Direct Current (DC) Circuits and Series Circuit Analysis Study Guide

Fundamentals of DC and AC Circuits

Direct Current (DC):
- Voltage and current remain constant with time in the ideal DC case.
- A battery-powered circuit is a primary real-world example of a DC system.
- On a voltage-versus-time graph, ideal DC appears as a straight horizontal line.
Alternating Current (AC):
- Voltage and/or current dynamically change with time.
- AC is commonly represented by a sinusoidal waveform.

Circuit Variables and Core Mathematical Laws

Basic Circuit Requirements:
- A functional basic circuit contains a voltage source, an electrical load (such as a resistor), and a closed continuous path for current flow.
Fundamental Circuit Variables and Units:
- Voltage ( $V$ ): Measured in volts ( $\text{V}$ ). Represents electrical potential difference across components. Across a resistor, it is referred to as a voltage drop ( $V_R$ ).
- Current ( $I$ ): Measured in amperes ( $\text{A}$ ). Represents the rate of flow of electric charge.
- Resistance ( $R$ ): Measured in ohms ( $\text{Ω}$ ). Represents opposition to current flow.

Ohm's Law:
- Expresses the linear relationship between voltage, current, and resistance:
  - $V = I \times R$
  - $I = \frac{V}{R}$
  - $R = \frac{V}{I}$
Relationship Between Current and Resistance:
- For a constant voltage source, current and resistance are inversely proportional:
  - $I \propto \frac{1}{R}$
- Increasing resistance decreases current; decreasing resistance increases current.
- An $I$ -versus- $R$ graph forms a decreasing inverse curve.
- Limiting behavior using a fixed $10\,\text{V}$ source:

    *   At  $R = 10\,\text{k}\Omega$ ,  $I = 1\,\text{mA}$ .
    *   At  $R = 10\,\Omega$ ,  $I = 1\,\text{A}$ .
    *   At  $R = 1\,\Omega$ ,  $I = 10\,\text{A}$ .
    *   Very small  $R$  produces a very large  $I$ .
    *   As  $R \rightarrow 0$ ,  $I \rightarrow \infty$  in an ideal circuit model.

Relationship Between Voltage and Resistance:
- If current is held constant, voltage drop and resistance are directly proportional:
  - $V \propto R$
- Follows the general linear equation $y = mx + b$ , where $y = V$ , $x = R$ , slope $m = I$ , and intercept $b = 0$ .
- Example for a constant current $I = 1\,\text{A}$ :
  - $R = 1\,\Omega \rightarrow V = 1\,\text{V}$
  - $R = 10\,\Omega \rightarrow V = 10\,\text{V}$
  - $R = 50\,\Omega \rightarrow V = 50\,\text{V}$
Watt's Law and Power Formulas:
- $P = V \times I$
- $P = I^2 \times R$
- $P = \frac{V^2}{R}$
Engineering Prefixes Standard:
- All quantitative electrical values must be expressed using standard engineering prefixes rather than raw decimals or unformatted scientific notation.
- Common prefixes: milli ( $10^{-3}$ ), micro ( $10^{-6}$ ), nano ( $10^{-9}$ ), pico ( $10^{-12}$ ), kilo ( $10^{3}$ ), mega ( $10^{6}$ ).
- Example: Write $10\,\text{mA}$ instead of $0.01\,\text{A}$ or $10 \times 10^{-3}\,\text{A}$ ; write $33\,\text{pA}$ instead of $0.000000000033\,\text{A}$ .

