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General physics

    Physics Lesson: Kirchhoff’s Laws and Circuit Analysis

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    🎯 Learning Objectives

    By the end of this lesson, you will be able to:

    • Use the formula for the combined resistance of two or more resistors in series.
    • Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in parallel.
    • Use the formula for the combined resistance of two or more resistors in parallel.
    • Use Kirchhoff’s laws to solve simple circuit problems.
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: By the end of this lesson, you will be able to:]
    • [Казахский перевод для: Use the formula for the combined resistance of two or more resistors in series.]
    • [Казахский перевод для: Derive, using Kirchhoff’s laws, a formula for the combined resistance of two or more resistors in parallel.]
    • [Казахский перевод для: Use the formula for the combined resistance of two or more resistors in parallel.]
    • [Казахский перевод для: Use Kirchhoff’s laws to solve simple circuit problems.]
    🗣️ Language Objectives

    By the end of this lesson, you will be able to:

    • Understand and use key vocabulary related to electric circuits, such as circuit, resistor, current, voltage, resistance, series connection, and parallel connection.
    • Discuss circuit diagrams and problem solutions using appropriate terminology.
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: By the end of this lesson, you will be able to:]
    • [Казахский перевод для: Understand and use key vocabulary related to electric circuits, such as circuit, resistor, current, voltage, resistance, series connection, and parallel connection.]
    • [Казахский перевод для: Discuss circuit diagrams and problem solutions using appropriate terminology.]
    🔑 Key Terms / Негізгі терминдер / Основные термины

    Familiarize yourself with these important terms. Pay attention to their translations.

    English TermRussian Translation (Русский перевод)Kazakh Translation (Қазақша аудармасы)
    Electric Current (I)Электрический токЭлектр тогы
    Voltage (V) / Potential DifferenceНапряжение / Разность потенциаловКернеу / Потенциалдар айырымы
    Resistance (R)СопротивлениеКедергі
    ResistorРезисторРезистор
    Series CircuitПоследовательная цепьТізбектей жалғау
    Parallel CircuitПараллельная цепьПараллель жалғау
    Kirchhoff’s Current Law (KCL)Первый закон Кирхгофа (закон токов)Кирхгофтың бірінші заңы (токтар заңы)
    Kirchhoff’s Voltage Law (KVL)Второй закон Кирхгофа (закон напряжений)Кирхгофтың екінші заңы (кернеулер заңы)
    Junction / NodeУзелТүйін
    LoopКонтурКонтур
    Electromotive Force (EMF, ε)Электродвижущая сила (ЭДС)Электр қозғаушы күші (ЭҚК)
    Power (P)МощностьҚуат
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Familiarize yourself with these important terms. Pay attention to their translations.]
    🃏 Flashcards: Test Your Knowledge!

    Use these flashcards to review the key concepts and terms. You can find many useful sets on Quizlet by searching for «Resistors in Series and Parallel» or «Kirchhoff’s Laws».

    Here is an example set (You might need to adjust the link or find a more specific one):

    (Source: Quizlet. If the embed doesn’t work, try this link: Electric Circuits Flashcards on Quizlet)

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Use these flashcards to review the key concepts and terms. You can find many useful sets on Quizlet by searching for «Resistors in Series and Parallel» or «Kirchhoff’s Laws». Here is an example set…]
    📖 Glossary / Глоссарий

    Circuit: A closed loop path that allows electric current to flow.

    Translation / Перевод / Аударма

    RU: Замкнутый путь, по которому может протекать электрический ток.

    KZ: Электр тогы ағатын тұйық жол.

    Resistor: An electrical component that implements electrical resistance as a circuit element.

    Translation / Перевод / Аударма

    RU: Электрический компонент, который реализует электрическое сопротивление как элемент цепи.

    KZ: Тізбек элементі ретінде электрлік кедергіні жүзеге асыратын электрлік компонент.

    Current (I): The rate of flow of electric charge. Measured in Amperes (A).

    Translation / Перевод / Аударма

    RU: Скорость потока электрического заряда. Измеряется в Амперах (А).

    KZ: Электр зарядының ағу жылдамдығы. Ампермен (А) өлшенеді.

    Voltage (V): The electric potential difference between two points. Measured in Volts (V).

    Translation / Перевод / Аударма

    RU: Разность электрических потенциалов между двумя точками. Измеряется в Вольтах (В).

    KZ: Екі нүкте арасындағы электрлік потенциалдар айырымы. Вольтпен (В) өлшенеді.

    Resistance (R): A measure of the opposition to current flow in an electrical circuit. Measured in Ohms (Ω).

    Translation / Перевод / Аударма

    RU: Мера противодействия протеканию тока в электрической цепи. Измеряется в Омах (Ω).

    KZ: Электр тізбегіндегі ток ағынына қарсылықтың өлшемі. Оммен (Ω) өлшенеді.

    Kirchhoff’s Current Law (KCL): The sum of currents entering a junction (or node) is equal to the sum of currents leaving the junction. This is a statement of charge conservation.

    Translation / Перевод / Аударма

    RU: Сумма токов, входящих в узел (или точку соединения), равна сумме токов, выходящих из узла. Это утверждение о сохранении заряда.

    KZ: Түйінге кіретін токтардың қосындысы түйіннен шығатын токтардың қосындысына тең. Бұл зарядтың сақталу заңының тұжырымы.

    Kirchhoff’s Voltage Law (KVL): The sum of the electromotive forces (EMFs) in any closed loop is equivalent to the sum of the potential drops in that loop. This is a statement of energy conservation.

