1. INDUCTORS, CAPACITORS AND ALTERNATING CURRENT
CHAPTER 2
INDUCTORS, CAPACITORS AND ALTERNATING CURRENT

2. LEARNING OUTCOMES
Upon completion of this chapter, students should be able to:
apply basic principles of inductors, capacitors and AC circuits
solve the combination problems

3. CONTENTS Inductors Capacitors AC basic circuits
R-L, R-C, R-L-C circuits

4. INDUCTORS
Also known as choke or coil – has characteristics against the change of the current
2 types of inductors used:
Fixed type Variable type

5. Unit : Henry
Symbol : L
Inductors – spiral coil of wire that creates magnetic field when current passes through it

6. inductors
The magnetic field is directed from left to right and is highly intensified
The changing in magnetic field gives interesting properties that known a inductance
Increase the current will expand the field that generates the emf in the coil

7. INDUCTORS The generated emf is opposes the applied voltage
The effects of inductors in the circuit:
Smooth wave ripples in the DC circuit
Improve the transmission characteristics of waves in the telephone line

8. INDUCTANCE
2 types:
Self inductance (L)
Mutual inductance (M)

9. SELF INDUCTANCE (L)
Occurs when a current flow in the coil causing the changing of flux in the winding
EMF generates
due to :
The changing of magnetic flux
The changing of current

10. L = N 𝑑∅ 𝑑𝑖 SELF INDUCTANCE (L) L = N 𝑑∅ 𝑑𝑡 · 𝑑𝑡 𝑑𝑖 , So :
Where; L = Self inductance
N = Number of turns
𝒅∅ 𝒅𝒕 = Flux change against time
𝒅𝒕 𝒅𝒊 = Current change against time
L = N 𝑑∅ 𝑑𝑖

11. MUTUAL INDUCTANCE (M) The ability of a first coil to produce 1 emf
in the nearest coil through induction or
when the current in the first coil is changing

12. MUTUAL INDUCTANCE (M) How they work??
When current flow in the first loop, flux will be produce in the first coil
The continuous current causes flux flow to the next coil and then generate emf in the second coil
Emf produced in second coil will cut the conductor and produce the voltage in second loop

13. Series inductors 𝑳 𝑻 = 𝑳 𝟏 + 𝑳 𝟐 + 𝑳 𝟑 +…+ 𝑳 𝒏 Total inductance 𝐿 𝑇
= sum of all values of inductance in the circuit
𝑳 𝑻 = 𝑳 𝟏 + 𝑳 𝟐 + 𝑳 𝟑 +…+ 𝑳 𝒏

14. PARALLEL inductors 𝟏 𝑳 𝑻 = 𝟏 𝑳 𝟏 + 𝟏 𝑳 𝟐 + …+ 𝟏 𝑳 𝒏
Total inductance 𝐿 𝑇
𝟏 𝑳 𝑻 = 𝟏 𝑳 𝟏 + 𝟏 𝑳 𝟐 + …+ 𝟏 𝑳 𝒏

15. inductors
Example :
Calculate the total inductance (LT) for the three coil when the value of each inductor is 0.02H, 44mH, 400 μH if the connection is in:
Series
Parallel

16. Inductive reactance, 𝑋 𝐿
Inductive reactance = the opposition to the current flow
Unit : Ω
Value of Inductive reactance depends on the inductance of the circuit due to the current change in the circuit

17. Inductive reactance, 𝑋 𝐿
Where; 𝑋 𝐿 = Inductive reactance (Ω) f = Frequency (Hz) L = Inductor (Henry)
𝑋 𝐿 = 2𝜋fL

18. Inductive reactance, 𝑋 𝐿
Example :
A coil with 0.2H connected with AC 200V, 50Hz. Calculate the inductive reactance in the circuit

19. Energy in Inductor
Unit : Joule (J)
E = LI²

20. A coil of 0. 75H is supplied with 3A current
A coil of 0.75H is supplied with 3A current. Calculate the energy in the circuit.

21. capAcitor Use : store electrical energy (charge) Unit : Farad (F)
Symbol : C

22. capAcitor
Capacitor or condenser built with two-conductors or plates arranged opposite each other
They are separated by insulator (dielectric)

23. capAcitor 2 plates  a plate has negative charge (-ve)
 a plate has positive charge (+ve)

24. capAcitor Effect of capacitor in the circuit:
Increasing the circuit power factor
Reducing the fireworks during the switch is on inside the circuit
Reduce radio interference test in the starter circuit pendaflour light
Strengthen the electric current
Store electrical charges

25. capacitance
Capacitance = Quantity or amount of electric charge needed to make difference between two plates
where, Q = Charge
V = Voltage
C = 𝑄 𝑉

26. capacitance 3 factors affect the value of capacitance Area of plate, A
Distance between 2 plates, d
Permeability, ε

