Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Resistive, Inductive and Capacitive(RIC) Circuits
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Resistive, Inductive and Capacitive(RIC) Circuits
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Subject: Radio Television and Electrical Work
Class: Senior Secondary 3
Term: 3rd Term
Week: 1
Theme: Electronic Devices And Circuits
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This topic introduces Senior Secondary 3 students to the fundamental characteristics of alternating current (AC) circuits containing resistors, inductors, and capacitors. Understanding these components and their interactions in AC circuits is crucial for aspiring technicians and engineers in Nigeria, as AC power is the primary form of electrical energy used in homes, industries, and electronic devices across the nation. This knowledge forms the bedrock for analyzing and designing various electrical and electronic systems, from simple home appliances to complex power distribution networks and communication systems.
Specific Learning Objectives for Students:
μ
F. Calculate the impedance of the circuit when connected to a 240 V, 50 Hz AC supply.
Solution: Given: R = 50 Ω L = 200 mH = 0.2 H C = 10 μF = 10 × 10−6 F f = 50 Hz First, calculate X_L: X_L = 2πfL X_L = 2 × 3.142 × 50 Hz × 0.2 H X_L = 62.84 Ω Next, calculate X_C: X_C = 1 / (2πfC) X_C = 1 / (2 × 3.142 × 50 Hz × 10 × 10−6 F) X_C = 1 / (0.003142) X_C = 318.27 Ω Now, calculate the impedance (Z): Z = √(R2 + (X_L - X_C)2) Z = √(502 + (62.84 - 318.27)2) Z = √(2500 + (-255.43)2) Z = √(2500 + 65244.4049) Z = √(67744.4049) Z = 260.28 Ω
Commentary: RLC circuits are prevalent in complex filter designs, tuned circuits, and power factor correction units used in Nigerian industrial settings. This section provides a detailed explanation of the core concepts necessary for teaching RIC circuits. 2.1 Alternating Current (AC)
Definition: Alternating Current (AC) is an electric current that periodically reverses direction and continuously changes its magnitude with time, in contrast to direct current (DC) which flows in only one direction.
Waveform: AC typically follows a sinusoidal waveform.
Frequency (f): The number of complete cycles of the waveform per second, measured in Hertz (Hz). The standard mains frequency in Nigeria is 50 Hz. Angular Frequency (ω): Related to frequency by the formula ω = 2πf, measured in radians per second (rad/s).
Phase: The position of a point in time (an instant) on a waveform cycle. When comparing two AC waveforms (e.g., voltage and current), they can be "in phase" (peak at the same time) or "out of phase" (peak at different times). 2.2 Resistive Circuits (Purely Resistive)
Component: Resistor (R). A component designed to oppose current flow and dissipate energy as heat.
Symbol: Standard resistor symbol (zigzag line).
Ohm's Law: In an AC resistive circuit, Ohm's law (V = IR) holds true, where V and I are instantaneous or RMS (Root Mean Square) values.
Phase Relationship: In a purely resistive AC circuit, the voltage across the resistor (V_R) and the current through it (I_R) are always in phase. This means they reach their peak values and zero values at the same time. 2.3 Inductive Circuits (Purely Inductive)
Component: Inductor (L). Typically a coil of wire that stores energy in a magnetic field when current flows through it.
Symbol: Coiled wire symbol.
Inductance (L): A measure of an inductor's ability to store energy in a magnetic field, measured in Henries (H). Inductive Reactance (X_L): Definition: The opposition offered by an inductor to the flow of alternating current. Unlike resistance, reactance does not dissipate energy.
Dependence: X_L is directly proportional to both the frequency of the AC supply (f) and the inductance (L) of the coil.