Series Circuit Characteristics and Theoretical Principles

Definition of a Series Circuit:
- Components are connected end-to-end in a single continuous loop.
- Test for Series Connection: Two or more elements are in series if and only if all current exiting the first element enters and flows through the second element and only the second element on its path back to the source.
- If a node allows current to split into alternate paths, those elements are not in series.
Current Consistency in Series:
- There is only one path for current to follow.
- Current is uniform throughout the entire loop:
  - $I_T = I_1 = I_2 = I_3 = \dots = I_n$
Voltage Division in Series:
- While current is identical across components, total voltage supplied by the source divides among all components in series.
Polarity Conventions:
- Voltage Sources: Current exits the positive terminal and returns to the negative terminal.
- Voltage Drops (Loads): Current enters the positive terminal and exits the negative terminal (opposite polarity convention to voltage sources).
Series Total Resistance:
- Resistances in series add directly to equal the total series resistance ( $R_T$ ):
  - $R_T = \sum_{i=1}^n R_i = R_1 + R_2 + R_3 + \dots + R_n$
Kirchhoff's Voltage Law (KVL):
- Statement: In any closed electrical loop, the algebraic sum of all voltages is zero, or the sum of voltage rises equals the sum of voltage drops.
  - $\sum V_{\text{rises}} = \sum V_{\text{drops}}$
  - $\sum V = 0$
- For a single source $V_S$ supplying three series resistors:
  - $V_S = V_{R1} + V_{R2} + V_{R3}$
  - $V_S - V_{R1} - V_{R2} - V_{R3} = 0$

Electrical Nodes, Nets, and Topology

Definition of an Electrical Node (Net):
- An electrical node is a unique electrical connection point, not a geographical, spatial, or topographical location on a schematic diagram.
- Points connected directly by ideal zero-ohm conductors with no intervening components are at the exact same electrical potential and form a single node.
- Component arrangement or visual wire shapes do not alter circuit functionality provided node connectivity is preserved.
Node Naming Standards:
- Nodes are designated using letters ( $A, B, C$ ) or numbers.
- Node $0$ is strictly reserved for the Reference Node (Ground).
Ground Reference Node:
- Ground establishes the zero-volt baseline ( $V_{\text{ref}} = 0\,\text{V}$ ).
- Connecting ground to a node forces that node's voltage potential to $0\,\text{V}$ .

