Alternating Current

This section introduces you to the fundamental concepts and applications of alternating current (AC), building on earlier knowledge of wave motion from Year 2. It explores how AC differs from direct current (DC) in behaviour and characteristics. You will study various AC waveforms, analyse sinusoidal signals using mathematical relationships, and investigate the roles of resistors, capacitors, and inductors in series AC circuits. The section also introduces the concepts of reactance, impedance, power factor, and the power triangle, helping learners to understand real-world applications such as energy transmission and power factor correction in industrial settings. Through hands-on activities, you will measure, interpret, and calculate AC parameters and appreciate their significance in electrical systems.

KEY IDEAS

· Capacitive reactance decreases with increasing frequency, allowing more current to flow through a capacitor in AC circuits.

· Current leads voltage by 90° in a purely capacitive circuit, due to the capacitor’s opposition to voltage change.

· Inductive reactance increases with increasing frequency, reducing the current flow through an inductor in AC circuits.

· In a purely resistive circuit, voltage and current are in phase, meaning they reach their peak values at the same time.

· Resonance occurs in an RLC circuit when inductive and capacitive reactances are equal, resulting in minimum impedance and maximum current.

· The power factor is the cosine of the phase angle between voltage and current, indicating how efficiently power is used in an AC circuit.

(AC) Welcome, Electrical Engineers-In-Training!

Think about how electricity powers your home, charges your phone, and lights up your school. Most of the electricity you use every day has an Alternating Current (AC)… but what exactly is AC, and how is it different from the Direct Current (DC) you might get from a battery?

Here, we will explore the fascinating world of AC, learning about its unique characteristics, different forms, and how we describe it mathematically. Get ready to understand the power that moves our modern world!

Alternating Current (AC)

You have already learned about Direct Current (DC). Remember how a battery pushes electrons in one continuous direction, like water flowing steadily through a pipe?

Alternating Current (AC) is different! Imagine the water in that pipe sloshing/ oscillating back and forth, continuously reversing its direction. That’s AC! The direction of the flow of electric charge (electrons) periodically reverses.

Why AC?

AC is incredibly important because it’s much more efficient to transmit over long distances than DC (from power plants to your home) and can be easily converted to different voltage levels using transformers.

Key Characteristics of AC

1. Periodic Variation

The voltage and current in an AC circuit are not constant. Their strength (magnitude) and direction continuously change, repeating the same pattern over and over again at regular intervals. This regular, repeating pattern is what we call “periodic variation.”

Period (T) is the time it takes for one complete cycle of the AC waveform to occur. It is the shortest time after which the waveform exactly repeats itself.

It’s measured in seconds (s).

2. Amplitude (Peak Value)

This is the maximum value that the voltage or current reaches during a cycle.

It represents the “highest point” or the “strongest” instantaneous value the wave attains. We denote peak voltage as Vₒ and peak current as Iₒ.

3. Instantaneous Value

In an AC circuit, the voltage and current are constantly changing. The value of the voltage or current at any specific moment in time (t) is called its instantaneous value.

a. For a sinusoidal AC, the value typically starts from zero, increases to a maximum (peak), then decreases back to zero. After passing zero, it reverses direction, increases to a maximum (peak) in the opposite direction, and then returns to zero, completing one cycle.

b. Instantaneous voltage is denoted as V(t) in volts (V), and instantaneous current as I(t) in amperes (A).

4. Waveform This is the shape of the graph you get when you plot voltage or current against time. It visually represents how the electrical quantity varies over time. The most common and efficient waveform for power distribution is the sinusoidal (sine) wave.

5. Time-varying Direction (and Analogy)

Unlike Direct Current (DC), where the flow is in one consistent direction, the direction of the current flow in AC constantly reverses.

Analogy for easy understanding Imagine the two pins of your phone charger (ignoring the earth pin if it has three). One is called the live pin, and the other is called the neutral pin.

At one instant, current flow into the live pin and out of the neutral pin. A fraction of a second later, this reverses: current flows out of the live pin and into the neutral pin. This back-and-forth flow happens repeatedly as long as the phone is plugged into power.

6. Frequency (f) Building on the idea of the current constantly reversing direction (as explained in points 1 and 5), Frequency (f) tells us how many complete cycles of the AC waveform happen every second. It’s measured in Hertz (Hz).

a. In Ghana, the frequency of mains electricity is 50 Hz. This means the current completes 50 full back-and-forth cycles every single second.

b. Since each complete cycle involves the current changing its direction twice (from positive to negative flow, and back again from negative to positive flow), a 50 Hz supply means the current actually reverses its direction 100 times every second.

Relationship between Frequency and Period Frequency and period are inversely related: f = 1/T. If the frequency is high, the period is short (many cycles per second). If the frequency is low, the period is long (fewer cycles per second). This is the same inverse relationship you’ve seen in other types of wave propagation.

Types of AC Waveforms

The waveform is the visual representation of how AC voltage or current changes over time. While the sine wave is ‘king’ for power, other shapes exist and are used in different electronic applications.

1. Sinusoidal Waveform (Sine Wave)

a. Shape: Smooth, continuous, and symmetrical oscillation. It looks like a mathematical sine curve.

b. Where it’s used: This is the standard waveform for electricity supplied to your homes and businesses by power plants. It’s efficient for long- distance transmission.

Figure 6.1: Sine wave

2. Square Waveform

a. Shape: Abruptly switches between two constant voltage (or current) levels, staying at each level for a period before switching again. It looks like a series of squares.

b. Where it is used: Very common in digital electronics (e.g., computers, microcontrollers) where signals are either “on” (high voltage) or “off” (low voltage). Also used in some power inverters to create an AC signal from DC.

Figure 6.2: Square wave

3. Triangular Waveform

a. Shape: Changes linearly with time, rising at a constant rate and then falling at a constant rate. Looks like a series of triangles.

b. Where it’s used: Found in some electronic circuits like sweep generators and in music synthesisers for creating certain sounds.

Figure 6.3: Triangular wave

4. Sawtooth Waveform

a. Shape: Rises linearly with time and then drops abruptly back to its starting level. Looks like the teeth of a saw.

b. Where it’s used: Commonly used in old television sets and oscilloscopes for scanning electron beams across the screen (sweep circuits), and also in some audio applications.

Figure 6.4: Sawtooth waveform Mathematical Relationships of AC Voltage and Current Since AC voltage and current vary continuously, we use mathematical equations to describe their instantaneous values at any given time. For sinusoidal AC (the most common type), these are sine functions:

Instantaneous Voltage: V(t) = Vₒ sinωt Instantaneous Current:: I(t) = Iₒ sinωt Where:

V(t) = Instantaneous voltage at time t I(t) = Instantaneous current at time t Vₒ= Peak (maximum) voltage (Amplitude) Iₒ= Peak (maximum) current (Amplitude) ω = 2πf = angular frequency (radians per second) t = time (seconds) These equations describe a smooth wave oscillating between +Vₒand −Vₒ(or +Iₒand −Iₒ) over time.

