Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Resonance Circuit
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Resonance Circuit
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Subject: Radio Television and Electrical Work
Class: Senior Secondary 2
Term: 1st Term
Week: 4
Theme: Electronic Communication Systems
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Watch on YouTubeThis topic introduces learners to resonance circuits, a fundamental concept in electronic communication systems. Resonance occurs in RLC (Resistor-Inductor-Capacitor) circuits when the inductive and capacitive reactances cancel each other out, leading to unique circuit behaviour at a specific frequency. Understanding resonance is critical for students to grasp how electronic devices like radios, televisions, and mobile phones are able to tune into specific channels or frequencies, blocking out unwanted signals.
= 1 / (2π√(LC)) (measured in Hertz, Hz) (
Note: For non-ideal parallel circuits with resistance in the inductive branch, the formula can be slightly more complex, but for SS2, the ideal case is usually sufficient).
Characteristics at Parallel Resonance: Impedance (Z): Maximum. At resonance, the currents through the inductor and capacitor are equal in magnitude and 180° out of phase. They circulate within the LC tank circuit, drawing very little current from the main source. This results in very high impedance seen by the source.
Current (I): Minimum. Due to maximum impedance, the total current drawn from the source is at its minimum.
Current in L and C: Large circulating currents flow between L and C, often much larger than the source current. Phase Angle (Φ): 0 degrees. The circuit behaves purely resistive to the source.
Power Factor: Unity (cos 0° = 1).
Worked Example 2: Parallel Resonance Circuit An ideal parallel LC circuit consists of an inductor L = 50 mH and a capacitor C = 20 n
F. Calculate the resonant frequency.
Solution: Given: L = 50 mH = 50 × 10−3 H = 0.05 H, C = 20 nF = 20 × 10−9 F Formula: f_r = 1 / (2π√(LC)) f_r = 1 / (2π√(0.05 H × 20 × 10−9 F)) f_r = 1 / (2π√(1 × 10−9)) f_r = 1 / (2π × √(10 × 10−10)) f_r = 1 / (2π × 3.162 × 10−5) f_r = 1 / (1.987 × 10−4) f_r ≈ 5032.9 Hz or 5.03 kHz 2.
5. Q-factor (Quality Factor) The Q-factor is a dimensionless parameter that describes the "quality" or "sharpness" of a resonant circuit. A higher Q-factor indicates a sharper, more selective resonance, meaning the circuit responds strongly to frequencies very close to the resonant frequency and rejects others effectively.
For a Series RLC Circuit: Q = XL / R = XC / R = (1/R) √(L/C)
For a Parallel RLC Circuit (ideal): Q = R / XL = R / XC = R √(C/L) (This assumes R is in parallel with L and C)
Worked Example 3: Q-factor Calculation Using the series RLC circuit from Example 1 (R = 10 Ω, L = 0.1 H, C = 1 × 10−7 F, f_r = 1591.55 Hz), calculate the Q-factor.
Solution: First, calculate XL at resonance: XL = 2πf_r L = 2π × 1591.55 Hz × 0.1 H ≈ 1000 Ω (or 1 kΩ) Q = XL / R Q = 1000 Ω / 10 Ω Q = 100 Alternatively, using Q = (1/R) √(L/C): Q = (1/10) √(0.1 / 1 × 10−7) Q = 0.1 √(1 × 106) Q = 0.1 1000 Q = 100 2.
6. Bandwidth (BW) The bandwidth of a resonant circuit is the range of frequencies over which the circuit's power output (or current for series, voltage for parallel) is at least half of its maximum value at resonance (i.e., between the half-power points or 3 dB points). It is a measure of the circuit's selectivity. * Formula: BW = f_r / Q (measured in Hertz, Hz) Where f_r is the resonant frequency and Q is the quality factor.
Worked Example 4: Bandwidth Calculation Using the series RLC circuit from Example 1 and 3 (f_r = 1591.55 Hz, Q = 100), calculate the bandwidth.
Solution:** Formula: BW = f_r / Q BW = 1591.55 Hz / 100 BW = 15.9155 Hz --- 2.