Voltage Division and Proportional Drop Analysis

Proportionality Rule:
- Source voltage divides across series resistors in direct proportion to each resistor's resistance relative to $R_T$ .
- A resistor comprising $X\%$ of the total resistance will drop $X\%$ of the total applied voltage.
Voltage Division Formula:
- $V_{Rx} = V_S \times \left(\frac{R_x}{R_T}\right)$
- The ratio $\frac{R_x}{R_T}$ represents the fractional portion of total resistance.
- Unit Analysis: The ohms units cancel in the ratio, leaving the final unit in volts.
Three-Resistor Worked Example ( $V_S = 10\,\text{V}$ , $R_1 = 100\,\Omega$ , $R_2 = 300\,\Omega$ , $R_3 = 600\,\Omega$ ):
- Total Series Resistance:
  - $R_T = 100\,\Omega + 300\,\Omega + 600\,\Omega = 1000\,\Omega = 1\,\text{k}\Omega$
- Proportional Computations:
  - $R_1$ is $10\%$ of $R_T \rightarrow V_{R1} = 10\% \times 10\,\text{V} = 1\,\text{V}\n * R_2 $is$ 30\% of $R_T \rightarrow V_{R2} = 30\% \times 10\,\text{V} = 3\,\text{V}
  - $R_3$ is $60\%$ of $R_T \rightarrow V_{R3} = 60\% \times 10\,\text{V} = 6\,\text{V}\n * Formula Substitution:\n * V_{R1} = 10\,\text{V} \times \left(\frac{100\,\Omega}{1000\,\Omega}\right) = 1\,\text{V}\n * V_{R2} = 10\,\text{V} \times \left(\frac{300\,\Omega}{1000\,\Omega}\right) = 3\,\text{V}\n * V_{R3} = 10\,\text{V} \times \left(\frac{600\,\Omega}{1000\,\Omega}\right) = 6\,\text{V}\n * KVL Check:\n * 1\,\text{V} + 3\,\text{V} + 6\,\text{V} = 10\,\text{V}\n\n# Circuit Analysis Methods: Ohm's Law vs. Voltage Division\n\n* **Analyzing Series Voltage Drops using Ohm's Law:**\n * Step 1: Determine total circuit current (I_T):\n * I_T = \frac{V_S}{R_T} = \frac{10\,\text{V}}{1000\,\Omega} = 10\,\text{mA}\n * Step 2: Calculate individual voltage drops (V_{Rx} = I_T \times R_x):\n * V_{R1} = 10\,\text{mA} \times 100\,\Omega = 1\,\text{V}\n * V_{R2} = 10\,\text{mA} \times 300\,\Omega = 3\,\text{V}\n * V_{R3} = 10\,\text{mA} \times 600\,\Omega = 6\,\text{V}\n * Reverse Current Verification:\n * I = \frac{V_{R1}}{R_1} = \frac{1\,\text{V}}{100\,\Omega} = 10\,\text{mA}\n\n* **Method Equivalence:**\n * Both Ohm's Law and the Voltage Division Formula yield identical results and reflect identical physical laws.\n\n# Calculating Node Voltages and Potential Differences\n\n* **Node Voltage Definition:**\n * The voltage potential measured from a given node to the reference node (Ground, 0\,\text{V}).\n\n* **Path Summation Method to Ground:**\n * Trace any path from the target node to ground, summing all voltage differences encountered.\n * Polarity Sign Rule: Add voltage (+ $) when entering a positive polarity; subtract voltage ($ -) when entering a negative polarity.\n * For the 10\,\text{V} $series circuit ($ V_S = 10\,\text{V} $, drops$ V_{R1}=1\,\text{V} $,$ V_{R2}=3\,\text{V} $,$ V_{R3}=6\,\text{V}):\n * **Node A (V_A):**\n * Path 1 (clockwise through drops): V_A = +1\,\text{V} + 3\,\text{V} + 6\,\text{V} = 10\,\text{V}\n * Path 2 (direct through source): V_A = +10\,\text{V}\n * **Node B (V_B):**\n * Path 1 (downward through drops): V_B = +3\,\text{V} + 6\,\text{V} = 9\,\text{V}\n * Path 2 (upward through source): V_B = -1\,\text{V} + 10\,\text{V} = 9\,\text{V}\n * **Node C (V_C):**\n * Path 1 (downward through R_3 $):$ V_C = +6\,\text{V}\n * Path 2 (upward through source): V_C = -3\,\text{V} - 1\,\text{V} + 10\,\text{V} = 6\,\text{V}\n * **Node 0 (V_0 $):** Ground baseline =$ 0\,\text{V}.