Reactance, Impedance, RLC Series Circuits and AC Power for SS 3

Class: SS 3
Term: First Term
Week: 2
Age: 17 years
Duration: 60 minutes
Subject: Physics
Previous Lesson: Model Transmission Systems and Communication Technology
Topic: SIMPLE A.C CIRCUITS
Subject Matter: Reactance and impedance: Series circuit containing resistance, inductance and capacitance; Power in a.c circuit

Specific Objectives

By the end of the lesson, pupils/students should be able to:

Cognitive Domain

• Define inductive reactance and capacitive reactance.
• Calculate inductive and capacitive reactance given frequency and inductance/capacitance.
• Define impedance in an AC circuit.
• Draw phasor diagrams for voltage and current in a series RLC circuit.
• Calculate the impedance of a series RLC circuit.
• Calculate the current flowing through a series RLC circuit.
• Explain the concept of power factor in an AC circuit.
• Calculate the average power dissipated in an AC circuit.

Affective Domain

• Appreciate the importance of reactance and impedance in electrical engineering.
• Demonstrate precision in drawing phasor diagrams and performing calculations.

Psychomotor Domain

• Construct simple phasor diagrams for RLC circuits.
• Solve numerical problems involving reactance, impedance, and power in AC circuits.

Social Domain

• Collaborate with peers to solve complex AC circuit problems.

Reference Materials

The following resources were used in planning this lesson:

• 2014 Senior Secondary Education Curriculum (SSEC)
• Relevant State Unified Scheme of Work
• New School Physics for Senior Secondary Schools
• FCT ERC/NAPPS Scheme of work

Instructional Materials

The teacher will teach this lesson with the aid of:

• Whiteboard/Chalkboard
• Markers/Chalk
• Projector (optional, for displaying diagrams and formulas)
• Calculators
• Diagrams of series RLC circuits and phasor diagrams

Rationale for the Lesson

Understanding reactance, impedance, and power in AC circuits is fundamental for students pursuing further studies in electrical engineering and electronics. This lesson provides the theoretical foundation and practical calculation skills necessary to analyse and design various AC systems. It helps students grasp how different components behave in AC circuits, which is crucial for real-world applications like power transmission and electronic device operation.

Prerequisite/Previous Knowledge

Students should have prior knowledge of basic AC circuits, including concepts of resistance, inductance, capacitance, frequency, peak and RMS values of voltage and current, and phase relationships in purely resistive, inductive, and capacitive circuits.

Lesson Content/Board Summary

Reactance, Impedance, RLC Series Circuits and AC Power

Reactance

Reactance is the opposition to the flow of alternating current (AC) due to inductance or capacitance in a circuit. It is measured in Ohms ((Omega)).

1. Inductive Reactance ((X_L)): This is the opposition offered by an inductor to the flow of AC. It increases with both inductance and frequency.

Formula:

(X_L = 2pi f L)

Where:

1. (X_L) = Inductive reactance (Ohms, (Omega))
2. (f) = Frequency of the AC supply (Hertz, Hz)
3. (L) = Inductance of the coil (Henries, H)

2. Capacitive Reactance ((X_C)): This is the opposition offered by a capacitor to the flow of AC. It decreases with both capacitance and frequency.

Formula:

(X_C = frac{1}{2pi f C})

Where:

1. (X_C) = Capacitive reactance (Ohms, (Omega))
2. (f) = Frequency of the AC supply (Hertz, Hz)
3. (C) = Capacitance of the capacitor (Farads, F)

Impedance ((Z))

Impedance is the total opposition to the flow of alternating current in a circuit containing resistance, inductance, and capacitance. It is the AC equivalent of resistance in a DC circuit and is also measured in Ohms ((Omega)). Impedance considers both resistance and reactance.

Series Circuit Containing Resistance, Inductance, and Capacitance (RLC Series Circuit)

In a series RLC circuit, a resistor (R), an inductor (L), and a capacitor (C) are connected in series to an AC voltage source.

Voltage and Current Relationships in Series RLC Circuits

In a series RLC circuit, the current ((I)) is the same through all components, but the voltage drops across each component are out of phase with each other and with the current:

1. Voltage across Resistor ((V_R)): In phase with the current ((I)). (V_R = IR).
2. Voltage across Inductor ((V_L)): Leads the current ((I)) by 90 degrees. (V_L = IX_L).
3. Voltage across Capacitor ((V_C)): Lags the current ((I)) by 90 degrees. (V_C = IX_C).

Phasor Diagram for Series RLC Circuit

The total voltage ((V)) across the series RLC circuit is the phasor sum of the individual voltage drops. Since (V_L) and (V_C) are 180 degrees out of phase, their resultant is (V_L – V_C) (or (V_C – V_L)).