27. capacitance Area of the plate, A C µ A
- Capacitance is directly proportional to the cross sectional area of the plates.
- Capacitance of a capacitor varies with the capacitor plate area.
- Area of large plates to accommodate many electrons and can save a lot of charge
C µ A

28. capacitance Distance between two plates, d
- Capacitance is inversely proportional to the distance between the plates.
Capacitance of a capacitor change when the distance between the plates changes.
Capacitance will increase when the plate is brought closer and less-plates removed.
C µ 𝟏 𝒅

29. capacitance Permeability, ε
- Capacitance is proportional to the permeability of the conductor
C µ ε

30. Series capacitors
𝟏 𝑪 𝑻 = 𝟏 𝑪 𝟏 + 𝟏 𝑪 𝟐 + 𝟏 𝑪 𝟑

31. Series capacitors 𝐶 𝑇 = 𝐶 1 𝐶 2 𝐶 3 𝐶 1 𝐶 2 + 𝐶 1 𝐶 3 + 𝐶 2 𝐶 3
It also can be;
𝐶 𝑇 = 𝐶 1 𝐶 2 𝐶 3 𝐶 1 𝐶 2 + 𝐶 1 𝐶 3 + 𝐶 2 𝐶 3

32. Series capacitors 𝑄 1 = 𝑄 2 = 𝑄 3 = 𝑄 𝑇 Where, 𝑄 𝑇 = 𝐶 𝑇 𝑉 𝑇
The charge for each capacitor;
𝑄 1 = 𝑄 2 = 𝑄 3 = 𝑄 𝑇
Where, 𝑄 𝑇 = 𝐶 𝑇 𝑉 𝑇

33. Series capacitors 𝑉 𝐶1 = 𝑄 𝑇 𝐶 1 𝑉 𝐶3 = 𝑄 𝑇 𝐶 3 Capacitance, C = 𝑄 𝑉
Voltage drop for each capacitor is
𝑉 𝐶2 = 𝑄 𝑇 𝐶 2
𝑉 𝐶3 = 𝑄 𝑇 𝐶 3
𝑉 𝐶1 = 𝑄 𝑇 𝐶 1

34. Series capacitors
𝟏 𝑪 𝑻 = 𝟏 𝑪 𝟏 + 𝟏 𝑪 𝟐 OR
𝑪 𝑻 = 𝑪 𝟏 𝑪 𝟐 𝑪 𝟏 + 𝑪 𝟐

35. Series capacitors 𝑉 𝐶1 = 𝐶 2 𝐶 1 + 𝐶 2 𝑉 𝑇 𝑉 𝐶2 = 𝐶 1 𝐶 1 + 𝐶 2 𝑉 𝑇
Voltage drop for each capacitor
𝑉 𝐶1 = 𝐶 2 𝐶 1 + 𝐶 𝑉 𝑇
𝑉 𝐶2 = 𝐶 1 𝐶 1 + 𝐶 𝑉 𝑇

36. Series capacitors Example
Two capacitors each value is 6μF and 10μF is connected in series with a 200V power supply. Calculate:
Total capacitance
Charge in each capacitor
Voltage across each capacitor

37. parallel capacitors
𝑪 𝑻 = 𝑪 𝟏 + 𝑪 𝟐 + 𝑪 𝟑

38. parallel capacitors 𝑉 𝐶1 = 𝑉 𝑐2 = 𝑉 𝐶3 = 𝑉 𝑇 𝑸 𝑪𝟏 = 𝑪 𝟏 𝑽 𝑻
Voltage drop at each capacitor is equal
𝑉 𝐶1 = 𝑉 𝑐2 = 𝑉 𝐶3 = 𝑉 𝑇
Charge can be calculated as
𝑸 𝑪𝟏 = 𝑪 𝟏 𝑽 𝑻
𝑸 𝑪𝟐 = 𝑪 𝟐 𝑽 𝑻
𝑸 𝑪𝟑 = 𝑪 𝟑 𝑽 𝑻

39. Example Series Parallel
Calculate the total capacitance of the three capacitors where the value of each capacitance is 120μF when it is connected in:
Series
Parallel

40. CAPACITANCE reactance, 𝑋 𝐶
Capacitance reactance = opposition to the flow of current
Unit : Ω
where ω=2𝜋𝑓
𝑋 𝐶 = 1 ω𝐶

41. CAPACITANCE reactance, 𝑋 𝐶SO
where 𝑋 𝐶 = Capacitance reactance (Ω)
f = Frequency (Hz)
C = Capacitor (F)
ω = Angular velocity ( 𝑟𝑎𝑑𝑠 −1 )
𝑋 𝐶 = 1 2𝜋𝑓𝐶

42. CAPACITANCE reactance, 𝑋 𝐶
Example 8μF capacitor connected to the supply of 240V, 50Hz. Calculate the value of capacitance reactance.

43. Energy in capacitor
Unit : Joule (J)
E = QV

44. Energy in capacitor Example
Capacitor with 8pF connected to the 600V power supply. Calculate the charge and energy that can be stored by the capacitor.