Notes for Chapters 1–4A: Circuit Variables, Elements, Simple Resistive Circuits, and Techniques of Circuit Analysis (Part A)

Page 1

• Topic: Quiz 1 Prequiz Review (Chapters 1–4A). This sets the scope for the early material and the kinds of concepts you should be comfortable with before Quiz 1.

Page 2

• Chapters covered: 1–4A
• Major subtopics:
  • Circuit Variables
  • Circuit Elements
  • Simple Resistive Circuits
  • Techniques of Circuit Analysis (Part A)
• This page serves as an index to the material and the analytical tools to be used across the chapters.

Page 3

Chapter 1. Circuit Variables

Main ideas

• Definitions:

  • Charge, denoted by
    q(t)
  • Voltage, denoted by
    v(t)
  • Current, denoted by
    i(t)
  • Power, denoted by
    p(t)
  • Energy, denoted by
    w(t)

• The ideal basic circuit element (the fundamental relation between voltage and current across an element).

• Polarity (positive & negative terminals).

• Passive Sign Convention (PSC): how to assign current direction and terminal polarity to determine the sign of power.

• Key relationships:

  • Charge and current:
    q(t) = \int i(t) \, dt
  • Energy and power:
    w(t) = \int p(t) \, dt\quad \text{and}\quad p(t) = v(t)i(t)

• PSC interpretation:

  • If current enters the positive terminal of an element, the element absorbs power:
    p(t) = v(t)i(t) \ge 0
  • If current enters the negative terminal, the element supplies power:
    p(t) < 0

• Notation recap:

  • Positive terminal refers to the labeled + node on the element.
  • Currents measured into the element follow the PSC sign convention.

Page 4

1.4 Voltage and Current Examples

Example 1.4

• Given the charge function for a point entering a circuit:
  q(t) = 5t \sin(4t)\ \,\text{mC}
• Find the current at time
  t = 0.5\ \,\text{s}.

Solution (method)

• Current is the time derivative of charge:
  i(t) = \dfrac{dq}{dt}.

• Differentiate:
  q(t) = 5t\sin(4t) \Rightarrow i(t) = 5\sin(4t) + 20t\cos(4t).

• Evaluate at
  t=0.5\ \,\text{s}:
  i(0.5) = 5\sin(2) + 20(0.5)\cos(2) = 5\sin(2) + 10\cos(2).

• Numerical (using radians):

  • \sin(2) \approx 0.9093,
  • \cos(2) \approx -0.4161,
  • Therefore,
    i(0.5) \approx 5(0.9093) + 10(-0.4161) \approx 4.5465 - 4.161 \approx 0.386\ \text{mA}.

• Note: The calculator should be in radians mode for the derivative with respect to t.

Page 5

1.6 Power and Energy (Examples)

Example: Power and energy absorbed by an element over the first 100 ms

• Given time-varying voltage and current (from the transcript, V(t) and I(t) are provided in a form that yields p(t) = V(t)I(t)); the example computes energy over the interval [0, 0.1 s].

• In the transcript, the voltage/current relationship is written as:

  • V(t) = 12 − 8t
  • I(t) = 5 − 8t
  • Hence, the instantaneous power is:
    p(t) = v(t)i(t) = (12 - 8t)(5 - 8t).

• Key results quoted in the transcript:

  • At t = 0.1 s, the instantaneous power is approximately
    p(0.1) \approx 12.11\ \text{W}.
  • Energy absorbed over the first 100 ms is
    W = \int_{0}^{0.1} p(t)\, dt \approx 2.99\ \text{J}.

• Summary of approach:

  • Compute p(t) via p = v i.
  • Evaluate p(t) at the final time to get instantaneous power.
  • Compute energy by integrating p(t) over the interval of interest.

Page 6

1.6 Power and Energy (Further Illustrations)

• Two cases of an element with absorbing power 12 W:

  • (a) p = 4 × 3 = 12 W
  • (b) p = 4 × 3 = 12 W

• Two cases of an element with supplying power 12 W:

  • (a) p = -4 × 3 = -12 W
  • (b) p = -4 × 3 = -12 W

• Takeaway: The sign of power depends on the PSC convention and the direction of current relative to the element’s polarity. Positive p means absorbed power; negative p means supplied power.