Root Mean Square (RMS) Values

When you hear about the voltage of a household outlet (like “240 V” in Ghana), that’s not the peak voltage. It’s the Root Mean Square (RMS) value!

The RMS value of AC voltage or current is a measure of its effective value. It’s the equivalent DC voltage or current that would deliver the same amount of power to a load. Because an AC voltage is constantly changing, its average value over a full cycle is zero (it goes positive and negative). RMS gives us a way to describe its “power equivalent.”

Formulae for RMS Values:

RMS Voltage (Vᵣₘₛ) = Vₒ_ √2 RMS current (Iᵣₘₛ) = Iₒ_ √2 RMS values are used for most practical applications

1. Specifying voltage ratings for household outlets.

2. Current ratings for circuit breakers and wires.

3. Calculating power in AC circuits (Power =VrmsIrms).

Multimeter When you use a multimeter to measure AC voltage or current, it typically displays the RMS value.

Activity 6.1 Comparing Alternating Current (AC) and Direct Current (DC) Objective: To help learners visually observe, compare, and understand the key differences between Alternating Current (AC) and Direct Current (DC) in terms of current direction, voltage behaviour, waveform shape, and typical uses.

What you need

1. Paper or notebook

2. Pen or pencil

3. Printed or digital copy of the voltage vs. time diagram showing AC and DC waveforms (like the provided image below)

Figure 6.5a Different types of AC signal What is Alternating Current or AC Current ElectroDuino Fig 6.5b: DC signal What is DC current ? Who Was inventor of dc current ?

What to do

1. Carefully look at the voltage versus time graphs: Different types of Alternating Current (AC) and one for Direct Current (DC).

2. For each graph, note down characteristics such as:

a. Direction of current flow (does it change direction or stay the same?)

b. How voltage changes over time (does it vary or stay constant?)

c. Shape of the waveform (e.g., sinusoidal, flat)

3. Use textbooks or the internet to find common examples where AC and DC are used (e.g., household power, batteries).

4. On your paper, draw a table with these columns

a. Sinusoidal Alternating Current (AC)

b. Square Alternating Current (AC)

c. Triangular Alternating Current (AC)

d. Sawtooth Alternating Current (AC)

e. Direct Current (DC)

5. Fill in the table based on your observations and research with entries such as:

a. Direction of flow

b. Voltage consistency

c. Typical applications

6. Check that your comparison captures the main differences between AC and DC clearly.

Activity 6.2 Deriving the Relationship Between Peak and RMS Values Objective: To understand and derive the mathematical relationship between peak value and RMS (root mean square) value for sinusoidal waveforms, and to apply this knowledge to calculate and compare electrical quantities in AC circuits.

What you need

1. Internet-enabled device or textbooks for research

2. Paper or notebooks

3. Pens or pencils

4. Calculator (optional) What to do

1. Partner up and gather your materials.

2. Use your textbooks, online resources, or videos to find out:

a. What is the definition of peak value?

b. What is the RMS value and why it is important?

c. How are peak (I₀or V₀) and RMS (Irms or Vrms) values related?

3. Look up how to derive the relationship between peak and RMS values for a sinusoidal waveform and follow the steps

4. Summarise the derivation and write down the final formula for RMS in terms of peak value.

5. Practice Conversions Using Simple Values

a. Given I₀=10 A, calculate Irms.

b. Given Irms =7 A, calculate I₀.

Repeat with other values to practice.

6. Take turns explaining the derivation and conversion to your partner to reinforce understanding.

Activity 6.3 Observing AC Waveform Using Oscilloscope and Multimeter Objective: To observe and understand the characteristics of an AC waveform using an oscilloscope and multimeter, identify its peak and RMS values, and compare measured and calculated RMS values for a sinusoidal signal.

What you need

1. Oscilloscope with appropriate probes

2. Digital multimeter set to AC mode

3. AC power source or function generator set to sinusoidal output

4. Connecting wires and test leads

5. Calculator

6. Paper and pen for recording observations What to do

1. Connect the AC power source output to the circuit where both the oscilloscope and multimeter can measure the same signal.

a. Connect one oscilloscope probe across the output to measure voltage or current waveform.

b. Connect the multimeter test leads in parallel to measure the voltage or in series for current, ensuring it is set to AC measurement mode.

2. Observe the Waveform on the Oscilloscope

a. Turn on the oscilloscope and adjust settings (time/div, volts/div) so you can clearly see at least one full cycle of the AC waveform.

b. Notice the waveform’s peak amplitude (maximum positive or negative value) on the screen.

c. Observe features such as the waveform’s time period (time for one complete cycle), frequency (= 1 / time period), and zero-crossings (points where the waveform crosses the time-axis).

3. Note the value displayed by the multimeter. This value represents the RMS (root mean square) voltage or current of the AC source.

4. Calculate Theoretical RMS Value

a. Use the peak value from the oscilloscope to calculate the theoretical RMS value by applying:

RMS value = Peak value_________ √2

b. Show your calculation on paper.

5. Compare and Analyse

a. Compare the RMS value displayed by the multimeter with the value you calculated theoretically.

b. Reflect on any differences and reasons why they might occur (e.g., instrument accuracy, waveform distortion).

6. Record Your Observations

a. Write down measured frequency, period, peak value, RMS reading, and your calculated RMS value.

b. Make notes on the waveform shape and key features you observed

Activity 6.4 Calculating Peak, RMS & Instantaneous Values in AC Signals Study the worked example below carefully before attempting the example questions that follow.

Worked Example 1

A sinusoidal AC voltage source has a frequency of 50 Hz and a peak voltage of 340 V.

a. What is the period (T) of this AC voltage?

b. What is the instantaneous voltage at t=0.005 s?

c. Calculate the RMS voltage of this source.

Solution

a. Period (T) We know that f = 1/T so T = 1/f .

Given f=50 Hz T = 1/50 = 0.02 s The period of the AC voltage is 0.02 s

b. Instantaneous voltage at t=0.005 s First, we need to find the angular frequency (omega).

ω = 2πf = 2π × 50 = 100π V(t) = 340sin(100π)(0.005) V(t) = 340 V

c. RMS voltage RMS Voltage (Vᵣₘₛ) = Vₒ_ √2 RMS Voltage (Vᵣₘₛ) = 340____ √2 = 240 V The RMS voltage of this source is approximately 240.42 V.