\n\n* **Ohm's Law Node Voltage Method:**\n * V_{\text{node}} = I_T \times R_{\text{node-to-ground}}\n * V_A = 10\,\text{mA} \times (100\,\Omega + 300\,\Omega + 600\,\Omega) = 10\,\text{V}\n * V_B = 10\,\text{mA} \times (300\,\Omega + 600\,\Omega) = 9\,\text{V}\n * V_C = 10\,\text{mA} \times 600\,\Omega = 6\,\text{V}\n\n* **Voltage Potential Difference Between Two Nodes (V_{AB}):**\n * The voltage across any two-terminal element or between any two nodes A $and$ B equals the difference of their node voltages:\n * V_{AB} = V_A - V_B\n * Examples:\n * V_{AB} = V_A - V_B = 10\,\text{V} - 9\,\text{V} = 1\,\text{V} = V_{R1}\n * V_{BC} = V_B - V_C = 9\,\text{V} - 6\,\text{V} = 3\,\text{V} = V_{R2}\n * V_{C0} = V_C - V_0 = 6\,\text{V} - 0\,\text{V} = 6\,\text{V} = V_{R3}\n * V_{AC} = V_A - V_C = 10\,\text{V} - 6\,\text{V} = 4\,\text{V}\n\n# Multiple Voltage Sources in Series\n\n* **Characteristics of Ideal Voltage Sources:**\n * Maintains a fixed output voltage regardless of current demand.\n * Internal impedance is zero ohms (0\,\text{Omega}, equivalent to a short circuit).\n\n* **Algebraic Combination of Series Sources:**\n * Voltage sources connected in series add algebraically:\n * V_T = \sum V_{\text{sources}}\n\n* **Determining Net Polarity using KVL Loop Analysis:**\n * Trace a loop direction to determine exiting terminal polarities.\n * Group sources by exiting polarity (positive-exiting vs. negative-exiting).\n * Sum each group. The net voltage magnitude is the difference between the two sums, taking the polarity of the larger sum.\n * Net current exits the positive terminal of the equivalent net voltage source.\n\n* **Comprehensive Worked Example (4 Sources, 4 Resistors):**\n * **Components:**\n * Voltage sources: V_1 = 10\,\text{V} $,$ V_2 = 12\,\text{V} $,$ V_3 = 7\,\text{V} $,$ V_4 = 14\,\text{V}.\n * Resistors: R_1 = 20\,\Omega $,$ R_2 = 10\,\Omega $,$ R_3 = 15\,\Omega $,$ R_4 = 5\,\Omega.\n * **Polarity Grouping:**\n * Positive-exiting sources: V_1 $($ +10\,\text{V} $) and$ V_4 $($ +14\,\text{V} $)$ \rightarrow $Sum =$ 24\,\text{V}.\n * Negative-exiting sources: V_2 $($ +12\,\text{V} $) and$ V_3 $($ +7\,\text{V} $)$ \rightarrow $Sum =$ 19\,\text{V}.\n * **Net Voltage Calculation:**\n * V_T = 24\,\text{V} - 19\,\text{V} = 5\,\text{V} $(polarity matches$ V_1 $and$ V_4).\n * **Total Resistance:**\n * R_T = 20\,\Omega + 10\,\Omega + 15\,\Omega + 5\,\Omega = 50\,\Omega\n * **Circuit Current:**\n * I_T = \frac{V_T}{R_T} = \frac{5\,\text{V}}{50\,\Omega} = 100\,\text{mA}\n * **Voltage Drops across Resistors:**\n * V_{R1} = 100\,\text{mA} \times 20\,\Omega = 2.0\,\text{V}\n * V_{R2} = 100\,\text{mA} \times 10\,\Omega = 1.0\,\text{V}\n * V_{R3} = 100\,\text{mA} \times 15\,\Omega = 1.5\,\text{V}\n * V_{R4} = 100\,\text{mA} \times 5\,\Omega = 0.5\,\text{V} = 500\,\text{mV}\n * Sum Check: 2.0\,\text{V} + 1.0\,\text{V} + 1.5\,\text{V} + 0.5\,\text{V} = 5.0\,\text{V}.\n * **Node Voltage Computations (Ground at Node G between V_3 $negative terminal and$ R_4):**\n * V_G = 0\,\text{V}\n * V_D = +1.5\,\text{V} + 7\,\text{V} = 8.5\,\text{V}\n * V_C = +12\,\text{V} + 1.5\,\text{V} + 7\,\text{V} = 20.5\,\text{V}\n * V_B = V_C + V_{R2} = 20.5\,\text{V} + 1.0\,\text{V} = 21.5\,\text{V}\n * V_A = V_B - V_1 = 21.5\,\text{V} - 10\,\text{V} = 11.5\,\text{V}\n * V_H = V_A + V_{R1} = 11.5\,\text{V} + 2.0\,\text{V} = 13.5\,\text{V}\n * V_E = V_G + V_3 = 0\,\text{V} + 7.0\,\text{V} = 7.0\,\text{V}\n * V_F = V_G + V_{R4} = 0\,\text{V} + 0.5\,\text{V} = 0.5\,\text{V}\n * **Node Voltage Difference Verifications:**\n * V_C - V_D = 20.5\,\text{V} - 8.5\,\text{V} = 12\,\text{V} = V_2\n * V_B - V_E = 21.5\,\text{V} - 7.0\,\text{V} = 14.5\,\text{V}\n * V_H - V_A = 13.5\,\text{V} - 11.5\,\text{V} = 2.0\,\text{V} = V_{R1}\n\n# Class Exercises and Practical Worked Examples\n\n* **Exercise 1: Full Series Analysis:**\n * **Given:** V_S = 9\,\text{V} $,$ R_1 = 100\,\Omega $,$ R_2 = 200\,\Omega $,$ R_3 = 300\,\Omega $, Ground at Node$ 0.