    Translation / Перевод / Аударма

    RU: Сумма электродвижущих сил (ЭДС) в любом замкнутом контуре эквивалентна сумме падений потенциала в этом контуре. Это утверждение о сохранении энергии.

    KZ: Кез келген тұйық контурдағы электр қозғаушы күштердің (ЭҚК) қосындысы сол контурдағы потенциалдардың төмендеуінің қосындысына эквивалентті. Бұл энергияның сақталу заңының тұжырымы.

    Қазақша аудармасы (Kazakh Translation for entire Glossary section)
    [Казахский перевод для всего раздела Глоссарий]
    💡 Theory: Resistors and Kirchhoff's Laws

    This section covers the fundamental principles of combining resistors and Kirchhoff’s laws for circuit analysis.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: This section covers the fundamental principles of combining resistors and Kirchhoff’s laws for circuit analysis.]

    1. Kirchhoff’s Laws

    Gustav Kirchhoff formulated two laws that are fundamental to circuit analysis. These laws are based on the conservation of charge and energy.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Gustav Kirchhoff formulated two laws that are fundamental to circuit analysis. These laws are based on the conservation of charge and energy.]

    a) Kirchhoff’s Current Law (KCL) — The Junction Rule

    KCL states that the algebraic sum of currents entering any junction (or node) in a circuit must equal the algebraic sum of currents leaving that junction. Essentially, what goes in must come out. This law is a consequence of the conservation of electric charge.

    ΣIin = ΣIout

    For example, if currents I1 and I2 enter a junction, and current I3 leaves it, then: I1 + I2 = I3.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: KCL states that the algebraic sum of currents entering any junction (or node) in a circuit must equal the algebraic sum of currents leaving that junction. Essentially, what goes in must come out. This law is a consequence of the conservation of electric charge. ΣIin = ΣIout. For example, if currents I1 and I2 enter a junction, and current I3 leaves it, then: I1 + I2 = I3.]

    b) Kirchhoff’s Voltage Law (KVL) — The Loop Rule

    KVL states that the algebraic sum of the potential differences (voltages) around any closed loop or path in a circuit must be zero. This law is a consequence of the conservation of energy.

    ΣΔV = 0 (around a closed loop)

    This means that the sum of electromotive forces (EMFs, e.g., from batteries) is equal to the sum of voltage drops across resistors and other components in the loop.

    Σε = ΣIR

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: KVL states that the algebraic sum of the potential differences (voltages) around any closed loop or path in a circuit must be zero. This law is a consequence of the conservation of energy. ΣΔV = 0 (around a closed loop). This means that the sum of electromotive forces (EMFs, e.g., from batteries) is equal to the sum of voltage drops across resistors and other components in the loop. Σε = ΣIR]

    2. Resistors in Series

    When resistors are connected in series, they are connected end-to-end, providing a single path for the current to flow.

    [Изображение: Резисторы соединенные последовательно]

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: When resistors are connected in series, they are connected end-to-end, providing a single path for the current to flow.]

    Key characteristics of a series circuit:

    • Current: The current is the same through each resistor: Itotal = I1 = I2 = I3 = …
    • Voltage: The total voltage across the combination is the sum of the individual voltage drops across each resistor: Vtotal = V1 + V2 + V3 + …
    • Resistance: The total (or equivalent) resistance RT is the sum of the individual resistances:

    RT = R1 + R2 + R3 + …

    Derivation (using Ohm’s Law V=IR and KVL):

    Consider resistors R1, R2, …, Rn in series. The total voltage VT is VT = V1 + V2 + … + Vn (from KVL).
    Since V1 = IR1, V2 = IR2, etc. (current I is the same),
    VT = IR1 + IR2 + … + IRn = I(R1 + R2 + … + Rn).
    If RT is the equivalent resistance, then VT = IRT.
    Therefore, IRT = I(R1 + R2 + … + Rn), which simplifies to RT = R1 + R2 + … + Rn.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Key characteristics of a series circuit… Derivation…]

    3. Resistors in Parallel

    When resistors are connected in parallel, they are connected across the same two points (junctions), providing multiple paths for the current to flow.

    [Изображение: Резисторы соединенные параллельно]

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: When resistors are connected in parallel, they are connected across the same two points (junctions), providing multiple paths for the current to flow.]

    Key characteristics of a parallel circuit:

    • Voltage: The voltage is the same across each resistor: Vtotal = V1 = V2 = V3 = …
    • Current: The total current is the sum of the currents through each branch (from KCL): Itotal = I1 + I2 + I3 + …
    • Resistance: The reciprocal of the total (or equivalent) resistance RT is the sum of the reciprocals of the individual resistances:

    1/RT = 1/R1 + 1/R2 + 1/R3 + …

    Derivation (using Ohm’s Law I=V/R and KCL):

    Consider resistors R1, R2, …, Rn in parallel. The total current IT is IT = I1 + I2 + … + In (from KCL).
    Since I1 = V/R1, I2 = V/R2, etc. (voltage V is the same across all),
    IT = V/R1 + V/R2 + … + V/Rn = V(1/R1 + 1/R2 + … + 1/Rn).
    If RT is the equivalent resistance, then IT = V/RT.
    Therefore, V/RT = V(1/R1 + 1/R2 + … + 1/Rn), which simplifies to 1/RT = 1/R1 + 1/R2 + … + 1/Rn.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Key characteristics of a parallel circuit… Derivation…]

    4. Solving Circuit Problems using Kirchhoff’s Laws

    For more complex circuits that cannot be simplified to simple series or parallel combinations, Kirchhoff’s laws are essential.