Page 7

Chapter 2. Circuit Elements

Main ideas

• Voltage sources and current sources.

• Ohm’s Law:
  v = iR

• Kirchhoff's Voltage Law (KVL) and Kirchhoff's Current Law (KCL).

• These form the basis for analyzing circuits with linear components and sources.

Page 8

2. Kirchhoff’s Laws (Example 2.8)

• Example 2.8 (Page 42):

  • a) Find 0 (some node/mesh quantity in the example).
  • b) Verify that the total power generated equals the power used.

• Sign convention note:

  • An electrical device or element can either absorb (positive power, p > 0) or supply (negative power, p < 0) power.
  • This is essential when checking power balance in a circuit.

Page 9

2.5 Circuits containing Dependent Sources (Example 2.11)

• Example 2.11 (Page 46):

  • a) Find 0 (some quantity such as node voltage or current).
  • b) Prove that power generated equals power used.

• Takeaway: Dependent sources obey the same KCL/KVL framework as independent sources; they introduce algebraic relationships that tie voltages/currents to other circuit variables.

Page 10

Chapter 3. Simple Resistive Circuits

Main ideas

• Resistors can be treated in series or in parallel.
• Voltage divider and current divider concepts for analyzing networks.
• Measuring current and voltage in circuits.
• General forms for combining resistors:
  • Parallel: for two resistors,
    R{eq} = \dfrac{R1 R2}{R1 + R_2}
  • More generally, for n parallel resistors,
    \dfrac{1}{R{eq}} = \sum{k=1}^n \dfrac{1}{R_k}.
• In a parallel network, the branch currents split; in a series network, the same current flows through all elements.

Page 11

Section: Example 3.7 (Current & Voltage Dividers)

• Method (as per the transcript):
  1) First determine the total current or the total voltage in the network.
  2) Use the current-divider formula to calculate the current through a chosen branch:

  • For two parallel resistors R1 and R2 with total current I, the current through R1 is
    i1 = I \frac{R2}{R1 + R2}.
    3) Find the branch voltage across the specified resistor (e.g., the 24-ohm resistor) using the current and Ohm’s law:
    V{branch} = i{branch} R_{branch}.
    4) Use the voltage divider formula on the remaining branch(es) to find the desired quantity (often another branch current or a voltage).

• This example illustrates how to combine current division and voltage division to solve for branch currents and voltages in a multi-branch circuit.

Page 12

Chapter 2. Circuit Elements – Homework Problem 2.35

• Problem context (as shown):

  • Involves resistors of values such as 8 kΩ and 12 kΩ, and a 200 V source.
  • The objective is to find unknown currents/voltages in a network containing both independent sources and resistors.

• Typical approach:

  • Apply Ohm’s Law and Kirchhoff’s laws to write equations for node voltages or branch currents.
  • Solve for the unknowns (often using node-voltage or mesh-current methods).

Page 13

Chapter 3. Simple Resistive Circuits – Homework Problem 3.19

• Problem context (as shown):

  • Involves a circuit with a 10 Ω resistor and a current of 2.4 A (example values visible in the transcript).
  • Objective is likely to find a missing current or voltage in the network.

• General strategy for these problems:

  • Identify series/parallel relationships.
  • Use Ohm’s Law and divider rules as appropriate.

Page 14

Chapter 4. Techniques of Circuit Analysis

Main ideas

• Terminology:

  • Node, Essential Node, Mesh, Branch, etc.

• Node-Voltage Method (also called the nodal analysis):

  • Uses Kirchhoff’s Current Law (KCL) at essential nodes.
  • Concept of a Supernode to handle voltage sources between non-reference nodes.

• Mesh-Current Method (also called mesh analysis):

  • Uses Kirchhoff’s Voltage Law (KVL) at meshes.
  • Introduces mesh currents as imaginary currents circulating around loops; branch currents are related to these mesh currents.

• The page hints at an example that leads to a set of simultaneous equations for node voltages or mesh currents (illustrative numbers show up in the equations).

Page 15

4.2 Introduction to the Node-Voltage Method

Key concepts and formulas

• KCL at a node: the algebraic sum of currents leaving (or entering) a node is zero.