1. Introduction to Resonance Resonance in an RLC circuit is a condition where the inductive reactance (XL) equals the capacitive reactance (XC) at a particular frequency. At this resonant frequency (f_r), the circuit behaves purely resistively, meaning the phase angle between the voltage and current is zero. This phenomenon is crucial for selecting specific frequencies in communication systems. 2.
2. Review of Reactance Inductive Reactance (XL): Opposition offered by an inductor to the flow of alternating current. It is directly proportional to the frequency (f) and inductance (L).
Formula: XL = 2πfL (measured in Ohms, Ω)
Capacitive Reactance (XC): Opposition offered by a capacitor to the flow of alternating current. It is inversely proportional to the frequency (f) and capacitance (C).
Formula: XC = 1 / (2πfC) (measured in Ohms, Ω) 2.
3. Series Resonance Circuit A series resonance circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series with an AC voltage source.
Condition for Resonance: Resonance occurs when XL = X
C. Derivation of Resonant Frequency (f_r): At resonance, XL = XC 2πf_r L = 1 / (2πf_r C) (2πf_r)^2 LC = 1 (2πf_r)^2 = 1 / (LC) 2πf_r = 1 / √(LC) f_r = 1 / (2π√(LC)) (measured in Hertz, Hz)
Characteristics at Series Resonance: Impedance (Z): Minimum. Since XL = XC, the reactive components cancel out, and the total impedance becomes equal to the resistance (Z = R).
Current (I): Maximum. Due to minimum impedance, the current flowing through the circuit is at its highest (I = V/R).
Voltage: The voltage across the inductor (VL) and capacitor (VC) can be very high, even higher than the source voltage, but they are 180° out of phase, so they cancel each other out across the reactive elements. The voltage across the resistor (VR) equals the source voltage (V). Phase Angle (Φ): 0 degrees. The circuit behaves purely resistive.
Power Factor: Unity (cos 0° = 1).
Worked Example 1: Series Resonance Circuit A series RLC circuit has a resistance R = 10 Ω, an inductance L = 100 mH, and a capacitance C = 0.1 μF. (a) Calculate the resonant frequency. (b) If the supply voltage is 20 V, calculate the current at resonance.
Solution: (a) Resonant Frequency (f_r): Given: R = 10 Ω, L = 100 mH = 100 × 10−3 H = 0.1 H, C = 0.1 μF = 0.1 × 10−6 F = 1 × 10−7 F Formula: f_r = 1 / (2π√(LC)) f_r = 1 / (2π√(0.1 H × 1 × 10−7 F)) f_r = 1 / (2π√(1 × 10−8)) f_r = 1 / (2π × 1 × 10−4) f_r = 1 / (6.2832 × 10−4) f_r ≈ 1591.55 Hz or 1.59 kHz (b) Current at Resonance (I_r): At resonance, Z = R = 10 Ω Given: V = 20 V Formula: I_r = V / R I_r = 20 V / 10 Ω I_r = 2 A 2.
4. Parallel Resonance Circuit (Anti-Resonance) A parallel resonance circuit typically consists of an inductor (L) and a capacitor (C) connected in parallel, often with a resistor (R) in parallel or in series with one of the components (for ideal analysis, R is sometimes neglected or considered purely resistive for the source). This configuration is also known as a "tank circuit." Condition for Resonance: For an ideal parallel LC circuit, resonance occurs when XL = X
C. Resonant Frequency (f_r): The formula for the resonant frequency for an ideal parallel LC circuit is the same as for a series RLC circuit: f_r = 1 / (2π√(LC)) (measured in Hertz, Hz) (
Note: For non-ideal parallel circuits with resistance in the inductive branch, the formula can be slightly more complex, but for SS2, the ideal case is usually sufficient).
Characteristics at Parallel Resonance: Impedance (Z): Maximum. At resonance, the currents through the inductor and capacitor are equal in magnitude and 180° out of phase. They circulate within the LC tank circuit, drawing very little current from the main source. This results in very high impedance seen by the source. * Current (I): Minimum.