    Steps for applying Kirchhoff’s Laws:

    1. Label currents: Assign a current direction to each branch of the circuit. If your assumed direction is wrong, the calculated current will be negative. Label each current (e.g., I1, I2, I3).
    2. Apply KCL: Write KCL equations for (N-1) junctions, where N is the number of junctions. This ensures independent equations.
    3. Apply KVL: Choose independent loops in the circuit. Traverse each loop, writing KVL equations.
      • When traversing a resistor in the direction of the assumed current, the potential change is -IR.
      • When traversing a resistor opposite to the direction of the assumed current, the potential change is +IR.
      • When traversing an EMF source from the negative to the positive terminal, the potential change is +ε.
      • When traversing an EMF source from the positive to the negative terminal, the potential change is -ε.
    4. Solve the system of equations: You will have a system of linear equations. Solve them simultaneously to find the unknown currents (and then voltages or resistances if needed).
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Solving Circuit Problems using Kirchhoff’s Laws… Steps for applying Kirchhoff’s Laws…]

    Theory Questions:

    1. Easy: If three resistors of 10 Ω, 20 Ω, and 30 Ω are connected in series, what is their total resistance?

    Answer

    For resistors in series, RTotal = R1 + R2 + R3.

    RTotal = 10 Ω + 20 Ω + 30 Ω = 60 Ω.

    2. Medium: Two resistors, R1 = 6 Ω and R2 = 3 Ω, are connected in parallel. This combination is then connected in series with a 4 Ω resistor and a 12V battery. What is the total current supplied by the battery?

    Answer

    First, find the equivalent resistance of the parallel combination (Rp):

    1/Rp = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

    So, Rp = 2 Ω.

    Now, this Rp is in series with the 4 Ω resistor. The total circuit resistance (RTotal) is:

    RTotal = Rp + 4 Ω = 2 Ω + 4 Ω = 6 Ω

    The total current supplied by the battery (ITotal) is given by Ohm’s Law (V = IR):

    ITotal = V / RTotal = 12V / 6 Ω = 2 A.

    3. Medium: State Kirchhoff’s Current Law and Kirchhoff’s Voltage Law. Explain the physical principle (conservation law) upon which each law is based.

    Answer

    Kirchhoff’s Current Law (KCL): The algebraic sum of currents entering a junction in an electrical circuit is equal to the algebraic sum of currents leaving the junction.

    Physical Principle: Conservation of electric charge. Charge cannot accumulate at a junction, so the rate at which charge enters must equal the rate at which it leaves.

    Kirchhoff’s Voltage Law (KVL): The algebraic sum of the changes in electric potential around any closed loop in an electrical circuit is zero.

    Physical Principle: Conservation of energy. If a charge moves around a closed loop and returns to its starting point, its net change in potential energy must be zero. Voltage is potential energy per unit charge.

    4. Hard (Critical Thinking): Imagine you have a complex circuit with multiple batteries and resistors. Why might it be impossible to simplify this circuit using only the rules for series and parallel combinations? How do Kirchhoff’s laws provide a more general approach to solving such circuits?

    Answer

    It might be impossible to simplify a complex circuit using only series and parallel rules if the components are not arranged in simple series or parallel configurations. For instance, circuits with multiple voltage sources in different branches, or bridge circuits, often cannot be broken down this way. Resistors might be connected such that they share a common point with several other resistors, but not all of them are across the same two points (for parallel) nor are they connected end-to-end in a single path (for series).

    Kirchhoff’s laws provide a more general approach because they are based on fundamental conservation principles (charge and energy) that apply to any circuit configuration, regardless of its complexity.

    KCL allows us to analyze how current splits and combines at various junctions.

    KVL allows us to track potential changes around any chosen loop.

    By applying these laws systematically, we can generate a set of simultaneous linear equations. Solving these equations allows us to determine all unknown currents and voltages in the circuit, even when simple series/parallel reduction is not feasible. They are powerful tools for analyzing any DC network.

    Қазақша аудармасы (Kazakh Translation for Theory Questions)
    [Казахский перевод для раздела Теоретические вопросы и ответы]
    🧠 Exercises: Memorize the Terms

    Activity 1: Match the Term with its Definition

    1. Current
    2. Voltage
    3. Resistance
    4. Series Circuit
    5. Parallel Circuit

    Definitions:

    • A) A measure of opposition to current flow.
    • B) A circuit where components are connected end-to-end, providing a single path for current.
    • C) The rate of flow of electric charge.
    • D) A circuit where components are connected across the same two points, providing multiple paths for current.
    • E) The electric potential difference between two points.
    Answer

    1. C) Current — The rate of flow of electric charge.

    2. E) Voltage — The electric potential difference between two points.

    3. A) Resistance — A measure of opposition to current flow.

    4. B) Series Circuit — A circuit where components are connected end-to-end, providing a single path for current.

    5. D) Parallel Circuit — A circuit where components are connected across the same two points, providing multiple paths for current.