• Choosing a reference node (ground): all node voltages are measured with respect to this reference.

• Node-voltage equations (for a node i connected to neighboring nodes j through resistors R{ij}): \sumj \frac{Vi - Vj}{R_{ij}} = 0.

• Sign convention:

  • Currents leaving the node are taken as positive; currents entering the node are negative.

• Example setup in the transcript shows node equations for two essential nodes (Nodes 1 and 2), illustrating the formation of KCL equations in practice.

• Practical notes:

  • When a voltage source is connected between a node and the reference, its effect is handled via supernode techniques or by directly setting the node voltage if the source is tied to ground.
  • The Node-Voltage Method yields a linear system that can be solved for node voltages, from which all currents and voltages in the network can be found.

Page 16

4.3 Example 4.3

• Tasks:

  • (a) Find a certain quantity (likely a node voltage or current).
  • (b) Show that the power supplied equals the power used (power balance check).

• The example demonstrates using node-voltage analysis to obtain results and then verifying power balance as a consistency check.

Page 17

4.5 Introduction to The Mesh Current Method

• Overview:

  • Define imaginary mesh currents in each independent loop, typically denoted as i1, i2, i_3, etc.
  • The actual currents in branches are linear combinations of the mesh currents (depending on how the meshes share components).

• Procedure:

  • Write KVL equations for each mesh, accounting for mutual (shared) resistances between meshes. The equations take the form:
    \text{(resistance in mesh)}\cdot i{mesh} + \text{(shared resistance)}\cdot (\pm i{other\;mesh}) = \text{applied} \; V.

• In the example, there are two meshes with currents i1 and i2. The branch currents derived from these mesh currents are labeled as 2 and 3 (as per the diagram in the transcript).

• Important relationships:

  • For a shared resistor between two meshes, the branch current is the difference of the mesh currents: e.g., if resistor is common to meshes 1 and 2, the branch current is
    i3 = i1 - i_2
  • This leads to three branch currents once the two mesh currents are found.

Page 18

4.5 Mesh Current Method – Worked Result

• Given the two-mesh system, solving the KVL equations yields:

  • Mesh current a:
    i_1 = 10\ \text{A}
  • Mesh current b:
    i_2 = 4\ \text{A}

• Corresponding branch currents:

  • i_1 = 10\ \text{A}
  • i_2 = 4\ \text{A}
  • Shared-branch current (i3): i3 = i1 - i2 = 6\ \text{A}

• This example illustrates the practical outcome of the Mesh Current Method: once the mesh currents are known, individual branch currents follow from their definitions and the diagram (shared resistors yield differences of mesh currents).

Summary of key formulas and concepts from Pages 3–18

• PSC and power sign conventions:
  • Power absorbed: p(t) = v(t)i(t) \ge 0
  • Power supplied: p(t) < 0
• Basic time-domain relations:
  • Charge: q(t) = \int i(t)\,dt
  • Energy: w(t) = \int p(t)\,dt
• Ohm’s Law: v = iR
• Resistors in parallel (two-resistor case):
  R{eq} = \dfrac{R1R2}{R1 + R_2}
• Current divider (two parallel resistors):
  i1 = I \dfrac{R2}{R1 + R2}, \quad i2 = I \dfrac{R1}{R1 + R2}
• Voltage divider: for series resistors R1 and R2 with input Vin, output across R2 is V{out} = V{in} \dfrac{R2}{R1 + R2}
• Node-Voltage Method (KCL):
  \sumj \dfrac{Vi - Vj}{R{ij}} = 0
  (currents considered leaving the node, with reference chosen for zero potential)
• Mesh-Current Method (KVL):
  • Solve for mesh currents (e.g., i1, i2, …), and obtain branch currents as combinations of mesh currents; for a shared resistor, i{shared} = i1 - i_2, etc.

Note: This page-by-page set of notes captures the core ideas and example structures presented in the transcript, including both the concepts and the numerical illustrations highlighted on the pages. If you have the original figures or diagrams, adding those references to these notes will help reinforce the visual understanding of node/mesh assignments and how currents flow through specific branches and resistors.