Lcr Circuit Graphical Solution Alternating Currents L-5
LCR Circuit Graphical Solution Alternating Currents
Lcr Circuit Graphical Solution Alternating Currents L-5
AC Circuit With Capacitive Load
A.C circuit with capacitive load.
Purely Resistive load: Current is in phase with voltage
Inductive load (L): Current lags behind voltage
Capacitive load (C) : Current leads the voltage
Lcr Circuit Graphical Solution Alternating Currents L-5
LC Circuit
E - voltage
L - inductor
I - current
c - capacitive
Lcr Circuit Graphical Solution Alternating Currents L-5
LC Circuit
The Ratio of Voltage Maximum to Current Maximum.
Resistive circuit IM=RVm
Inductive load : Inductive Resistance
XL=ωL
Im=XLVm=ωLVm
Capacitive Load : XC=ωC1
Im=XCVm=Vm.ωC
Lcr Circuit Graphical Solution Alternating Currents L-5
Phase
V(t)=Vmsinωt
I(t)=Imsin(ωt+ϕ)
ϕ - phase by which current leads voltage
Resistive : ϕ = 0
Inductive : ϕ=2−π
Lcr Circuit Graphical Solution Alternating Currents L-5
Impedance
V(t)=Vmeiωt=Vm[cosωt+isinωt]
I(t) =Imei(ωt+Φ)
Complex Impedance: Z=I(t)V(t)=ImVm⋅e−ιφ
For resistive circuit: φ=0,Z=R
For inductive circuit: φ=−π/2, Z=ωLeiπ/2=iωL
For a capacitive: Z=ωc1e−iπ/2=(−i)ωc1=iωc1.
Lcr Circuit Graphical Solution Alternating Currents L-5
Impedance
Z=R+i[ωL−ωC1]
=R+i[XL−XC]
|Z| =R2+(XL−XC)2
tanφ=RXC−XL
Impedance triangle: ∣XC−XL∣
Lcr Circuit Graphical Solution Alternating Currents L-5
Power
Average for a resistive circuit
2ImR : rms current
2Im=Irms
P≡Irms2R
Inductors and Capacitors
Lcr Circuit Graphical Solution Alternating Currents L-5
LCR Circuit (Series)
V(t)−VR−VL−Vc=0
V(t)=IR+LdtdI+CQ
Current must be the same through the three elements.
I=Imsin(ωt+φ)
Voltage is in series
V=Vmsinωt
Lcr Circuit Graphical Solution Alternating Currents L-5
LCR Circuit (Series)
I=Imsin(ωt+ϕ).
∣OA∣=(XC−XL)Im
Vm2=(Vmm)2+(VCm−VLm)2
=(ImR)2+Im2(XC−XL)2
Vm=ImR2+(XC−XL)2=ImZ
Z=R2+(XC−XL)2
Lcr Circuit Graphical Solution Alternating Currents L-5
Impedance
Z⟶R
Z⟶XC,XL
Z=R2+(XC−XL)2
Example: r.m.s current of 2A is passing through the circuit.
Lcr Circuit Graphical Solution Alternating Currents L-5
Problem
Z =R2+(XC−XL)2
=802+602=100Ω
Vrms=Irms×Z=200V−rms
VR=IR=2×80=160V(rms)
VC=2XC=200V(rms)
VL=2XL=80V(rms)
Lcr Circuit Graphical Solution Alternating Currents L-5
Impedance
VR = 160 V(rms)
VL = 80 V
VC = 200 V
Resulting voltage lags by φ
tanφ=160120=43
φ=370 (0.64 rods)
Lcr Circuit Graphical Solution Alternating Currents L-5
Inductive circuit
Time Lag Between Current Max and Voltage Max
ωφ=4000.64 = 1.6 ms.
what happens when ω increases.
φ→0 for capacitive circuit.
tanφ=RXC−XL capacitive behaves lika a conductor.
For inductive circuit
Lcr Circuit Graphical Solution Alternating Currents L-5
Average Power Delivered
⟨P⟩=Irms2⋅R
=4×80
=320 watts
Example:- (RC Circuit)
R=3Ω,C=2.5×10−4F
ω=1000rad/s
vmax=5 V

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Lcr Circuit Graphical Solution Alternating Currents L-5
Average Power Delivered
XC=ωC1=1000×2.5×10−41
V=5sinωt=4Ω
I=Imsin(ωt+φ)
Z=R2+XC2=5Ω,Im=1A
vnmax=3V,vCmax=ωCI=4V
Lcr Circuit Graphical Solution Alternating Currents L-5
Impedance
Lcr Circuit Graphical Solution Alternating Currents L-5
Example
XL=ωL=2000Ω
XC=ωC1=5000Ω
Z=R2+(XC−XL)2=1800Ω
Im=1800140=0.078A,Irms=55MA
VRmax=78V
VCmax=Imax.XC=39V
VVmax=156V
tanφ=RXC−XL=φ=−560
Lcr Circuit Graphical Solution Alternating Currents L-5
Thank You
Lcr Circuit Graphical Solution Alternating Currents L-5 LCR Circuit Graphical Solution Alternating Currents $\rightarrow$ $\rightarrow$ LCR Circuit Graphical Solution Alternating Currents $\rightarrow$ AC Circuit With Capacitive Load $\rightarrow$ LC Circuit