Formula: X_L = 2πfL = ωL Where: X_L = Inductive Reactance in Ohms (Ω) π (pi) ≈ 3.142 f = Frequency in Hertz (Hz) L = Inductance in Henries (H)
Phase Relationship: In a purely inductive AC circuit, the voltage across the inductor (V_L) leads the current through it (I_L) by 90 degrees (or π/2 radians). This is often remembered as ELI the ICE man: Voltage (E) Leads Current (I) in an Inductor (L). Worked Example 2.3.1 (Inductive Reactance): A choke coil used in a Nigerian workshop has an inductance of 150 m
H. Calculate its inductive reactance when connected to a 240 V, 50 Hz AC mains supply.
Solution: Given: L = 150 mH = 150 × 10−3 H = 0.15 H f = 50 Hz Formula for inductive reactance: X_L = 2πfL Substitute the values: X_L = 2 × 3.142 × 50 Hz × 0.15 H X_L = 47.13 Ω
Commentary: This calculation shows how a common component like a choke coil (inductor) used in a local workshop or for fluorescent lighting exhibits reactance to AC current. 2.4 Capacitive Circuits (Purely Capacitive)
Component: Capacitor (C). A component that stores electrical energy in an electric field between two conductive plates separated by a dielectric material.
Symbol: Two parallel lines.
Capacitance (C): A measure of a capacitor's ability to store charge, measured in Farads (F). Capacitive Reactance (X_C): Definition: The opposition offered by a capacitor to the flow of alternating current. Like X_L, it does not dissipate energy.
Dependence: X_C is inversely proportional to both the frequency of the AC supply (f) and the capacitance (C) of the capacitor.
Formula: X_C = 1 / (2πfC) = 1 / (ωC)
Where: X_C = Capacitive Reactance in Ohms (Ω) π (pi) ≈ 3.142 f = Frequency in Hertz (Hz) C = Capacitance in Farads (F)
Phase Relationship: In a purely capacitive AC circuit, the current through the capacitor (I_C) leads the voltage across it (V_C) by 90 degrees (or π/2 radians). This is remembered as ELI the is inversely proportional to both the frequency of the AC supply (f) and the capacitance (C) of the capacitor.
Formula: X_C = 1 / (2πfC) = 1 / (ωC)
Where: X_C = Capacitive Reactance in Ohms (Ω) π (pi) ≈ 3.142 f = Frequency in Hertz (Hz) C = Capacitance in Farads (F)
Phase Relationship: In a purely capacitive AC circuit, the current through the capacitor (I_C) leads the voltage across it (V_C) by 90 degrees (or π/2 radians).
This is remembered as ELI the ICE man: Current (I) Leads Voltage (C) in a Capacitor (E). Worked Example 2.4.1 (Capacitive Reactance): A capacitor used in a power supply filter circuit for a local radio workshop has a capacitance of 10 μ
F. Calculate its capacitive reactance when connected to a 50 Hz AC supply.
Solution: Given: C = 10 μF = 10 × 10−6 F f = 50 Hz Formula for capacitive reactance: X_C = 1 / (2πfC)
Substitute the values: X_C = 1 / (2 × 3.142 × 50 Hz × 10 × 10−6 F) X_C = 1 / (0.003142) X_C = 318.27 Ω
Commentary: This demonstrates how capacitors, crucial in filtering out unwanted AC ripples in power supplies for electronics commonly repaired in Nigeria, present opposition to AC current. 2.5 Series AC Circuits (RC, RL, LC, RLC) In AC circuits containing combinations of R, L, and C components, their individual oppositions (resistance and reactances) cannot be simply added arithmetically due to their differing phase relationships. Instead, vector addition is used to find the total opposition, known as Impedance (Z).
Impedance (Z): The total opposition to current flow in an AC circuit, combining resistance and reactance. It is measured in Ohms (Ω). 2.5.1 Series RC Circuit (Resistor and Capacitor in Series)
Components: A resistor (R) and a capacitor (C) connected in series across an AC supply.
Vector Diagram: The voltage across the resistor (V_R) is in phase with the current (I). The voltage across the capacitor (V_C) lags the current (I) by 90°.