Worked Example 2

An AC voltage is described by the equation: V(t)=325sin(100πt)

a. What is the peak voltage?

b. Determine the frequency of the AC voltage.

c. What is the rms voltage?

Solution

a. Peak voltage, 325 V

b. ω = 100π = 2πf, f = 100π____ 2π = 50 Hz.

c. Vᵣₘₛ = ⱽ⁰_ √2 = 325_____ √2 = 229.81 V Practice problems Now, using the worked examples as guide, solve the following problems individually or in groups.

1. An AC voltage has a frequency of 60 Hz and a peak voltage of 325 V.

a. Find the period.

b. Calculate the RMS voltage.

2. An alternating current is given by the expression: I(t) = 8sin(314t). Calculate the frequency of the current.

a. What is the peak current?

b. Determine the rms current.

c. How much time does it take for the current to reach its first peak after t=0?

Welcome back, future electrical innovator! You have already built a solid foundation in understanding the fundamental characteristics of alternating current and voltage. We’ve explored how AC differs from the steady flow of DC, delved into the various types of AC waveforms that power our world, and even mastered the mathematical relationships that define these constantly changing currents and voltages.

Here, we’re taking that knowledge a step further. Having understood the nature of AC, we’re now ready to see how common electronic components such as resistors, inductors, and capacitors interact with these dynamic waves when connected in series. Get ready to uncover how the combination of these simple components affects the operation of AC circuits. Let us now seek further understanding!

The Dynamic Trio: Resistors, Inductors and Capacitors in AC Circuits

Figure 6.6: The blue ring is a toroidal inductor, which stores energy in a magnetic field.

Next to it is a capacitor, used to store electrical energy and smooth out voltage in circuits.

Imagine these three components as characters with distinct personalities in an AC circuit. Let us get to know them!

Component Symbol Unit Description

Resistor R Ω (Ohm) The Current Opposer: Just like in DC circuits as you learnt in year 2, a resistor opposes the flow of current. It converts electrical energy into heat.

In AC, the voltage and current are always in phase, meaning they rise and fall together in magnitude. Think of it like two friends climbing and descending hills and valleys together at the same time.

Capacitor C F (Farad) The Charge Storer:

A capacitor stores electrical energy in an electric field. We learnt this in year 2; that as current flows through the capacitor, it stores charges which can be released later.

· In AC, it opposes changes in voltage. Whether the voltage is increasing or decreasing, it opposes it, seeking to let the voltage remain at a constant value.

· Because of this, the current, which is free to change, always increases or decreases well ahead of the voltage; we say the current leads the voltage by 90°.

· So, like the two friends, when the current is at the peak of the hill, the voltage is halfway up, climbing the hill behind.

· This property of opposing the change in voltage is called capacitive reactance, XC and decreases as frequency increases.

Component Symbol Unit Description

Inductor L H (henry) The Current Change Resister:

An inductor (often a coil of wire) stores electrical energy in a magnetic field.

Its fundamental job is to oppose any change in the current flowing through it.

Here’s how it acts in an AC circuit:

· When the AC current is increasing: As the current tries to get stronger, the inductor fights back. It creates its own induced voltage (sometimes called “back-EMF”) that pushes against the incoming current, trying to keep it from increasing. It’s like a strong water current pushing against a door trying to open it wider, and the inductor pushing back to slow that opening down.

· When the AC current is decreasing: If the current tries to get weaker or stop, the inductor again fights back. Its magnetic field starts to collapse, and this collapse generates an induced voltage that tries to maintain the current flow. It’s like the inductor giving the current a push from behind to keep it going.

· In both situations, this “pushing back” or “pushing forward” by the induced voltage represents energy being stored in, or released from, the coil’s magnetic field. This is why an inductor can even make current flow briefly in a circuit even after the main AC source has been disconnected – it’s releasing its stored energy.

Component Symbol Unit Description

· Since the coil opposes only the current from increasing or decreasing, it means the voltage is free to increase and decrease ahead of the current and is said to lead the current by 90°.

· Still, think of the two friends; while voltage is at the peak of the hill, current is halfway through its uphill journey behind.

· The opposition to the change in the current is called inductive reactance and increases as frequency increases.

So, in summary we can say that

1. For a resistor, the current and voltage are in phase, and the phase difference is 0°.

2. For a capacitor, the current and the voltage are 90° out of phase, with the current leading, the phase difference being 90°.

3. For an inductor, the current and the voltage are 90° out of phase, with the voltage leading, the phase difference being 90°.

Why are these phase differences so important?

They are the key to understanding how these components interact in AC circuits, especially when combined!

Let us look at each of the three components individually when they are in an AC circuit.

Figure 6.7: A purely resistive circuit contains only one or more resistors Purely Resistive Circuit What is it? A circuit with only resistors.

Behaviour in AC

Simple! The voltage and current are perfectly synchronised.

Mathematically Ohm’s Law still rules! V=IR holds for;

Instantaneous voltage; V(t) = I(t) R Peak voltage; Vₒ = Iₒ R RMS voltage; Vᵣₘₛ = Iᵣₘₛ R Also, Instantaneous Current: I(t) = Iₒ sinωt Instantaneous Voltage: V (t) = Vₒ sinωt Notice how V and I have the same sin(ωt) part – they are in phase!

Instantaneous Power: P(t) = I(t) × V(t) = Iₒ sinωt × Vₒ sinωt = I₀ V₀ sin²ωt Average Power: P = Iᵣₘₛ Vᵣₘₛ = ^(Io)_ √2 × ^(Vo)_ √2 = Iₒ Vₒ___ 2 All the power is converted into heat, useful work!

Visualising the Action

Phasor Diagram

It’s a “Vector” (like an arrow): Each AC voltage or current is drawn as an arrow, called a “phasor.”

Length of the arrow: Represents the strength or maximum value (amplitude) of the voltage or current. A longer arrow means a higher voltage or current.

Angle of the arrow: Shows its phase or timing relative to other voltages/currents in the circuit. Remember how current can lead or lag voltage? This angle visually represents that. We usually imagine these arrows rotating counter-clockwise, like the hands of a clock in reverse.

The Reference

Usually, we pick one quantity (like the current in a series circuit, since it’s the same everywhere) and draw its phasor horizontally, pointing to the right. This is our “reference point.” All other phasors are then drawn at an angle relative to this reference.

1. If a voltage/current leads the reference, its arrow will be drawn counter- clockwise (ahead) of the reference arrow. (Think of Capacitors, where current leads voltage.)

2. If a voltage/current lags the reference, its arrow will be drawn clockwise (behind) the reference arrow. (Think of Inductors, where current lags voltage, or voltage leads current, which is the same thing).