    Activity 2: Fill in the Blanks

    1. Kirchhoff’s Current Law is based on the conservation of ______.
    2. In a series circuit, the ______ is the same through all components.
    3. In a parallel circuit, the ______ is the same across all components.
    4. The unit of resistance is the ______.
    5. Kirchhoff’s Voltage Law states that the sum of ______ around a closed loop is zero.
    Answer
    1. Kirchhoff’s Current Law is based on the conservation of charge.
    2. In a series circuit, the current is the same through all components.
    3. In a parallel circuit, the voltage is the same across all components.
    4. The unit of resistance is the Ohm (Ω).
    5. Kirchhoff’s Voltage Law states that the sum of potential differences (or voltages) around a closed loop is zero.
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Exercises: Memorize the Terms, Activity 1, Activity 2, and their answers.]
    📺 Watch & Learn: Video Resources

    Watch this video to get a visual understanding of Kirchhoff’s Laws and how they are applied:

    (If the video doesn’t load, you can watch it here: Kirchhoff’s Laws on YouTube by The Organic Chemistry Tutor — This is an example, you might find others more suitable for Cambridge AS/A Level specifically)

    Another useful video on Series and Parallel Circuits:

    (If the video doesn’t load, you can watch it here: Series and Parallel Circuits by The Organic Chemistry Tutor)

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Watch this video to get a visual understanding of Kirchhoff’s Laws and how they are applied… Another useful video on Series and Parallel Circuits…]
    🧮 Solved Examples: Putting Theory into Practice

    Here are some examples demonstrating how to solve circuit problems.

    Қазақша аудармасы (Kazakh Translation for introduction)
    [Казахский перевод для: Here are some examples demonstrating how to solve circuit problems.]

    Problem 1: Series-Parallel Combination

    Consider the circuit shown below. Find:

    a) The total resistance of the circuit.

    b) The total current supplied by the battery.

    c) The current through the 6 Ω resistor.

    [Изображение: Схема для Задачи 1]

    (Imagine a 12V battery connected to a 4 Ω resistor, which is then connected to a parallel combination of a 6 Ω resistor and a 3 Ω resistor.)

    Textual Pronunciation of Solution Steps (Текстовое произношение решения)

    Part a: Total Resistance

    1. First, identify the parallel combination of resistors (six Ohm and three Ohm).
    2. Calculate the equivalent resistance of this parallel pair using the formula: one over R-parallel equals one over R-one plus one over R-two.
    3. For the six Ohm and three Ohm resistors: one over R-parallel equals one over six plus one over three.
    4. Find a common denominator, which is six. So, one over R-parallel equals one over six plus two over six, which equals three over six, or one half.
    5. Therefore, R-parallel equals two Ohms.
    6. Next, this equivalent parallel resistance (two Ohms) is in series with the four Ohm resistor.
    7. Calculate the total resistance by adding the series resistances: R-total equals R-series plus R-parallel.
    8. So, R-total equals four Ohms plus two Ohms, which equals six Ohms.

    Part b: Total Current

    1. Use Ohm’s Law (V equals I times R) for the entire circuit to find the total current.
    2. Total current I-total equals total voltage V-total divided by total resistance R-total.
    3. Given V-total is twelve Volts and we found R-total is six Ohms.
    4. So, I-total equals twelve Volts divided by six Ohms, which equals two Amperes.

    Part c: Current through the 6 Ω resistor

    1. The total current (two Amperes) flows through the four Ohm resistor.
    2. Then, this current splits when it reaches the parallel combination.
    3. To find the current through the six Ohm resistor, first find the voltage across the parallel combination.
    4. Voltage across parallel (V-parallel) equals total current times R-parallel.
    5. So, V-parallel equals two Amperes times two Ohms, which equals four Volts.
    6. This four Volts is the voltage across both the six Ohm and the three Ohm resistor.
    7. Now, use Ohm’s Law for the six Ohm resistor: I-six-Ohm equals V-parallel divided by six Ohms.
    8. So, I-six-Ohm equals four Volts divided by six Ohms, which equals four over six, or two-thirds of an Ampere (approximately zero point six seven Amperes).

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Textual Pronunciation of Solution Steps for Problem 1]

    [/su_spoiler]

    Solution (Решение)

    a) Total Resistance (RTotal)

    The 6 Ω and 3 Ω resistors are in parallel. Let their equivalent resistance be Rp.

    1/Rp = 1/6Ω + 1/3Ω = 1/6 + 2/6 = 3/6 = 1/2

    Rp = 2 Ω

    This Rp is in series with the 4 Ω resistor.

    RTotal = 4Ω + Rp = 4Ω + 2Ω = 6 Ω

    b) Total Current (ITotal)

    Using Ohm’s Law for the entire circuit: V = ITotal * RTotal

    ITotal = V / RTotal = 12V / 6Ω = 2 A

    c) Current through the 6 Ω resistor (I)

    First, find the voltage across the parallel combination (Vp). This is the voltage across Rp.

    Vp = ITotal * Rp = 2A * 2Ω = 4 V

    This voltage Vp is the same across both the 6 Ω and 3 Ω resistors.

    Now, for the 6 Ω resistor:

    I = Vp / 6Ω = 4V / 6Ω = 2/3 A ≈ 0.67 A

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Solution for Problem 1]

    [/su_spoiler]

    Problem 2: Applying Kirchhoff’s Laws

    For the circuit shown below, use Kirchhoff’s laws to find the currents I1, I2, and I3.

    [Изображение: Схема для Задачи 2]

    (Imagine a circuit with two loops. Loop 1: 10V battery, R1=2Ω, R2=3Ω. Loop 2: 5V battery, R2=3Ω, R3=4Ω. R2 is common to both loops. Assume I1 flows from 10V through R1, I2 flows through R2 downwards, I3 flows from 5V through R3. Junction A above R2, Junction B below R2.)