\n * **Solution Steps:**\n * 1. Total Resistance: R_T = 100\,\Omega + 200\,\Omega + 300\,\Omega = 600\,\Omega\n * 2. Total Current: I_T = \frac{9\,\text{V}}{600\,\Omega} = 15\,\text{mA}\n * 3. Voltage Drops:\n * V_{R1} = 15\,\text{mA} \times 100\,\Omega = 1.5\,\text{V}\n * V_{R2} = 15\,\text{mA} \times 200\,\Omega = 3.0\,\text{V}\n * V_{R3} = 15\,\text{mA} \times 300\,\Omega = 4.5\,\text{V}\n * 4. Node Voltages:\n * V_A = 9.0\,\text{V}\n * V_B = 3.0\,\text{V} + 4.5\,\text{V} = 7.5\,\text{V}\n * V_C = 4.5\,\text{V}\n * V_0 = 0\,\text{V}\n\n* **Exercise 2: Series Circuit Design Problem:**\n * **Design Requirements:**\n * Three resistors (R_1, R_2, R_3) in series.\n * Two voltage sources in series (V_{S1} = 9\,\text{V} $and$ V_{S2} = -4\,\text{V} connected in opposing polarity).\n * Ground connected to the negative terminal of the negative source.\n * Target Total Current: I_T = 500\,\text{mA}.\n * Target Voltage Drops: V_{R1} = 1.0\,\text{V} $,$ V_{R2} = 2.5\,\text{V} $,$ V_{R3} = 1.5\,\text{V}.\n * **Design Steps:**\n * 1. Net Voltage: V_T = 9\,\text{V} + (-4\,\text{V}) = 5\,\text{V}\n * 2. Required Total Resistance: R_T = \frac{V_T}{I_T} = \frac{5\,\text{V}}{500\,\text{mA}} = 10\,\Omega\n * 3. Resistor Design Values (R_x = \frac{V_{Rx}}{I_T}):\n * R_1 = \frac{1.0\,\text{V}}{500\,\text{mA}} = 2\,\Omega\n * R_2 = \frac{2.5\,\text{V}}{500\,\text{mA}} = 5\,\Omega\n * R_3 = \frac{1.5\,\text{V}}{500\,\text{mA}} = 3\,\Omega\n * 4. Verification Check:\n * 2\,\Omega + 5\,\Omega + 3\,\Omega = 10\,\Omega (Valid)\n * 1.0\,\text{V} + 2.5\,\text{V} + 1.5\,\text{V} = 5.0\,\text{V} (Valid)\n\n# Circuit Simulation and Analysis in MultiSIM\n\n* **Software Setup and Component Placement:**\n * Open component window: `Place` \rightarrow `Component`.\n * Resistors: Group `Basic` \rightarrow `Resistor` (place on sheet and edit default values).\n * Sources & Ground: Group `Sources` \rightarrow $`Power Sources`$ \rightarrow `DC Power` and `Ground`.\n\n* **Mandatory Sheet Properties Configuration:**\n * Right-click empty schematic sheet \rightarrow $`Properties`$ \rightarrow $`Net Names`$ \rightarrow Select **Show All**.\n * Showing all net names (node names) is strictly required to identify node designations assigned by SPICE.\n\n* **Professional Simulation Rules:**\n * **Prohibition on Virtual Instruments:** Virtual sidebar meters are strictly for introductory lab instruction and receive zero academic/professional credit for simulation deliverables.\n * **Professional Standard Workflow:**\n * Navigate to `Simulate` \rightarrow `Analyses and Simulation`.\n * Select `DC Operating Point` analysis.\n\n* **Measuring Resistor Voltage Drops in MultiSIM:**\n * DC Operating Point analysis outputs individual node voltages and branch currents, but does not directly output component voltage drops.\n * To measure voltage drop across a resistor, formulate an expression using node voltage differences:\n * Click `Add Expression`.\n * Select terminal nodes: V_{R1} = V_A - V_B $,$ V_{R2} = V_B - V_C $,$ V_{R3} = V_C - V_D$$.
- Executing the simulation produces a full table confirming steady-state node voltages, custom component voltage drop expressions, and branch currents.