As we consider the phasor diagrams of these components individually, we will learn how their phasors are drawn.

If they are in phase (like voltage and current across a resistor), their arrows point in the same direction, as show below

Figure 6.8: Phasor diagram for resistive voltage and current in phase Since the voltage and current are in phase, their phasors align with each other on the reference axis.

Note that their magnitudes are different, though; however, they both achieve maximum magnitude at the same time, and that’s what ‘in phase’ means.

Waveform Diagram

If you plot the voltage and current over time, their sinusoidal waves will overlap perfectly as depicted below.

Figure 6.9: Waveforms of the resistive voltage and current in phase.

Purely Inductive Circuit

What is it? A circuit with only an inductor.

Figure 6.10: A purely inductive circuit contains only one or more inductors/coils Behaviour in AC This is where things get interesting! The inductor resists changes in current. This resistance to change causes the current to lag the voltage by 90°.

Mathematically, Inductive Reactance (X_(L)): This is the inductor’s opposition to AC flow. It’s also measured in ohms (Ω), just like resistance, but it’s not the same!

X_(L) = Vₒ__ Iₒ but Vₒ = ωL Iₒ (deriving this is beyond the scope of this book) X_(L) = ωLIₒ____ Iₒ = ωL Also ω = 2πf X_(L) = 2πfL Where ω is the angular frequency (radians/second) f is the frequency (Hz) L is the inductance (Henry, H) You will learn how to derive some of the equations if you pursue Electrical Engineering at the university.

Voltage-Current Relationship

V_(L)=IX_(L) (Similar to Ohm’s Law, but using reactance)

1. Instantaneous Voltage: Voltage V (t) = Vₒ sinωt

2. Instantaneous Current: Current (t) = Iₒ sin(ωt − π__

2) = Iₒ cosωt , The current is “behind” the voltage by 90° or π__ 2 radians .)

3. Instantaneous power: P(t) = I(t) × V(t) = Iₒ cosωt × Vₒ sinωt = Iₒ Vₒ_____ 2 sin2ωt

4. Average Power: Pav = 0 Surprise! A pure inductor does not consume average power. It stores energy during one half-cycle and returns it to the source during the next.

Visualising the Action

Phasor Diagram

Since in an inductor the voltage leads, the voltage phasor points upwards (or leads), and the current phasor points to the right on the reference axis (or lags by 90°).

Figure 6.11: Phasor diagram of inductive voltage leading current by 90° Waveform Diagram The voltage wave reaches its peak before the current wave. The instantaneous power oscillates symmetrically above and below zero, visually confirming that no net power is consumed.

Figure 6.12: Waveforms of inductive voltage and current 90° out of phase.

Purely Capacitive Circuit

What is it? A circuit with only a capacitor.

Figure 6.13: Purely capacitive circuit Behaviour in AC The capacitor resists changes in voltage. This leads to the current leading the voltage by 90°.

Mathematically, Capacitive Reactance (X_(C)): This is the capacitor’s opposition to AC flow. Also measured in Ohms (Ω).

But I_(O) = ωC Vₒ (deriving this beyond the scope of this book) X_(c) = Vₒ____ ωCVₒ = 1___ ωC Also ω = 2πf X_(c) = 1/2πfC Where ω is the angular frequency (radians/second) f is the frequency (Hz) C is the capacitance (Farad, F) Key Takeaway The faster the AC (higher frequency), the less the capacitor opposes current flow!

1. Voltage-Current Relationship: V_(C) = IX_(C)

2. Instantaneous Voltage: Voltage V (t) = Vₒ sinωt

3. Instantaneous Current: Current I(t) = Iₒ sin(ωt − π__

2) = Iₒ cosωt

4. The current is “ahead” of the voltage!

5. Average Power: Pₐᵥ = 0 Like inductors, pure capacitors also do not consume average power. They store and release energy.

Visualising the Action

Phasor Diagram

Since the current leads the voltage, the current phasor points upwards (or leads), and the voltage phasor points to the right on the reference axis (or lags by 90°).

Figure 6.14: Capacitive current leads capacitive voltage by 90° Waveform Diagram The current wave reaches its peak before the voltage wave. The instantaneous power also oscillates symmetrically about zero.

Figure 6.15: Waveforms of capacitive current leading voltage Reactance verses Impedance: The Total Opposition Now that we understand how the individual components oppose AC, let’s look at the bigger picture when they are combined.

Reactance (X) It is the opposition to AC current due to energy storage elements (inductors and capacitors). It’s all about how these components react to changes in current or voltage by storing energy. Unit: Ohms (Ω) Net Reactance (X) When you have both inductors and capacitors, their reactance partially cancels each other out because their actions are opposite. Xₙₑₜ = X_(L)−X_(C)

1. If X_(L)>X_(C): The circuit is primarily inductive. The current will lag the voltage.

2. If X_(C) >X_(L) : The circuit is primarily capacitive. The current will lead the voltage.

3. If X_(L) =X_(C) : This is a special condition called resonance, where the net reactance is zero.

We will explore this exciting concept later!

Impedance (Z) It is the total opposition to the flow of alternating current in an AC circuit. It combines the effects of both resistance (which dissipates energy) and reactance (which stores and releases energy). Unit: Ohms (Ω) Think of it as the AC version of total/equivalent resistance!

Impedance in Series R-L-C Circuits

When resistors, inductors, and capacitors are connected in series, the total opposition (impedance) is found by considering their phase relationships. We cannot just add them up directly like in DC circuits! Their sum is found by using vector addition as follows;

Z = √_________________ (R² + (+( X_(L)−X_(C)))²) Z = √R²+ X²This can be represented in other ways such as the following Z = R + jX (Rectangular form, where j is the imaginary unit) Z = |Z|∠θ (Polar form) In complex notation, R is said to be the real component, while X is called the imaginary component, signified by the presence of j. (In some university courses you will learn in engineering maths about complex numbers) Where Z is the impedance (Ω) R is the resistance (Ω) X is the net reactance (Ω) Graphical Representation (Impedance Triangle) Just like voltages, we can represent impedance using a right-angled triangle.

The horizontal side is Resistance (R).

The vertical side is Net Reactance (X).

The hypotenuse is Impedance (Z).

Figure 6.16: Impedance triangle Phase Angle (θ) The phase angle of impedance tells us the overall phase difference between the total voltage and total current in the circuit.

A positive θ means the circuit is inductive (current lags voltage).

A negative θ means the circuit is capacitive (current leads voltage).

A θ of 0° means the circuit is purely resistive or at resonance (current and voltage are in phase).