    Textual Pronunciation of Solution Steps (Текстовое произношение решения)
    1. Label Currents and Directions:
      Assume I-one flows out of the positive terminal of the ten Volt battery, through R-one (two Ohms).
      Assume I-three flows out of the positive terminal of the five Volt battery, through R-three (four Ohms).
      Assume I-two flows downwards through the middle resistor R-two (three Ohms).
      Identify a junction, let’s call it ‘A’, where I-one enters, and I-two and I-three (or a combination) are related. Let’s refine this: Assume I1 flows from the 10V battery. At junction A, I1 splits into I2 (through R2) and another current that goes towards the 5V battery. It’s better to define currents in each branch clearly.
      Let’s redefine: I1 = current from 10V battery through R1. I2 = current through R2 (downwards). I3 = current from 5V battery through R3.
      At Junction A (top of R2): I1 enters. I2 leaves. Current going into the branch with 5V battery and R3 is I3. So, at Junction A: I1 = I2 + I3 (if I3 is defined as flowing to the right from junction A).
      Let’s use standard current definitions for each branch and apply KCL at a node.
      Let I1 be current through R1 (from 10V source). Let I2 be current through R3 (from 5V source). Let I3 be current through R2 (downwards).
      KCL at top node (Junction A): I1 enters, I3 enters, I2 leaves. So, I1 + I3 = I2. (Equation 1)
    2. Apply KVL to Loops:

      Loop 1 (Left loop, clockwise): Start from the negative terminal of the ten Volt battery, go up through it (+10V). Go through R-one (two Ohms) in the direction of I-one (-2 times I-one). Go down through R-two (three Ohms) in the direction of I-three (-3 times I-three). Return to the start.
      Equation: positive ten minus two I-one minus three I-three equals zero. (Equation 2 — *Correction: If I3 is current through R2, then it should be -3*I3. Let’s stick to the initial I1, I2, I3 definition where I2 is through R2. KCL: I1 = I2 + I3′ where I3′ is current through R3 and 5V battery. This gets complicated. Let’s use the textbook approach: define branch currents and apply KCL at one node, then KVL to two independent loops.)

      Let’s restart KVL with clear current definitions for each distinct branch:
      I1: current flowing from 10V battery, through R1.
      I2: current flowing downwards through R2.
      I3: current flowing from 5V battery, through R3, towards the junction.
      KCL at Junction A (top of R2): I1 = I2 + I3 (Equation 1)

      KVL Loop 1 (Left loop, containing 10V, R1, R2, clockwise):
      Start below 10V battery, move up: +10V.
      Move through R1 (2Ω) in direction of I1: -2*I1.
      Move down through R2 (3Ω) in direction of I2: -3*I2.
      Back to start.
      Equation: 10 — 2*I1 — 3*I2 = 0 (Equation 2)

      KVL Loop 2 (Right loop, containing 5V, R3, R2, counter-clockwise for this example to make 5V positive):
      Start below 5V battery, move up: +5V.
      Move through R3 (4Ω) in direction of I3: -4*I3.
      Move up through R2 (3Ω) opposite to direction of I2: +3*I2.
      Back to start.
      Equation: 5 — 4*I3 + 3*I2 = 0 (Equation 3)

    3. Solve the System of Equations:
      We have three equations:
      1) I1 = I2 + I3
      2) 10 — 2*I1 — 3*I2 = 0 => 2*I1 + 3*I2 = 10
      3) 5 — 4*I3 + 3*I2 = 0 => 4*I3 — 3*I2 = 5

      Substitute (1) into (2):
      2*(I2 + I3) + 3*I2 = 10
      2*I2 + 2*I3 + 3*I2 = 10
      5*I2 + 2*I3 = 10 (Equation 4)

      Now we have a system of two equations with I2 and I3:
      4) 5*I2 + 2*I3 = 10
      3) -3*I2 + 4*I3 = 5 (rearranged)

      Multiply (4) by 2: 10*I2 + 4*I3 = 20 (Equation 5)
      Subtract (3) from (5):
      (10*I2 + 4*I3) — (-3*I2 + 4*I3) = 20 — 5
      10*I2 + 4*I3 + 3*I2 — 4*I3 = 15
      13*I2 = 15
      I2 = 15/13 Amperes (approximately one point one five Amperes).

      Substitute I2 into (4):
      5*(15/13) + 2*I3 = 10
      75/13 + 2*I3 = 10
      2*I3 = 10 — 75/13 = 130/13 — 75/13 = 55/13
      I3 = 55/26 Amperes (approximately two point one two Amperes).

      Substitute I2 and I3 into (1):
      I1 = I2 + I3 = 15/13 + 55/26
      I1 = 30/26 + 55/26 = 85/26 Amperes (approximately three point two seven Amperes).

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Textual Pronunciation of Solution Steps for Problem 2]

    [/su_spoiler]

    Solution (Решение)

    Let’s define the currents and their assumed directions:

    — I1: current flowing from the 10V battery, through R1 (2Ω).

    — I2: current flowing downwards through R2 (3Ω).

    — I3: current flowing from the 5V battery, through R3 (4Ω), towards the junction with R2.

