DET2033-Electric Circuit Chapter 2

1. R-L-C Circuits
DET 2033

2. How vector is represented?

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 ωt
is applied.

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 ωt) / R
But Vm / R =Im (Maximum current)
Therefore i = Im sin ωt

5. 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.

6. Phase Relationship And Vector Diagram

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 ωt)(Im sin ωt)
= Vm Im sin² ωt
= ((Vm Im)/2) (1-cos2 ωt)
=[(Vm Im)/2] - [((Vm Im)/2) (cos2 ωt)]

8. 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.

9. A Purely Inductive AC Circuit
•Pure inductor is said to be pure if it contains no
resistance.
Purely inductive ac circuit

10. 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.

11. i = Im sin(ωt – π/2)
v = Vm sin ωt
Thus current flowing through purely inductive
circuit lags by 90°.
Phase Relationship And Vector Diagram

12. 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.

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. 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°.

15. Here i will be maximum when sin(ωt + π/2)=1
Xc = 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

16. 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.

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. •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°.
φ
ω
LV
I
RV
V
Phasor diagram

19. .∙. 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

20. 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)

21. POWER IN R-L SERIES CIRCUIT
P = VI
In 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)

22. 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.

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. •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°.
φ
ω
CV
I
RV
V
Phasor diagram

25. .∙. 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

26. 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)

27. POWER IN R-C SERIES CIRCUIT
P = VI
In 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)

28. 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.

29. 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

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. •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. Case-1:When VL > VC the phasor diagram would
be;
V would be the vector sum of VR and (VL - VC)

33. • 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)

34. 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)

35. •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

36. 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)

37. 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)

38. •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

39. END
of
CHAPTER 2