Series Connected AC Circuits: R-C, R-L and R-L-C

When components are in series, the current (I) flowing through each element is the same, but the voltages across them will have different magnitudes and phases.

Let’s consider the individual series combinations of resistor, capacitor and inductor in pairs.

Series R-C Circuits (Resistor and Capacitor)

Figure 6.16: A circuit containing a resistor (R) and a capacitor (C) in series.

1. Current (I): Same through R and C.

2. Voltage across Resistor (VR): In phase with current.

3. Voltage across Capacitor (VC): Lags current by 90°.

4. Total Voltage (V): The phasor sums of VR and VC given as V²= V_(R) ²+ V_(C) ²V²= (IR)²+ (I X_(c))²V2 = I²R²+ I²X_(C) ²V²= I²(R²+ X_(C) ²) V = I √(R²+ X_(C) ²) Impedance (Z) is given as Z = V/I = √________ (R2 + X_(C) ²) Z = √(R²+ X_(C) ²) Phase Angle (θ) is given as tanϕ = V_(C)_ V_(R) = I X_(C)_ IR = X_(C)_ R ϕ = tan⁻¹(X_(C)_ R ) (This will be negative because it’s capacitive, current leads voltage)

Note

· ϕ is the phase angle between the overall current and overall voltage of the circuit.

· θ is the phase angle between the current and voltage in each individual component.

Series R-L Circuit (Resistor and Inductor)

1. Current (I): Same through R and L.

2. Voltage across Resistor (VR): In phase with current.

3. Voltage across Inductor (VL): Leads current by 90°.

4. Total Voltage (V): The phasor sums of VR and VL ; V = √( V_(R) ²+ V_(L) ²)

5. Impedance (Z): Z = √(R²+ X_(L) ²)

6. Phase Angle (θ): θ = tan⁻¹(X_(L)___ R ) (Positive because it’s inductive, current lags voltage) Series R-L-C Circuit (Resistor, Inductor, and Capacitor)

Figure 6.17: A Resistor (R), inductor (L) and capacitor (C) in series.

This is the most general and exciting case!

1. Current (I): Same through all elements.

2. Voltages

a. V_(R): In phase with current.

b. V_(L): Leads current by 90°.

c. V_(C): Lags current by 90°.

3. Net Reactive Voltage: Since VL and VC are both 90° ahead and behind the current, respectively, they are 180° apart/out of phase. So they directly oppose each other. The net reactive voltage is VL−VC .

4. Total Voltage (V): The phasor sum of VR and the net reactive voltage. V = √___________ V_(R) ²+ ( V_(L) ²− V_(C) ²) Impedance (Z): As derived earlier, Z = √______________ (R² + (X_(L)−X_(C))²)

5. Phase Angle (θ): θ = tan⁻¹(( X_(L)−X_(C))________ R )

Figure 6.18 – A: Phasors of inductive (V_(L)) and capacitive (V_(C)) voltages 180° apart and 90° out of phase with the resistive voltage (V_(R)). B: Effective reactive voltage leading V_(R)by 90°– B.

The Magic of Resonance

A truly special condition occurs in an R-L-C series circuit when: Inductive Reactance (X_(L)) = Capacitive Reactance (X_(C)) At resonance

1. X_(L)−X_(C) = 0, so the net reactance is zero.

2. The impedance Z = √(R² + (0)²) = R. This means the impedance is at its minimum value, equal to the resistance!

3. The current in the circuit will be at its maximum for a given voltage (since I=V/Z, and Z is minimum).

4. The phase angle θ=0°, meaning the circuit behaves like a purely resistive circuit, with voltage and current in phase.

Resonance Frequency (f₀)

This is the specific frequency at which resonance occurs.

Inductive reactance = Capacitive reactance i.e. X_(C) = X_(L) and f = fₒ 1/2π fₒ C = 2π f₀ L 4π2f₀ ²LC = 1 f₀ ²= 1/4π²LC f₀ = 1/2π √__ LC Resonance is a fundamental concept used in radio tuning, filters, and many other electronic applications! However, you do not need to know the details of this.

The Power Triangle And Power Factor: Efficiency In Ac Circuits In AC circuits, power is not as simple as P=VI. Due to the phase differences, we talk about three types of power.

1. Active Power (P)/Real Power/True Power

a. What it is: The actual power that is consumed by the circuit and converted into useful work (like heat in a resistor, light from a bulb, or mechanical energy in a motor).

b. Where it’s dissipated: Only by resistive components.

c. Unit: Watts (W)

d. Formula: P=V_(RMS) I_(RMS) cosθ

2. Reactive Power (Q)

a. What it is: The power that is exchanged between the source and the reactive components (inductors and capacitors). It’s stored in their magnetic and electric fields during one part of the cycle and then returned to the source in the next. It does not do useful work.

b. Why it’s needed: Essential for the operation of inductive and capacitive devices.

c. Unit: Volt-Ampere Reactive (VAR)

d. Formulae

i. Q_(L) = V_(L)I = I2X_(L) (Inductive reactive power, conventionally positive)

ii. Q_(C)= V_(C)I = I²X_(C) (Capacitive reactive power, conventionally negative)

iii. Q =V_(RMS) I_(RMS) sinθ (Net reactive power)

3. Apparent Power (S)

a. What it is: The total power supplied by the source, comprising both active and reactive power. It’s the product of the total voltage and total current, without considering phase.

b. Unit: Volt-Amperes (VA)

c. Formula: S=V_(RMS)I_(RMS) The Power Triangle This is a powerful visual tool! It’s a right-angled triangle where:

1. The horizontal side represents Active Power (P).

2. The vertical side represents Reactive Power (Q).

3. The hypotenuse represents Apparent Power (S).

4. The angle between P and S is the phase angle (θ) we discussed earlier!

Figure 6.19: The power triangle follows Pythagoras’ theorem to conveniently represent power in an AC circuit.

From the Pythagorean theorem: S²= P²+ Q²Power Factor (PF)

a. What it is: The Power Factor tells us how efficiently electrical power is being converted into useful work. It’s the ratio of active power to apparent power.

b. Formula: PF = P/s = cosθ .

This is why the phase angle θ is so important!

c. Range: The power factor is a dimensionless number between 0 and 1.

Interpreting the Power Factor

1. PF = 1 (Unity Power Factor) The ideal scenario! This happens in purely resistive circuits or at resonance in RLC circuits. Voltage and current are perfectly in phase, so all the apparent power is active power. This is the most efficient use of power.

2. PF < 1 (Lagging Power Factor) Occurs predominantly in inductive circuits (like motors, transformers, fluorescent lights). The current lags the voltage. A lower lagging PF means a larger portion of the apparent power is reactive, not useful.

