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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Energy in Inductor Unit : Joule (J) E = 1 2 LI²

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 2 𝑉 𝑇 𝑉 𝐶2 = 𝐶 1 𝐶 1 + 𝐶 2 𝑉 𝑇

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

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

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

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

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

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

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

Energy in capacitor Unit : Joule (J) E = 1 2 QV

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.