Due Teacher Activities: Introduction (10 minutes): Review basic concepts of AC circuits, resistors, inductors, and capacitors. Ask students to brainstorm how radios tune into different stations (activating prior knowledge). Introduce resonance as the underlying principle for this "tuning." Explanation of Key Concepts (30 minutes): Use whiteboard or projector to clearly define XL, XC, and their dependence on frequency. Draw and explain the circuit diagrams for series and parallel RLC circuits. Derive the resonant frequency formula step-by-step for the series RLC circuit on the board, emphasizing the condition XL = XC. Explain the characteristics (Impedance, Current, Voltage, Phase Angle) at resonance for series and parallel circuits, using analogies (e.g., a swing reaching maximum amplitude with correct timing for resonance). Present Worked Examples 1 and 2, solving them meticulously on the board, encouraging questions.
Advanced Concepts: Q-factor and Bandwidth (20 minutes): Define Q-factor and explain its significance for selectivity using graphical representations of response curves (sharp vs. broad peaks). Introduce the formula for Q-factor for both series and parallel circuits. Define bandwidth and explain its relation to Q-factor and selectivity. Present Worked Examples 3 and 4, solving them on the board. Practical Demonstration/Simulation (if resources permit) (15 minutes): If available, demonstrate a simple RLC circuit using a function generator, oscilloscope, and actual components. Show how varying the frequency causes the current/voltage across components to change, peaking at resonance. Alternatively, use an online circuit simulator (e.g., PhET simulations, EveryCircuit) to demonstrate resonance.
Group Activity/Discussion (10 minutes): Divide students into small groups. Provide each group with a scenario (e.g., "designing a filter for a local radio station") and ask them to discuss how resonance circuits would be applied. Encourage groups to think about the implications of high vs. low Q-factors in their scenarios.
Guided Practice (15 minutes): Distribute guided practice questions. Circulate among students, providing support and clarifications.
Student Activities: Active Listening and Note-Taking: Pay attention to explanations and record key definitions, formulae, and characteristics in their notebooks.
Participation: Answer questions posed by the teacher, contribute to discussions.
Problem Solving: Attempt to solve worked examples as they are presented or immediately after.
Observation: Carefully observe any practical demonstrations or simulations, noting the changes in circuit behaviour.
Group Collaboration: Work with peers during group activities, discussing concepts and applying them to given scenarios.
Guided Practice: Work individually or in pairs on the guided practice questions to solidify understanding. --- Question 1: A series RLC circuit has L = 25 mH, C = 400 nF, and R = 50 Ω. (a) Calculate the resonant frequency of the circuit. (b) Calculate the inductive reactance (XL) and capacitive reactance (XC) at resonance. (c) What is the total impedance of the circuit at resonance?
Solution 1: (a) Resonant Frequency (f_r): Given: L = 25 mH = 25 × 10−3 H; C = 400 nF = 400 × 10−9 F; R = 50 Ω Formula: f_r = 1 / (2π√(LC)) f_r = 1 / (2π√(25 × 10−3 H × 400 × 10−9 F)) f_r = 1 / (2π√(10000 × 10−12)) f_r = 1 / (2π√(10−8)) f_r = 1 / (2π × 10−4) f_r ≈ 1591.55 Hz
Commentary: This calculation shows the direct application of the resonant frequency formula for a series RLC circuit.* (b) Inductive Reactance (XL) and Capacitive Reactance (XC) at resonance: At resonance, XL = XC. XL = 2πf_r L XL = 2π × 1591.55 Hz × 25 × 10−3 H XL ≈ 250 Ω XC = 1 / (2πf_r C) XC = 1 / (2π × 1591.55 Hz × 400 × 10−9 F) XC ≈ 250 Ω
Commentary: Verifies that at resonance, the inductive and capacitive reactances are indeed equal in magnitude, as expected.* (c)
Total Impedance (Z) at resonance: At series resonance, Z =
R. Z = 50 Ω
Commentary: Highlights the key characteristic of series resonance where impedance is at its minimum and purely resistive.* Question 2: A parallel LC tank circuit is designed to receive a local Nigerian radio station broadcasting at 96.5 MHz. If the inductor used has an inductance of 0.2 μH, calculate the capacitance required for tuning.