    1. Apply KCL at Junction A (top of R2):

    I1 = I2 + I3 (Equation 1)

    2. Apply KVL to Loop 1 (Left loop: 10V, R1, R2 — traversed clockwise):

    +10V — I1R1 — I2R2 = 0

    10 — 2I1 — 3I2 = 0 (Equation 2)

    3. Apply KVL to Loop 2 (Right loop: 5V, R3, R2 — traversed counter-clockwise from bottom of 5V source):

    +5V — I3R3 + I2R2 = 0 (Note: +I2R2 because we traverse R2 against assumed I2 direction)

    5 — 4I3 + 3I2 = 0 (Equation 3)

    4. Solve the system of equations:

    Substitute Eq. (1) into Eq. (2):

    10 — 2(I2 + I3) — 3I2 = 0

    10 — 2I2 — 2I3 — 3I2 = 0

    10 — 5I2 — 2I3 = 0 => 5I2 + 2I3 = 10 (Equation 4)

    Now we have Eq. (3) and Eq. (4):

    3I2 — 4I3 = -5 (Rearranging Eq. 3)

    5I2 + 2I3 = 10

    Multiply the second of these by 2:

    10I2 + 4I3 = 20

    Add this to the rearranged Eq. (3):

    (3I2 — 4I3) + (10I2 + 4I3) = -5 + 20

    13I2 = 15 => I2 = 15/13 A ≈ 1.15 A

    Substitute I2 into 5I2 + 2I3 = 10:

    5(15/13) + 2I3 = 10

    75/13 + 2I3 = 10

    2I3 = 10 — 75/13 = (130 — 75)/13 = 55/13

    I3 = 55/26 A ≈ 2.12 A

    Substitute I2 and I3 into Eq. (1):

    I1 = I2 + I3 = 15/13 + 55/26 = 30/26 + 55/26 = 85/26 A

    I1 = 85/26 A ≈ 3.27 A

    All currents are positive, so the assumed directions were correct.

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Solution for Problem 2]

    [/su_spoiler]

    Қазақша аудармасы (Kazakh Translation for entire Solved Examples section)
    [Казахский перевод для всего раздела Решенные примеры]

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    🔬 Research Task: PhET Simulation

    Explore the PhET Interactive Simulation "Circuit Construction Kit: DC" to build and analyze circuits.

    (If the simulation doesn't load, visit: PhET Circuit Construction Kit: DC)

    Tasks:

    1. Build a simple series circuit with a battery and three resistors (e.g., 10Ω, 20Ω, 30Ω). Use the voltmeter to measure the voltage across each resistor and the ammeter to measure the current through them. Verify that RTotal = R1 + R2 + R3 and that current is the same through all.
    2. Build a simple parallel circuit with a battery and two resistors (e.g., 10Ω, 20Ω). Measure the voltage across each resistor and the current through each branch. Verify that 1/RTotal = 1/R1 + 1/R2 and that voltage is the same across both.
    3. Try to build the circuit from "Solved Example 1". Measure the currents and voltages and compare them to the calculated values.
    Expected Observations & Answers

    1. Series Circuit:

    • You should observe that the current measured by the ammeter is the same at all points in the series circuit.
    • The sum of the voltage drops across the individual resistors (V1 + V2 + V3) should equal the battery voltage.
    • The total resistance calculated from Vbattery / Itotal should match R1 + R2 + R3.

    2. Parallel Circuit:

    • You should observe that the voltage measured across each parallel resistor is the same and equal to the battery voltage (assuming no other series components).
    • The total current from the battery should be equal to the sum of the currents flowing through each parallel branch (I1 + I2).
    • The total resistance calculated from Vbattery / Itotal should match the formula 1/RTotal = 1/R1 + 1/R2.

    3. Solved Example 1 Circuit:

    • The measurements from the simulation should closely match the calculated values:
      • Total current &approx; 2A.
      • Voltage across 4Ω resistor &approx; 8V (Itotal * 4Ω).
      • Voltage across the parallel combination (6Ω || 3Ω) &approx; 4V.
      • Current through 6Ω resistor &approx; 0.67A.
      • Current through 3Ω resistor &approx; 1.33A (4V / 3Ω).

    (Note: Slight discrepancies may occur due to the simulation’s internal model or measurement precision.)

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Research Task: PhET Simulation, Tasks, and Expected Observations & Answers.]
    🤝 Pair/Group Work: Collaborative Learning

    Work with a partner or in a small group on the following activity:

    Activity: Circuit Challenge on LearningApps.org

    Explore circuit building or problem-solving activities on LearningApps.org. Here’s an example category (you may need to search for a specific relevant app):

    Physics — Electricity on LearningApps.org

    Instructions:

    1. Go to the link above or search for «series parallel circuits quiz» or «Kirchhoff’s laws quiz» on LearningApps.org or Quizizz.com.
    2. Choose an activity that involves calculating equivalent resistance, current, or voltage in series and parallel circuits, or applying Kirchhoff’s laws.
    3. Work together to solve the problems presented in the app. Discuss your approaches and reasoning.
    4. If you choose a quiz, try to achieve the highest score as a team.

    Alternatively, use Quizizz: Series and Parallel Circuits on Quizizz

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Pair/Group Work: Collaborative Learning, Activity, Instructions.]
    📝 Individual Work: Structured Questions

    Answer the following questions. Show all your working.

    Қазақша аудармасы (Kazakh Translation for introduction)
    [Казахский перевод для: Answer the following questions. Show all your working.]

    Question 1: A 12.0 V battery is connected to three resistors (R1=4.0 Ω, R2=8.0 Ω, R3=12.0 Ω) connected in series.

    a) Calculate the total resistance of the circuit.

    b) Calculate the current flowing from the battery.

    c) Calculate the potential difference across each resistor.

    Answer

    a) RTotal = R1 + R2 + R3 = 4.0 Ω + 8.0 Ω + 12.0 Ω = 24.0 Ω

    b) I = V / RTotal = 12.0 V / 24.0 Ω = 0.50 A

    c) V1 = IR1 = 0.50 A * 4.0 Ω = 2.0 V

    V2 = IR2 = 0.50 A * 8.0 Ω = 4.0 V

    V3 = IR3 = 0.50 A * 12.0 Ω = 6.0 V

    (Check: 2.0V + 4.0V + 6.0V = 12.0V, which is the battery voltage, as expected from KVL)

    Question 2: The same three resistors (R1=4.0 Ω, R2=8.0 Ω, R3=12.0 Ω) are now connected in parallel to the 12.0 V battery.

    a) Calculate the total resistance of the circuit.

    b) Calculate the total current flowing from the battery.

    c) Calculate the current flowing through each resistor.