3. PF < 1 (Leading Power Factor) Occurs in predominantly capacitive circuits. The current leads the voltage.

Less common for typical loads, but capacitors are often used to correct lagging power factors.

4. PF = 0 Occurs in purely inductive or purely capacitive circuits. No active power is consumed; all power is reactive.

Why is a High-Power Factor Important?

A low power factor (meaning a large phase difference between voltage and current) is bad news for electricity generation and distribution!

1. Increased Current: To deliver the same amount of useful (active) power, a low PF circuit draws more total current (I = S/V).

2. Higher Losses: More current means more energy lost as heat (I2R losses) in transmission lines and equipment.

3. Larger Equipment: Utilities need to install larger, more expensive wires, transformers, and generators to handle the extra current.

4. Higher Bills: Industrial consumers are often charged penalties for maintaining a low power factor because it costs the utility more to deliver the power.

Power Factor Correction

To combat low power factors (especially from inductive loads like motors), capacitors are often connected in parallel with the load. These capacitors provide leading reactive power that cancels out the lagging reactive power, bringing the power factor closer to unity and improving overall system efficiency!

You have now taken a significant step into understanding the complex yet incredibly important world of AC circuits! These concepts are the bedrock for advanced electrical engineering and play a vital role in the technology that powers our modern world. Keep exploring!

Activity 6.5 Voltage and Current vs. Time Graphs for Pure Circuit Elements Objective: To help learners sketch and understand how voltage and current waveforms behave in pure resistor, capacitor, and inductor circuits, with a focus on identifying and explaining the phase relationships and their physical meanings.

What you need

1. Paper or notebook

2. Pen or pencil

3. Ruler (optional, for neater graphs) What to do

1. Make three sets of axes labelled “Voltage/Current” (vertical) and “Time” (horizontal), one set for each: pure resistor, pure capacitor, and pure inductor.

2. Sketch the Waveforms and Mark Phase Relationships

a. Resistor (R): Sketch two sine waves—one for voltage and one for current. Show them perfectly overlapping (in phase): their peaks, troughs, and zero-crossings are at the same times.

Annotate: “Voltage and current are in phase. Peaks and zero- crossings happen together.”

b. Capacitor (C): Draw voltage as a sine wave. Sketch the current sine wave so that it reaches its peak a quarter cycle (90°) before the voltage does (current LEADS voltage).

i. Mark and label the time difference: “Current leads voltage by 90° (¼ cycle).”

ii. Annotate: “At each zero-voltage crossing, current is at a peak.

Physically, the fastest charging/discharging happens when voltage passes zero (fully changing).”

c. Inductor (L): Draw voltage as a sine wave. Sketch the current sine wave so that it reaches its peak a quarter cycle (90°) after the voltage does (current LAGS voltage).

i. Mark and label the time difference: “Current lags voltage by 90° (¼ cycle).”

ii. Annotate: “When voltage crosses zero, current is at a peak.

The inductor resists changes, so peak current comes after peak voltage.”

3. On all your graphs, use arrows or markers to indicate where peaks/troughs and zero-crossings occur for both the voltage and the current waveforms.

4. Next to each graph or underneath, write a short explanation of what the phase relationship means for energy flow in that element:

a. In resistors, energy is dissipated as heat instantly.

b. In capacitors, energy is stored and released in the electric field, with current showing the fastest flow as voltage changes most rapidly.

c. In inductors, energy is stored and released in the magnetic field, with current building up or decreasing with a delay due to inductance.

Activity 6.6 - Deriving Frequency Dependence of Current in Capacitor and Inductor Circuits Objective: To derive and understand how the current in capacitor and inductor circuits depends on the frequency of an AC supply, by calculating and analysing the frequency dependence of capacitive and inductive reactance using given data and standard formulas.

What you need

1. Paper or notebook

2. Pen or pencil

3. Calculator

4. Provided exemplar data set (see below)

Example of dataset:

At a known AC frequency f=50 Hz:

Capacitor circuit: Current I_(C)=0.5 A when voltage V=10 V, capacitance C=100 μF Inductor circuit: Current I_(L)=0.2 A when voltage V=10 V, inductance L=0.05 H.

What to do

1. Write down the values of voltage, current, frequency, capacitance, and inductance provided for the two circuits.

2. Calculate Capacitive Reactance Xc. Use the formula: X_(c) = V__ I_(c) to calculate XC from the given voltage and capacitor current.

3. Recall Capacitive Reactance Formula. Research or remember that X_(c) = 1/2πfC Notice the inverse relationship between X_(C) and frequency f.

4. Using the calculated XC, verify it matches the formula and conclude that the capacitive reactance decreases as frequency increases, causing current to increase.

5. Calculate Inductive Reactance XL using similar steps:

X_(L) = V __ I_(L) Compute X_(L)using the voltage and inductor current data.

6. Recall Inductive Reactance Formula: X_(L) = 2πfL

Note the direct relationship between X_(L)and frequency f.

7. Compare the calculated XL with the formula and understand that inductive reactance increases with frequency, causing current to decrease.

8. Summarise Your Result. Write down the two standard expressions clearly:

I_(c) = V __ X_(c) = V × 2πfC I_(L) = V___ X_(L) = V/2πfL Emphasising the inverse frequency dependence for capacitors and direct frequency dependence for inductors.

Activity 6.7 Constructing the Power Triangle and Calculating Power Factor Linked to Phasor Diagrams Objective: To help learners construct and interpret the power triangle, calculate power factor, and relate it to phasor diagrams, in order to understand how true power, reactive power, and apparent power are connected and how the phase difference between current and voltage affects power usage in AC circuits.

What you need

1. Paper or notebook

2. Pen or pencil

3. Calculator

4. Provided values for apparent power (S), true power (P), and reactive power (Q)

5. Reference materials or images of power triangles and phasor diagrams (optional) Provided Values Apparent Power, S=5000 VA True Power, P=4000 W Reactive Power, Q=3000 VAR What to do

1. Understand the Components. Recall that

a. S (apparent power) is the hypotenuse of the power triangle

b. P (true or real power) is the adjacent side (horizontal)

c. Q (reactive power) is the opposite side (vertical)

2. Draw the power triangle

a. On graph paper, draw a right-angled triangle.

b. Label the hypotenuse as S=5000 VA.

c. Label the adjacent side as P=4000 W.

d. Label the opposite side as Q=3000 VAR.

3. Calculate the angle θ between P and S using:

cosθ = P/S

4. Calculate the Power Factor (PF) using:

PF = P/s = cosθ

5. Link to the Phasor Diagram

a. Understand that in the phasor diagram, the vector for voltage is usually the reference (horizontal axis).

b. The current vector lags or leads voltage by angle θ depending on whether the load is inductive or capacitive.

c. The reactive power Q is associated with this phase difference.

d. Sketch simple phasor vectors for voltage and current showing this angle θ.