The total voltage can be found using the Pythagorean theorem on the phasor diagram:

(V = sqrt{V_R^2 + (V_L – V_C)^2})

Impedance of a Series RLC Circuit

Dividing the voltage equation by the current (I) (since (V_R = IR), (V_L = IX_L), (V_C = IX_C)), we get the impedance equation:

Formula:

(Z = sqrt{R^2 + (X_L – X_C)^2})

Where:

1. (Z) = Impedance ((Omega))
2. (R) = Resistance ((Omega))
3. (X_L) = Inductive reactance ((Omega))
4. (X_C) = Capacitive reactance ((Omega))

The current in the circuit is then given by Ohm’s Law for AC circuits:

(I = frac{V}{Z})

Example 1: Calculating Impedance and Current in an RLC Circuit

Question: A series RLC circuit has a resistance of (100 , Omega), an inductance of (0.5 , H), and a capacitance of (20 , mu F). It is connected to an AC supply of (200 , V) (RMS) at a frequency of (50 , Hz). Calculate the impedance of the circuit and the RMS current flowing through it.

Solution:

Step 1: Calculate inductive reactance ((X_L)).

(X_L = 2pi f L = 2 times 3.142 times 50 , Hz times 0.5 , H = 157.1 , Omega)

Step 2: Calculate capacitive reactance ((X_C)).

(X_C = frac{1}{2pi f C} = frac{1}{2 times 3.142 times 50 , Hz times 20 times 10^{-6} , F} = frac{1}{0.006284} = 159.2 , Omega)

Step 3: Calculate the impedance ((Z)).

(Z = sqrt{R^2 + (X_L – X_C)^2})

(Z = sqrt{(100)^2 + (157.1 – 159.2)^2})

(Z = sqrt{10000 + (-2.1)^2})

(Z = sqrt{10000 + 4.41})

(Z = sqrt{10004.41} = 100.02 , Omega)

Step 4: Calculate the RMS current ((I)).

(I = frac{V}{Z} = frac{200 , V}{100.02 , Omega} = 1.9996 , A approx 2.00 , A)

Answer: The impedance of the circuit is approximately (100.02 , Omega), and the RMS current is approximately (2.00 , A).

Power in A.C Circuit

In AC circuits, power can be instantaneous or average. Due to phase differences between voltage and current, not all the power delivered by the source is dissipated as heat or work.

Average Power ((P_{avg}))

The average power dissipated in an AC circuit is the actual power used by the circuit, primarily by the resistive component. Inductors and capacitors store and release energy, but ideally do not dissipate average power.

Formula:

(P_{avg} = V_{rms} I_{rms} cos phi)

Where:

1. (P_{avg}) = Average power (Watts, W)
2. (V_{rms}) = RMS voltage (Volts, V)
3. (I_{rms}) = RMS current (Amperes, A)
4. (cos phi) = Power factor

Power Factor ((cos phi))

The power factor is the cosine of the phase angle ((phi)) between the voltage and current in an AC circuit. It represents the fraction of the total apparent power that is actually consumed or dissipated. For a series RLC circuit, the phase angle (phi) can be found from the impedance triangle:

(cos phi = frac{R}{Z})

Where:

1. (R) = Resistance ((Omega))
2. (Z) = Impedance ((Omega))

A power factor of 1 (unity) indicates a purely resistive circuit where voltage and current are in phase, and all power is dissipated. A power factor of 0 indicates a purely reactive circuit (ideal inductor or capacitor) where no average power is dissipated.

Example 2: Calculating Average Power and Power Factor

Question: Using the data from Example 1 ( (R = 100 , Omega), (X_L = 157.1 , Omega), (X_C = 159.2 , Omega), (Z = 100.02 , Omega), (V_{rms} = 200 , V), (I_{rms} = 2.00 , A)), calculate the power factor and the average power dissipated by the circuit.

Solution:

Step 1: Calculate the power factor ((cos phi)).

(cos phi = frac{R}{Z} = frac{100 , Omega}{100.02 , Omega} = 0.9998)

Step 2: Calculate the average power ((P_{avg})).

(P_{avg} = V_{rms} I_{rms} cos phi)

(P_{avg} = 200 , V times 2.00 , A times 0.9998)

(P_{avg} = 399.92 , W approx 400 , W)

Answer: The power factor is approximately (0.9998), and the average power dissipated is approximately (400 , W).

Teaching Methods/Instructional Techniques

Discussion, Explanation, Question and Answer, Problem Solving, Guided Practice, Demonstration

Instructional Procedures

Step 1: Introduction

Time: 5 minutes

Teaching Skill: Recalling/Engaging

Teacher’s Activity: The teacher greets the students and reviews previous knowledge on basic AC circuits, asking questions like “What is an AC circuit?” and “What are the basic components in an AC circuit?”. The teacher then introduces the topic of reactance and impedance as extensions of resistance in AC circuits.

Pupils’ Activity: Pupils respond to questions and listen attentively to the introduction of the new topic.

Learning Point: AC circuit components review

Step 2: Reactance (Inductive and Capacitive)

Time: 10 minutes

Teaching Skill: Explanation/Calculation

Teacher’s Activity: The teacher explains the concept of reactance, differentiating between inductive reactance ((X_L)) and capacitive reactance ((X_C)). The teacher writes down the formulas for (X_L) and (X_C) on the board, explaining each variable and its unit. A simple calculation example for each is provided.

Pupils’ Activity: Pupils listen, ask questions for clarification, and copy the formulas and examples into their notebooks.

