = Capacitive Reactance
VC= i XC= Voltage drop across C (lagging i by 900)
Impedance Triangle
3
Series RLC circuit
V= RMS of applied voltage
VR= i R= Voltage drop across R ( in phase with i )
VL= i XL(leading i by 90O)
VC= i XC= drop across C (lagging i by 900)
4
Continued…
V= Vm Sin(ωt)
tan ɸ= (XL-XC )/ R
i = Im sin(ωt ± ϕ)
= X/R
When XL>XC use - sign
When XC >XL use + sign 5
1. A capacitor of 8 µF takes a current on 1 A when the applied voltage
is 230V. Find
a) Frequency of applied voltage
b) Resistor that needs to be added in series to capacitor for reducing
current to 0.5 A.( same frequency)
c) Phase angle of resulting circuit.
Solution Z= V/i
= 230/0.5= 460Ω
XC= V/i= 230/1= 230Ω Z2= R2+XC2
R= 398Ω
XC= 1/cω = 1/ C.2π F
tan ɸ = XC/R
F= 86.5 Hz = tan-1 (XC/R)
= 300
6
2. A series RLC circuit with resistance of 12Ω, inductance of 0.15H and
capacitance of 100µF is connected across a voltage of 100V, 50 Hz
Find the following.
a) Impedance
b) Current
c) Voltage across R, L and C
d) Phase difference between v and i
Solution
i = V/Z = 100/19.4 = 5.15 A
F= 50 Hz and ω= 2π F
VR= i R= 5.15x12=61.8V
XL= Lω= L (2π F)= 47.1 Ω
VL= i XL=242.5V
XC= 1/Cω = 1/ C(2π F)
= 31.85 Ω
VC= i XC=164V
Z2= R2+ (XL-XC)2
Z = 19.4Ω tan ϕ = (XL-XC)/R , ϕ=51⁰ 7