6. Resolve Components of the Power Triangle. Using Pythagoras theorem, verify S²= P²+ Q².

Calculate the left-hand side and right-hand side values and compare.

7. Summarise your findings

a. Explain how the power triangle represents the relationship among P,Q,S, and power factor.

b. Explain how power factor relates to the phase difference shown in phasor diagrams.

Activity 6.8 Building and Measuring R-C, R-L and R-L-C Series Circuits with AC Power Objective: To build and measure R-C, R-L, and R-L-C series circuits using AC power, and to observe and analyse how resistors, capacitors, and inductors affect current, voltage, and phase relationships, in order to understand the behaviour of AC circuits with different components.

What you need

1. Resistors (various values)

2. Capacitors (known capacitance)

3. Inductors (known inductance)

4. AC power source or function generator with adjustable frequency

5. Digital multimeter (set for AC voltage and current)

6. Oscilloscope with probes

7. Connecting wires and breadboard or circuit board

8. Paper and pen for recording observations What to do

1. Work together in small groups to complete the activity.

2. Build the circuits

a. Construct an R-C series circuit using a resistor and capacitor in series connected to the AC power source.

b. Build an R-L series circuit with resistor and inductor in series.

c. Build an R-L-C series circuit by combining resistor, inductor, and capacitor in series.

3. Set Up measurement instruments

a. Connect the multimeter to measure current in the circuit (series connection).

b. Connect multimeter or oscilloscope probes across each component to measure voltage.

c. Use the oscilloscope to observe voltage and current waveforms.

4. Take measurements

a. Record the voltage across each component and the total current using the multimeter.

b. On the oscilloscope, observe and note the waveforms of voltage and current. Pay attention to their phase relationships.

5. Analyse phase relationships

a. Observe if current and voltage are in phase or out of phase for each component.

b. Explore how adding capacitors or inductors shifts the phase between current and voltage.

6. Discuss effects of components

a. In your group, discuss how the resistor, capacitor, and inductor each affect current magnitude and phase in the circuit.

b. Consider how the combination of all three components influences the overall circuit behaviour.

7. Write down your measured values, waveform observations, and group conclusions about the components’ impact on current and phase.

Activity 6.9 - Exploring R-C & R-L Series Circuits Using PhET Virtual Lab Objective: To explore and compare how capacitors and inductors behave in AC circuits by building R-C and R-L series circuits using the PhET virtual lab, and to observe the effects on voltage and current, leading to a better understanding of the roles of these components in alternating current systems.

What you need

1. Computer or tablet with internet access

2. Access to the PhET Circuit Construction Kit: AC virtual lab (Link: https://phet.colorado.edu/en/simulation/circuit-construction-kit- ac-virtual-lab)

3. Paper or notebook

4. Pen or pencil What to do

1. Open the PhET Circuit Construction Kit: AC

Go to the link and launch the virtual lab in your browser.

2. Build an R-C Series Circuit

a. Select a resistor and capacitor from the component options.

b. Connect them in series to the AC voltage source by dragging wires accordingly.

c. Make sure the circuit is complete with the power source.

3. Measure Voltages Across Components

a. Use the built-in voltmeter tool in the simulation to measure the voltage across the resistor.

b. Repeat for the capacitor.

c. Record these voltage values in your notebook.

4. Use the ammeter tool to measure and record the current flowing through the entire circuit.

5. Build an R-L Series Circuit

a. Replace the capacitor with an inductor, keeping the resistor and AC source connected in series.

b. Repeat the voltage measurement across the resistor and inductor, recording your results.

c. Measure and record the total current again.

6. Observe and Compare

a. Notice how voltage is divided differently in the R-C and R-L circuits.

b. Observe any changes in current magnitude.

c. Reflect on how capacitors and inductors influence the circuit compared to resistors.

7. Summarise how the presence of capacitors and inductors affects voltage and current in AC circuits based on your measurements.

Activity 6.10 Calculating Current, Power and Reactance in AC Circuits Study the worked example below carefully before attempting the example questions that follow.

Worked Example 1

An AC voltage of 120 V (rms) at 60 Hz is applied across a purely resistive load of 15 Ω. Find:

a. The current in the circuit.

b. The power consumed.

Solution

a. Current, I = V/R = 120/15 = 8 A

b. Power consumed, P = V × I = 120 × 8 = 960 W

Worked Example 2

An AC source of 100 V (rms) at 50 Hz is connected to a capacitor of capacitance 100 μF.

Find the current in the circuit.

Solution

Capacitive reactance: X_(C) = 1/2πfC = 1/2π × 50 × 100 × 10⁻⁶≈ 31.8 Ω Current: I = V__ X_(C) = 100/31.8 ≈ 3.14 A

Worked Example 3

An inductor of 0.2H is connected to a 240V (rms), 60Hz AC source.

Find the current in the circuit.

Solution

Inductive reactance: X_(L) = 2πfL = 2π × 60 × 0.2 = 75.4 Ω Current: I = V__ X_(L) = 240/75.4 ≈ 3.18 A

Worked Example 4

In a series R-L-C circuit, R = 10Ω, L = 0.1 H, and C = 100μF, connected to a 120V (rms), 50Hz source. Find the current in the circuit.

Solution

Step 1: find the net reactance X_(L) = 2πfL = 2π × 50 × 0.1 = 31.4 Ω X_(C) = 1/2πfC = 1/2π × 50 × 100 × 10⁻⁶= 31.8 Ω Net reactance, X = X_(L) − X_(C) = 31.4 − 31.8 = − 0.4 Ω Impedance, Z = √R²+ X²Z = √__________ 10²+ (− 0.4)²≈ 10.01 Ω

Step 2: use ohm’s law to find the current Current, I = V/Z = ¹²⁰__(10.01) ≈ 11.99 A

Worked Example 5

A device draws 10 A at 230 V with a power factor of 0.8 lagging.

Calculate:

a. Real power

b. Reactive power

c. Apparent power

Solution

a. S = V × I = 230 × 10 = 2300 VA

b. P = S × cos(θ ) = 2300 × 0.8 = 1840 W

c. Q = √(S² − P²) = √(2300² − 1840²) ≈ 1380 VAR

Worked Example 6

A load has a real power consumption of 1500 W and draws 2000 VA of apparent power.

Find the power factor and state whether it is lagging or leading if the load is inductive.