Impedance Formula: Z = √(R2 + X_C2) Phase Angle (φ): The angle by which the total voltage lags the total current. tan φ = X_C /
R. Worked Example 2.5.1 (Series RC Circuit): An RC circuit in a locally assembled inverter has a resistor of 150 Ω and a capacitor of 20 μF connected in series. Calculate the impedance of the circuit when connected to a 240 V, 50 Hz AC supply.
Solution: Given: R = 150 Ω C = 20 μF = 20 × 10−6 F f = 50 Hz First, calculate the capacitive reactance (X_C): X_C = 1 / (2πfC) X_C = 1 / (2 × 3.142 × 50 Hz × 20 × 10−6 F) X_C = 1 / (0.006284) X_C = 159.13 Ω Now, calculate the impedance (Z): Z = √(R2 + X_C2) Z = √(1502 + 159.132) Z = √(22500 + 25322.4969) Z = √(47822.4969) Z = 218.68 Ω
Commentary: RC circuits are fundamental in filter designs for inverter power supplies, common in Nigeria due to power fluctuations. 2.5.2 Series RL Circuit (Resistor and Inductor in Series)
Components: A resistor (R) and an inductor (L) connected in series across an AC supply.
Vector Diagram: The voltage across the resistor (V_R) is in phase with the current (I). The voltage across the inductor (V_L) leads the current (I) by 90°.
Impedance Formula: Z = √(R2 + X_L2) Phase Angle (φ): The angle by which the total voltage leads the total current. tan φ = X_L /
R. Worked Example 2.5.2 (Series RL Circuit): A motor winding in a grinding machine in a Nigerian market can be modeled as a series RL circuit with a resistance of 30 Ω and an inductance of 100 m
H. Calculate the impedance of the winding when connected to a 240 V, 50 Hz AC supply.
Solution: Given: R = 30 Ω L = 100 mH = 0.1 H f = 50 Hz First, calculate the inductive reactance (X_L): X_L = 2πfL X_L = 2 × 3.142 × 50 Hz × 0.1 H X_L = 31.42 Example 2.5.2 (Series RL Circuit): A motor winding in a grinding machine in a Nigerian market can be modeled as a series RL circuit with a resistance of 30 Ω and an inductance of 100 m
H. Calculate the impedance of the winding when connected to a 240 V, 50 Hz AC supply.
Solution: Given: R = 30 Ω L = 100 mH = 0.1 H f = 50 Hz First, calculate the inductive reactance (X_L): X_L = 2πfL X_L = 2 × 3.142 × 50 Hz × 0.1 H X_L = 31.42 Ω Now, calculate the impedance (Z): Z = √(R2 + X_L2) Z = √(302 + 31.422) Z = √(900 + 987.2164) Z = √(1887.2164) Z = 43.44 Ω
Commentary: RL circuits are good models for electric motor windings and transformers, common in Nigerian industries and households. 2.5.3 Series LC Circuit (Inductor and Capacitor in Series)
Components: An inductor (L) and a capacitor (C) connected in series across an AC supply. Often, ideal LC circuits assume zero resistance.
Vector Diagram: V_L leads I by 90°, and V_C lags I by 90°. This means V_L and V_C are 180° out of phase with each other.
Impedance Formula: Z = |X_L - X_C| (The absolute difference between the reactances).
Resonance: A crucial phenomenon where X_L = X_C. At resonance, the impedance Z becomes minimum (ideally 0 Ω if R=0), and the circuit allows maximum current flow for a given voltage. The resonance frequency (f_r) is given by f_r = 1 / (2π√(LC)). Worked Example 2.5.3 (Series LC Circuit): An antenna tuning circuit in an old Nigerian radio consists of an inductor of 200 mH and a capacitor of 5 μF in series. Calculate the impedance of the circuit at a frequency of 50 Hz.