    Answer

    a) 1/RTotal = 1/R1 + 1/R2 + 1/R3 = 1/4.0 + 1/8.0 + 1/12.0

    1/RTotal = (6 + 3 + 2) / 24.0 = 11/24.0 Ω-1

    RTotal = 24.0 / 11 Ω &approx; 2.18 Ω

    b) ITotal = V / RTotal = 12.0 V / (24.0/11 Ω) = 12.0 * 11 / 24.0 A = 11/2 A = 5.5 A

    c) Voltage across each resistor is 12.0 V.

    I1 = V / R1 = 12.0 V / 4.0 Ω = 3.0 A

    I2 = V / R2 = 12.0 V / 8.0 Ω = 1.5 A

    I3 = V / R3 = 12.0 V / 12.0 Ω = 1.0 A

    (Check: 3.0A + 1.5A + 1.0A = 5.5A, which is the total current, as expected from KCL)

    Question 3 (Analysis): A circuit consists of a 24V battery. Resistor A (6Ω) and Resistor B (12Ω) are connected in parallel. This parallel combination is then connected in series with Resistor C (2Ω).

    a) Draw the circuit diagram.

    b) Calculate the total resistance of the circuit.

    c) Calculate the current flowing through Resistor C.

    d) Calculate the current flowing through Resistor A and Resistor B.

    e) Calculate the power dissipated by Resistor C.

    Answer

    a) Circuit Diagram: [Student should draw a diagram showing a 24V battery, then Resistor C (2Ω) in series with a parallel branch containing Resistor A (6Ω) and Resistor B (12Ω)].
    [Изображение: Схема для Вопроса 3]

    b) Resistance of parallel A & B (RAB):
    1/RAB = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4. So, RAB = 4Ω.
    Total Resistance RTotal = RC + RAB = 2Ω + 4Ω = .

    c) Current through Resistor C (IC) is the total current ITotal:
    ITotal = Vbattery / RTotal = 24V / 6Ω = 4A. So, IC = 4A.

    d) Voltage across the parallel combination (VAB):
    VAB = ITotal * RAB = 4A * 4Ω = 16V.
    Current through Resistor A (IA): IA = VAB / RA = 16V / 6Ω = 8/3 A &approx; 2.67A.
    Current through Resistor B (IB): IB = VAB / RB = 16V / 12Ω = 4/3 A &approx; 1.33A.
    (Check: IA + IB = 8/3 + 4/3 = 12/3 = 4A = ITotal)

    e) Power dissipated by Resistor C (PC):
    PC = IC2 * RC = (4A)2 * 2Ω = 16 * 2 = 32W.
    (Alternatively, VC = ICRC = 4A * 2Ω = 8V. PC = VCIC = 8V * 4A = 32W)

    Question 4 (Synthesis — Kirchhoff’s Laws):
    Consider the circuit below. Use Kirchhoff’s laws to determine the magnitude and direction of the current in the 5 Ω resistor.

    [Изображение: Схема для Вопроса 4]

    (Example setup: Left loop: 20V battery, 2Ω resistor. Right loop: 10V battery (opposing 20V if in series aiding), 3Ω resistor. Middle branch connecting the two loops: 5Ω resistor.)

    Answer

    Let I1 be current from 20V source (e.g., clockwise in left loop).
    Let I2 be current from 10V source (e.g., counter-clockwise in right loop, so it flows into the junction with the 5Ω resistor).
    Let I3 be current through the 5Ω resistor (e.g., downwards).

    Assume directions:
    I1 flows from 20V source, through 2Ω resistor.
    I2 flows from 10V source, through 3Ω resistor.
    I3 flows downwards through 5Ω resistor.

    KCL at top junction: I1 = I2 + I3 (if I2 flows towards junction from right, and I1 from left, I3 down)
    Let’s redefine for clarity:
    I1: current through 2Ω resistor (left branch, downwards after 20V battery)
    I2: current through 3Ω resistor (right branch, downwards after 10V battery)
    I3: current through 5Ω resistor (middle branch, downwards)

    KCL at top node (P): I1 + I2 = I3 (assuming I1 and I2 flow towards P from their batteries, and I3 flows down from P)
    This setup is ambiguous without a clear diagram. Let’s use a standard diagram:
    A 20V source and 2Ω resistor on the left. A 10V source and 3Ω resistor on the right. A 5Ω resistor connecting the midpoints.

    Let I1 be the current flowing clockwise in the left loop.
    Let I2 be the current flowing clockwise in the right loop.
    The current through the 5Ω resistor will be (I1 — I2) downwards, or (I2 — I1) upwards.