Solution

Power factor = P_ S = 1500_ 2000 = 0.75 Since the load is inductive, the power factor is lagging Practice Problems

1. An electric heater rated at 230 V (rms) draws a current of 10 A from an AC supply.

a. Calculate the resistance of the heater. [

b. Find the power consumed by the heater.

2. A 60 Hz AC voltage of 120 V (rms) is applied across a 50 μF capacitor.

a. Calculate the capacitive reactance.

b. Find the current in the circuit.

3. A coil of inductance 0.1 H is connected to a 100 V (rms), 50 Hz AC supply.

a. Calculate the inductive reactance.

b. Determine the current through the coil.

4. A series RLC circuit contains a resistor (R = 40 Ω), an inductor (L = 0.2 H), and a capacitor (C = 60 μF). It is powered by a 120 V (rms), 50 Hz AC source.

a. Calculate the inductive and capacitive reactance.

b. Determine the net reactance and the total impedance.

c. Calculate the current in the circuit.

5. A device draws a current of 15 A at 220 V (rms) with a power factor of 0.75 lagging.

a. Calculate the real power, apparent power, and reactive power.

b. Determine the required reactive power compensation to raise the power factor to 0.95.

c. Suggest a component that could be added to achieve this.

Activity 6.11 Case Study Improving Poor Power Factor

Objective: To understand the impact of poor power factor in industrial systems and explore how installing capacitor banks can correct it, improve energy efficiency, and reduce electricity costs.

What you need

1. Case study description (provided below)

2. Paper or notebook

3. Pen or pencil

4. Calculator (optional)

5. Access to research materials (textbooks, internet) for background on power factor and capacitors What to do

1. You are presented with an industrial plant experiencing poor power factor (e.g., around 0.68). This has resulted in high reactive power demand, increased current flow, energy loss, and extra charges on the electricity bill.

2. Research why poor power factor causes

a. Inefficient energy use

b. Higher demand from the power supplier

c. Increased energy losses and higher electricity costs

3. Research Capacitor Use in Power Factor Correction. Learn how capacitors supply reactive power locally, reducing the total current drawn from the utility and improving the power factor.

4. Based on your research, suggest installing appropriately sized capacitor banks at the plant. Explain how these capacitors will:

a. Supply reactive power

b. Reduce the load on the utility

c. Improve the plant’s power factor to a more efficient level (e.g., above 0.95)

d. Result in reduced energy losses and electricity bills

5. Using your understanding or sample calculations, explain how improving power factor reduces electricity demand charges and lowers monthly costs.

Consider the payback period due to savings on energy bills.

6. Write a short report or presentation summarising:

a. The problem of poor power factor in the plant

b. How capacitors correct the power factor

c. The benefits in terms of efficiency and cost savings

d. Any other factors to consider (e.g., maintenance, monitoring)

Activity 6.12 Phasor Diagrams for R, R-L and R-C series circuits

1. R Series Circuit (Pure Resistor)

Steps

a. Draw a horizontal line to the right — this is the voltage phasor (V).

b. Draw the current phasor (I) in phase with voltage — it lies on the same line as voltage.

2. R-L Series Circuit (Resistor + Inductor) Steps

a. Draw the current phasor (I) horizontally to the right — this is your reference.

b. Draw the voltage across the resistor (V_R) in phase with current — on the same horizontal line.

c. Draw the voltage across the inductor (V_L) straight up — it leads the current by 90°.

d. Draw the total voltage (V) as the vector sum of VRV_R and VLV_L

— use the tip-to-tail method to complete the right triangle.

3. R-C Series Circuit (Resistor + Capacitor) Steps

a. Draw the current phasor (I) horizontally to the right — this is your reference.

b. Draw the voltage across the resistor (V_R) in phase with current — on the same horizontal line.

c. Draw the voltage across the capacitor (V_C) straight down — it lags the current by 90°.

d. Draw the total voltage (V) as the vector sum of VRV_R and VCV_C

— complete the triangle.

4. Now, use Pythagoras’s Theorem to calculate the impedance for each of these circuits. Also, use trigonometry to find the phase angle (angle of impedance).

Review Question 6.1

1. A sinusoidal alternating current has a peak value of 5 A. Calculate:

a. The RMS value of the current.

b. The peak-to-peak current.

2. An AC voltage is represented by the equation: V (t ) = 240sin (100πt ).

Determine the

a. Peak voltage.

b. RMS voltage.

c. Frequency of the AC source.

3. A resistor of resistance 20 Ω is connected to an AC source with an RMS voltage of 120 V. Calculate:

a. The RMS current through the resistor.

b. The peak current.

c. The average power dissipated in the resistor.

4. An oscilloscope displays a periodic voltage waveform that repeats every 4 milliseconds. The peak voltage is 15 V. Calculate:

a. The frequency of the waveform.

b. The RMS value of the voltage.

Review Question 6.2

1. An AC voltage of 125 V (rms) at 50 Hz is applied across a purely resistive load of 25 Ω.

Find

a. The current in the circuit.

b. The power consumed.

2. A 60 Hz AC source of 84 V (rms) is connected to a capacitor of capacitance 100 μF. Find the current in the circuit.

3. An inductor of 0.3H is connected to a 230V (rms), 50Hz AC source.

Find the current in the circuit.

4. In a series RLC circuit, R = 20 Ω, L = 0.05 H, and C = 150 μF, connected to a 100 V (rms), 60 Hz source. Find the current in the circuit.

5. An industrial sewing machine draws 10 A at 150 V with a power factor of 0.8 lagging.

Calculate the

a. Real power

b. Reactive power

c. Apparent power

6. A load has a real power consumption of 1800 W and draws 3000 VA of apparent power.

Find the power factor and state whether it is lagging or leading if the load is inductive.

7. A series RLC circuit in a radio set has R=30 Ω, L=0.2 H, and a variable capacitor C.

If the circuit is driven by a 100 V (rms) source at 60 Hz, determine the value of C that causes resonance, and compute the current in the circuit at resonance.

8. An AC motor rated at 5 kW operates with a lagging power factor of 0.7 and draws 8.5 kVA

a. Calculate the reactive power.

b. Determine how much reactive power must be cancelled (using a capacitor) to improve the power factor to 0.9.

c. What will be the new apparent power after correction?

9. A factory in Kasoa operates with several inductive machines and has a total real power consumption of 200 kW and an average power factor of 0.65 lagging. The management wants to improve the power factor to 0.95 to reduce losses and avoid penalties.

a. Calculate the existing reactive power demand.

b. Calculate the required reactive power compensation.

c. Determine the capacitance (in μF) of a bank of capacitors connected in parallel to the supply (at 50 Hz) that would provide the required compensation.

d. Discuss how this power factor improvement affects the total current drawn and energy costs over time.