Solution: Given: L = 200 mH = 0.2 H C = 5 μF = 5 × 10−6 F f = 50 Hz First, calculate X_L: X_L = 2πfL X_L = 2 × 3.142 × 50 Hz × 0.2 H X_L = 62.84 Ω Next, calculate X_C: X_C = 1 / (2πfC) X_C = 1 / (2 × 3.142 × 50 Hz × 5 × 10−6 F) X_C = 1 / (0.001571) X_C = 636.54 Ω Now, calculate the impedance (Z): Z = |X_L - X_C| Z = |62.84 Ω - 636.54 Ω| Z = |-573.7 Ω| Z = 573.7 Ω
Commentary: LC circuits are essential in radio receivers and transmitters for tuning to specific frequencies. Understanding resonance is key here. 2.5.4 Series RLC Circuit (Resistor, Inductor, and Capacitor in Series)
Components: A resistor (R), an inductor (L), and a capacitor (C) connected in series across an AC supply.
Vector Diagram: V_R is in phase with
I. V_L leads I by 90°. V_C lags I by 90°. Thus, V_L and V_C are in opposition. Net Reactance (X_net): The combined effect of inductive and capacitive reactance is X_net = X_L - X_C (or X_C - X_L, depending on which is larger; the square will handle the sign later).
Impedance Formula: Z = √(R2 + (X_L - X_C)2) * Phase Angle (φ): tan φ = (X_L - X_C) / R. If X_L > X_C, the circuit is inductive (voltage leads current). If X_C > X_L, the circuit is capacitive (current leads voltage). If X_L = X_C (resonance), then φ = 0°, and the circuit behaves purely resistively. Worked Example 2.5.4 (Series RLC Circuit): A power supply filter in an industrial control system in Lagos has a series RLC circuit with R = 50 Ω, L = 200 mH, and C = 10 μ
F. Calculate the impedance of the circuit when connected to a 240 V, 50 Hz AC supply.
Solution:** Given: R = 50 Ω L = 200 mH = 0.2 H C = 10 μF = 10 × 10−6 F f = 50 Hz First, calculate X_L: X_L = 2πfL X_L = 2 × 3.142 × 50 Hz × 0.2 H X_L = 62.84 Ω Next, calculate X_C: X_C = 1 / (2πfC) X_C = 1 / (2 × 3.142 × 50 Hz × 10 × 10−6 F) X_C = 1 / (0.003142)
Power Factor Correction in Nigerian Industries and Homes: Context: Industrial loads (e.g., electric motors in factories, grinding mills in markets, large refrigerators in homes) are often inductive, meaning current lags voltage. This results in a "poor power factor," leading to inefficient power usage and higher electricity bills from DISCOs (Distribution Companies) due to penalties for reactive power.
Application: Capacitors are connected in parallel with these inductive loads to improve the power factor. The capacitive reactance cancels out some of the inductive reactance, bringing the current and voltage closer in phase, thus making power usage more efficient. Students can relate this to PHCN/DISCO bills and energy efficiency initiatives in Nigeria.
Tuning Circuits in Radios and Televisions: Context: Radios and TVs are ubiquitous in Nigeria, from locally repaired sets to modern smart TVs. These devices rely heavily on RLC circuits for tuning.
Application: LC circuits are used as "tuned circuits" to select a specific radio or TV station frequency while rejecting others. When the frequency of the incoming signal matches the resonant frequency (where X_L = X_C) of the LC circuit, the circuit offers minimum impedance (in series) or maximum impedance (in parallel), allowing strong signal reception at that specific frequency. This explains how listeners in different parts of Nigeria can tune their radios to local stations like FRCN, Channels Television, or Cool F
M. Power Supply Filtering in Electronic Devices (Inverters, Phone Chargers): Context: Many Nigerian homes and businesses use inverters, UPS, and various phone chargers due to unreliable power supply. The output of rectified AC (which is DC with ripples) needs to be smoothed.
Application: Capacitors and inductors, often in combination (LC filters or RC filters), are critical components in power supply units. Capacitors smooth out voltage ripples, while inductors can oppose sudden changes in current. This ensures a stable, clean DC power output for sensitive electronic devices, protecting them from damage and ensuring proper operation. Students can relate this to the power adapters of their phones or the inverters in their homes.