    Loop 1 (left, clockwise): 20 — 2I1 — 5(I1 — I2) = 0
    => 20 — 2I1 — 5I1 + 5I2 = 0
    => 20 — 7I1 + 5I2 = 0 (Eq. 1)

    Loop 2 (right, clockwise): -10 — 3I2 — 5(I2 — I1) = 0 (Note: -10V if traversing clockwise against 10V battery, and current through 5Ω is I2-I1 if I2 is dominant downwards in middle)
    Let’s assume 10V battery also pushes current clockwise in its own loop.
    Loop 2 (right, clockwise): 10 — 3I2 — 5(I2 — I1) = 0
    => 10 — 3I2 — 5I2 + 5I1 = 0
    => 10 — 8I2 + 5I1 = 0 (Eq. 2)

    From Eq. 1: 7I1 — 5I2 = 20
    From Eq. 2: 5I1 — 8I2 = -10

    Multiply Eq.1 by 8: 56I1 — 40I2 = 160
    Multiply Eq.2 by 5: 25I1 — 40I2 = -50
    Subtract second from first: (56-25)I1 = 160 — (-50)
    31I1 = 210 => I1 = 210/31 A &approx; 6.77 A

    Substitute I1 into 5I1 — 8I2 = -10:
    5(210/31) — 8I2 = -10
    1050/31 — 8I2 = -10
    8I2 = 1050/31 + 10 = 1050/31 + 310/31 = 1360/31
    I2 = 1360 / (31 * 8) = 170/31 A &approx; 5.48 A

    Current through 5Ω resistor is I = I1 — I2 (assuming downwards direction for I1 contribution)
    I = 210/31 — 170/31 = 40/31 A &approx; 1.29 A.
    Since the result is positive, the assumed direction (I1 > I2, so net current downwards) is correct.
    Magnitude: 40/31 A. Direction: Downwards (if top of 5Ω is connected to junction of 2Ω and 20V, and bottom to junction of 3Ω and 10V, and I1, I2 are loop currents).
    (Note: Solution depends heavily on the exact circuit diagram and assumed current directions. The placeholder diagram is not specific enough for a definitive single solution without making assumptions. Students should draw their diagram and clearly label currents and loops.)

    Question 5 (Critical Analysis & Design): You are given three identical resistors, each with resistance R.

    a) How many different values of total resistance can you obtain by connecting these three resistors in various combinations (using all three each time)?

    b) Draw the circuit diagram for each combination and calculate its total resistance in terms of R.

    c) If R = 10Ω, and you need a total resistance of 15Ω using these three resistors, which combination would you use? Explain if it’s possible.

    Answer

    a) There are 4 possible distinct combinations using all three resistors:

    1. All three in series.
    2. All three in parallel.
    3. Two in parallel, and this combination in series with the third.
    4. Two in series, and this combination in parallel with the third.

    b) Circuit Diagrams and Total Resistance:

    1. All three in series:

      Diagram: R — R — R

      RTotal = R + R + R = 3R
    2. All three in parallel:

      Diagram: Three resistors connected across the same two points.

      1/RTotal = 1/R + 1/R + 1/R = 3/R

      RTotal = R/3
    3. Two in parallel, in series with the third: (R || R) + R

      Diagram: (R || R) — R

      Resistance of two in parallel (Rp): 1/Rp = 1/R + 1/R = 2/R => Rp = R/2.

      RTotal = Rp + R = R/2 + R = 3R/2 (or 1.5R)
    4. Two in series, in parallel with the third: (R + R) || R

      Diagram: (R — R) || R

      Resistance of two in series (Rs): Rs = R + R = 2R.

      1/RTotal = 1/Rs + 1/R = 1/(2R) + 1/R = 1/(2R) + 2/(2R) = 3/(2R)

      RTotal = 2R/3

    c) If R = 10Ω, we need a total resistance of 15Ω.

    • Combination 1: 3R = 3 * 10Ω = 30Ω (No)
    • Combination 2: R/3 = 10Ω/3 &approx; 3.33Ω (No)
    • Combination 3: 3R/2 = (3 * 10Ω)/2 = 30Ω/2 = 15Ω (Yes!)
    • Combination 4: 2R/3 = (2 * 10Ω)/3 = 20Ω/3 &approx; 6.67Ω (No)

    Yes, it is possible. You would use the combination where two resistors are connected in parallel, and this parallel group is connected in series with the third resistor. This gives a total resistance of 3R/2, which is 1.5R. If R=10Ω, then 1.5 * 10Ω = 15Ω.

    Қазақша аудармасы (Kazakh Translation for Individual Work section)
    [Казахский перевод для всего раздела Индивидуальная работа: Структурированные вопросы и ответы]
    🔗 Further Study & Resources

    For more detailed explanations, examples, and practice problems, check out these excellent resources:

    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Further Study & Resources and the list of links.]
    🤔 Reflection / Рефлексия

    Take a few moments to reflect on your learning:

    1. What was the most challenging concept for you in this lesson? What steps can you take to understand it better?
    2. Which part of the lesson did you find most interesting or helpful? Why?
    3. How confident do you feel about using Kirchhoff’s laws to solve circuit problems now? (Scale of 1-5, where 5 is very confident)
    4. Can you explain, in your own words, why the total resistance decreases when resistors are added in parallel?
    5. What is one question you still have about series circuits, parallel circuits, or Kirchhoff’s laws?
    Қазақша аудармасы (Kazakh Translation)
    [Казахский перевод для: Take a few moments to reflect on your learning:]
    1. [Казахский перевод для: What was the most challenging concept for you in this lesson? What steps can you take to understand it better?]
    2. [Казахский перевод для: Which part of the lesson did you find most interesting or helpful? Why?]
    3. [Казахский перевод для: How confident do you feel about using Kirchhoff’s laws to solve circuit problems now? (Scale of 1-5, where 5 is very confident)]
    4. [Казахский перевод для: Can you explain, in your own words, why the total resistance decreases when resistors are added in parallel?]
    5. [Казахский перевод для: What is one question you still have about series circuits, parallel circuits, or Kirchhoff’s laws?]