Learning Point: Reactance definition and calculation

Step 3: Impedance

Time: 5 minutes

Teaching Skill: Explanation

Teacher’s Activity: The teacher defines impedance as the total opposition to AC flow in a circuit containing resistance and reactance. The teacher explains that impedance combines the effects of resistance, inductive reactance, and capacitive reactance.

Pupils’ Activity: Pupils listen and understand the concept of impedance as the total opposition.

Learning Point: Impedance concept

Step 4: Series RLC Circuit Analysis

Time: 15 minutes

Teaching Skill: Explanation/Demonstration

Teacher’s Activity: The teacher draws a series RLC circuit diagram and explains the phase relationships between voltage and current for each component (resistor, inductor, capacitor). The teacher then demonstrates how to draw a phasor diagram for the total voltage in a series RLC circuit, leading to the formula for total impedance (Z = sqrt{R^2 + (X_L – X_C)^2}).

Pupils’ Activity: Pupils observe the diagrams, ask questions, and attempt to sketch the phasor diagrams in their notes.

Learning Point: RLC circuit impedance formula

Step 5: Calculating Current in RLC Circuits (Activity)

Time: 10 minutes

Teaching Skill: Guided Practice/Problem Solving

Teacher’s Activity: The teacher leads the students through a worked example (Example 1 from the Board Summary) on how to calculate the impedance and current in a simple RLC series circuit. The teacher guides them step-by-step, ensuring they understand the application of the formulas.

Pupils’ Activity: Pupils actively participate in solving the problem, performing calculations with their calculators, and asking questions where they encounter difficulties.

Learning Point: RLC circuit current calculation

Step 6: Power in AC Circuits

Time: 5 minutes

Teaching Skill: Explanation/Calculation

Teacher’s Activity: The teacher explains the concept of average power in AC circuits and introduces the power factor ((cos phi)). The teacher presents the formula (P_{avg} = V_{rms} I_{rms} cos phi) and explains how to calculate the power factor using (cos phi = R/Z). A worked example (Example 2) is discussed briefly.

Pupils’ Activity: Pupils listen, copy the formulas, and follow the explanation of power in AC circuits.

Learning Point: AC power and power factor

Step 7: Evaluation/Review

Time: 5 minutes

Teaching Skill: Questioning/Assessment

Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. Define inductive reactance and state its formula.
2. What is the total opposition to current flow in an RLC circuit called?
3. How is the impedance of a series RLC circuit calculated?
4. What is the significance of the power factor in an AC circuit?

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Reactance, impedance, power understanding

Step 8: Note-Taking

Time: 10 minutes

Teaching Skill: Guided Writing

Teacher’s Activity: The teacher guides pupils/students to copy the essential Board Summary notes, including definitions, formulas, and worked examples for reactance, impedance, RLC circuits, and AC power, into their notebooks.

Pupils’ Activity: Pupils/students copy the notes carefully into their notebooks.

Learning Point: Key concepts recording

Step 9: Conclusion

Time: 5 minutes

Teaching Skill: Summarising

Teacher’s Activity: The teacher briefly summarises the main points of the lesson, reiterating the importance of reactance, impedance, and power factor in understanding AC circuits. The teacher encourages students to practice more problems.

Pupils’ Activity: Pupils listen and ask any final questions.

Learning Point: Lesson consolidation

Continuous Assessment/Further Study

Type: Homework

Instruction: Solve the following problems in your notebook:

1. An inductor of (0.2 , H) is connected to a (240 , V), (60 , Hz) AC supply. Calculate its inductive reactance and the current flowing through it.
2. A capacitor of (50 , mu F) is connected to a (240 , V), (60 , Hz) AC supply. Calculate its capacitive reactance and the current flowing through it.
3. A series RLC circuit has (R = 50 , Omega), (L = 0.3 , H), and (C = 40 , mu F). If it is connected to a (120 , V), (50 , Hz) AC source, calculate:
   1. The inductive reactance.
   2. The capacitive reactance.
   3. The impedance of the circuit.
   4. The RMS current in the circuit.
   5. The power factor.
   6. The average power dissipated.

Lesson Keywords

• Reactance – Opposition to AC flow by inductors or capacitors.
• Inductive Reactance ((X_L)) – Opposition by an inductor, proportional to frequency and inductance.
• Capacitive Reactance ((X_C)) – Opposition by a capacitor, inversely proportional to frequency and capacitance.
• Impedance ((Z)) – Total opposition to AC flow in a circuit with resistance and reactance.
• RLC Circuit – An electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or parallel.
• Power Factor ((cos phi)) – The cosine of the phase angle between voltage and current in an AC circuit, indicating the fraction of real power.
• Average Power ((P_{avg})) – The actual power dissipated in an AC circuit.

Differentiation

For struggling learners: Provide additional simpler problems focusing only on calculating (X_L) or (X_C) before combining them. Use visual aids like colour-coded diagrams for phasor relationships. Offer step-by-step guidance for each calculation.
For advanced learners: Challenge them to consider the conditions for resonance in an RLC circuit or to analyse the circuit when the frequency changes. Encourage them to explore the concept of apparent power and reactive power.

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