Solution 2: Given: f_r = 96.5 MHz = 96.5 × 106 Hz; L = 0.2 μH = 0.2 × 10−6 H Formula: f_r = 1 / (2π√(LC))
Rearrange to find C: f_r2 = 1 / ((2π)2LC) C = 1 / ((2π)2Lf_r2) C = 1 / ((2π)2 × 0.2 × 10−6 H × (96.5 × 106 Hz)2) C = 1 / (39.478 × 0.2 × 10−6 × 9312.25 × 1012) C = 1 / (39.478 × 0.2 × 9312.25 × 106) C = 1 / (73507.8 × 106) C ≈ 1.359 × 10−11 F C ≈ 13.59 pF (picofarads)
Commentary: This example directly links resonance to practical radio tuning in Nigeria, demonstrating how component values determine the receivable frequency. It also shows algebraic manipulation of the resonant frequency formula.* Question 3: For a series RLC circuit with R = 20 Ω, L = 200 mH, and C = 2 μF, calculate: (a) The Q-factor of the circuit. (b) The bandwidth of the circuit.
Solution 3: (a)
Q-factor: First, calculate resonant frequency (f_r): L = 200 mH = 0.2 H; C = 2 μF = 2 × 10−6 F f_r = 1 / (2π√(0.2 H × 2 × 10−6 F)) f_r = 1 / (2π√(4 × 10−7)) f_r = 1 / (2π × 6.324 × 10−4) f_r ≈ 251.65 Hz Now, calculate XL at resonance: XL = 2πf_r L = 2π × 251.65 Hz × 0.2 H XL ≈ 316.22 Ω Q = XL / R Q = 316.22 Ω / 20 Ω Q ≈ 15.81
Commentary: This solution requires a multi-step calculation, first finding f_r, then XL, then
Q. It emphasizes the sequential nature of problem-solving in circuit analysis.* (b)
Bandwidth (BW): Formula: BW = f_r / Q BW = 251.65 Hz / 15.81 BW ≈ 15.92 Hz
Commentary: This demonstrates how Q-factor directly influences the bandwidth, indicating the selectivity of the circuit. A lower Q-factor (compared to example 3 Q=100) results in a wider bandwidth.* ---
Radio and Television Tuning (e.g., NTA, FRCN, Wazobia FM): Resonance circuits are at the heart of every radio and television receiver. When a user tunes to a specific station (e.g., NTA 2 Channel 5 Lagos or Wazobia FM 95.1 MHz), they are adjusting a variable capacitor or inductor within an RLC circuit in the receiver. This adjustment changes the resonant frequency of the circuit to match the frequency of the desired broadcast signal. At resonance, the circuit has maximum sensitivity to that specific frequency, allowing the receiver to pick up that station clearly while effectively rejecting other stations broadcasting at different frequencies. This is crucial for clear reception of news, entertainment, and educational programs across Nigeria. Frequency Filtering (e.g., Noise Reduction in Communication): Resonance circuits are used as filters to selectively pass or block certain frequency bands. For instance, in mobile phone base stations and local internet service providers' equipment, resonant circuits act as band-pass filters, allowing only the designated communication frequencies to pass through while attenuating noise and interference from other sources. This ensures the clarity and reliability of phone calls and internet data transmission, vital for Nigeria's growing digital economy. Metal Detectors (e.g., Security, Mineral Exploration): Resonance circuits are employed in metal detectors commonly seen at Nigerian airports, public buildings, and sometimes used in informal mineral prospecting. A metal detector typically contains an LC circuit that oscillates at a specific resonant frequency. When the coil of the detector passes over a metallic object, the metal induces eddy currents which subtly change the inductance of the coil. This change in inductance shifts the resonant frequency of the circuit, causing a detectable change in current or voltage, which is then converted into an audible alarm. This application demonstrates resonance beyond just communication. ---