R-L-C CircuitsDET 2033

How vector is represented?

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 ωtis applied.

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 ωt) / R

But Vm / R =Im (Maximum current)

Therefore i = Im sin ωt

Since

v = Vm sin ωt and

i = Im sin ωt 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.

Phase Relationship And Vector Diagram

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 ωt)(Im sin ωt)

= Vm Im sin² ωt

= ((Vm Im)/2) (1-cos2 ωt)

=[(Vm Im)/2] - [((Vm Im)/2) (cos2 ωt)]

Constant part=(Vm Im)/2 A fluctuating part=((Vm Im)/2) (cos2 ωt)The average value of the fluctuating part of the

power [((Vm Im)/2) (cos2 ωt)] 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.

A Purely Inductive AC Circuit

•Pure inductor is said to be pure if it contains no resistance.

Purely inductive ac circuit

Due to the self inductance of coil, there will be back emf produced which opposes change in current.

v = Vm sin ωt = -(-L di/dt)di = (Vm/ L) sin ωt dt

Integrating we get;i = -(Vm / L ω) cos ωt

= (Vm / L ω) sin(ωt – π/2)

Here i will be maximum when sin(ωt – π/2) = 1 Thus Im = Vm / L ω = Vm / XL

Where XL = Lω, this is known as inductive reactance.

i = Im sin(ωt – π/2) v = Vm sin ωtThus current flowing through purely inductive

circuit lags by 90°.

Phase Relationship And Vector Diagram

Power In A Purely Inductive Circuit

P = vi = (Vm sin ωt)(Im sin(ωt – π/2))

= (Vm sin ωt)(- Im cos ωt)

= -(2Vm Im sin ωt cos ωt) /2

= -(Vm Im /2) sin2 ωt

The average power consumed by a purely inductive circuit is zero.

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

We know q = CV

q = C Vm sin ωt The current is given by rate of change of charge i = dq/dt

= d/dt (C Vm sin ωt )

= ω C Vm cos ωt

= [Vm / (1/ ω C)] sin(ωt + π/2)Thus in purely capacitive circuit current leads

voltage by 90°.

Here i will be maximum when sin(ωt + π/2)=1Xc = 1 / ω C is called capacitive resistance and it

is measured in ohms. Im = Vm / (1/ ω C)

i = Im sin(ωt + π/2)

= Im cos ωt

Phase Relationship And Vector Diagram

Power In A Purely Capacitive Circuit

P = vi = (Vm sin ωt)(Im cos ωt)

=(2Vm Im sin ωt cos ωt) /2 =(Vm Im /2) sin2 ωt

The average power consumed in a purely capacitive circuit is zero.

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

• 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°.

ω

LV

IRV

V

Phasor diagram

.∙. V = √(VR² + VL²)

= √(IR)² + (IXL)²

= I √(R)² + (XL)² = IZ

Where Z=√(R)² + (XL)²

(Impedance)² = (Resistance)² + (Reactance)²

From figure it can be said that voltage leads current by an angle Ø such that,

tan θ = VL / VR

= IXL / IR

= XL / R

θ = tanˉ¹(XL / R)

Thus in R-L series circuit current lags voltage by an angle θ If V= Vm sin ωt

I = Im sin(ωt- θ)Where θ = tanˉ¹(XL / R)

POWER IN R-L SERIES CIRCUIT

P = VIIn R-L series circuit,V = Vm sin ωt

I = Im sin(ωt- )

P = [Vm sin ωt][Im sin(ωt- θ)]

= [(Vm Im)/2](2 sin ωt sin(ωt- θ))

= [(Vm Im)/2] cos θ - [(Vm Im)/2] cos(2ωt- θ)

Constant part = [(Vm Im)/2] cos θ Variable part = [(Vm Im)/2] cos(2ωt- θ)Average of variable power component over a

complete cycle is zero.Thus average power over complete cycle is

given by,

Pav = ½ Vm Im cos θ = Vrms Irms cos θcos θ is known as Power Factor.

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

• 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°.

ω

CV

IRV

V

Phasor diagram

.∙. V = √(VR² + VC²)

= √(IR)² + (IXC)²

= I √(R)² + (XC)² = IZ

Where Z=√(R)² + (XC)²

(Impedance)² = (Resistance)² + (Capacitance)²

From figure it can be said that voltage lags current by an angle Ø such that,

tan θ = VC / VR = IXC / IR = XC / R

θ = tanˉ¹(XC / R)

Thus in R-L series circuit current leads voltage by an angle θ If

V= Vm sin ωt

I = Im sin(ωt+ θ)Where θ = tanˉ¹(XC / R)

POWER IN R-C SERIES CIRCUIT

P = VIIn R-C series circuit,

V = Vm sin ωt

I = Im sin(ωt + θ)

P = [Vm sin ωt][Im sin(ωt + θ)]

= [(Vm Im)/2](2 sin ωt sin(ωt + θ))

= [(Vm Im)/2] cos θ - [(Vm Im)/2] cos (2ωt + θ)

Constant part = [(Vm Im)/2] cos θ Variable part = [(Vm Im)/2] cos(2ωt + θ)Average of variable power component over a

complete cycle is zero.Thus average power over complete cycle is

given by,

Pav = ½ Vm Im cos θ = Vrms Irms cos θ

**cos θ is known as Power Factor.

R-L-C Series Circuit

•Consider a circuit consisting of RΩ 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

• 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.

•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

Case-1:When VL > VC the phasor diagram would be;

V would be the vector sum of VR and (VL - VC)

• V² = (VR)² + (VL - VC)² = (IR)² + (IXL - IXC)²

= I² [R²+ (XL - XC)²] V = I √ [R²+ (XL - XC)²]

I = V / √ [R²+ (XL - XC)²]•Z = √ [R²+ (XL - XC)²]• tan θ = (VL – VC) / VR

•θ = tanˉ¹((VL – VC) / VR)

When VL > VC current lags voltage by angle θ V = Vm sin ωt I = Im sin(ωt - θ) Im = Vm / Z = V / √ [R²+ (XL - XC)²]

Power consumed

P = VI= [Vm sin ωt][Im sin(ωt- θ)] = [(Vm Im)/2](2 sin ωt sin(ωt- θ)) = [(Vm Im)/2] cos θ - [(Vm Im)/2] cos(2ωt- θ)

•Average of variable power component over a complete cycle is zero.

•Thus average power over complete cycle is given by,

Pav = ½ Vm Im cos θ

= Vrms Irms cos θ

Case-2:When VL < VC

V² = (VR)² + (VC - VL)² = (IR)² + (IXC - IXL)² = I² [R²+ (XC - XL)²] V = I √ [R²+ (XC - XL)²]

I = V / √ [R²+ (XC - XL)²]• Z = √ [R²+ (XC - XL)²]• tan θ = (VC – VL) / VR

• θ = tanˉ¹((VC – VL) / VR)

When VL < VC current leads voltage by angle θ V = Vm sin ωt I = Im sin(ωt + θ) Im = Vm / Z = V / √ [R²+ (XC - XL)²]

Power consumed P = VI = [Vm sin ωt][Im sin(ωt + θ)] = [(Vm Im)/2](2 sin ωt sin(ωt + θ)) = [(Vm Im)/2] cos θ - [(Vm Im)/2] cos(2ωt + θ)

•Average of variable power component over a complete cycle is zero.

•Thus average power over complete cycle is given by,

Pav = ½ Vm Im cos θ

= Vrms Irms cos θ

END of

CHAPTER 2