(09/01) Lecture Parallel Resistor Proofs, Current Division, and Circuit Analysis by Reduction

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Proof: Equivalent Resistance of Resistors in Parallel

Objective and Setup

• Equivalence Question: The proof evaluates whether a circuit with $n$ resistors connected in parallel is equivalent to a circuit with a single equivalent resistor ($R_{eq}$). Equivalence diagrams indicate this comparison with a question mark over the equivalence symbol ($\stackrel{?}{=}$).

• Parallel Connection Criteria: An inspection of $n$ resistors shows that all $n$ resistors share identical top and bottom node connections. Therefore, by definition, all $n$ resistors are in parallel with each other.

• Physical Implications of Parallel Resistors:

  • The voltage ($V$) across every parallel resistor is identical.

  • The total entering current ($I$) divides proportionally across each branch to form branch currents $I_1, I_2, \dots, I_n$.

Theoretical Foundation

• Conservation Law: While series resistor proofs exploit the Law of Conservation of Energy, parallel resistor proofs exploit the Law of Conservation of Matter (matter in the universe is neither created nor destroyed).

• System Constraints:

  1. Voltage $V$ must remain identical in both the original and equivalent circuit diagrams.

  2. Total current $I$ must remain identical in both the original and equivalent circuit diagrams.

Mathematical Derivation

• Current Summation: By the Law of Conservation of Matter, the total current $I$ entering the parallel network equals the sum of all individual branch currents:

$I = I_1 + I_2 + \dots + I_n$

• Ohm's Law Recasting: Applying Ohm's Law ($V = I \times R$) solved for current gives:

$I = \frac{V}{R}$

• Branch Current Substitutions: Substitute Ohm's Law into each branch current term:

$I_1 = \frac{V}{R_1}$

$I_2 = \frac{V}{R_2}$

$I_n = \frac{V}{R_n}$

• Summation Substitution:

$I = \frac{V}{R_1} + \frac{V}{R_2} + \dots + \frac{V}{R_n}$

• Factoring Voltage: Factoring out the common voltage term $V$ yields:

$I = V \times \left(\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\right)$

• Equivalent Circuit Equation: For the equivalent circuit consisting of a single resistor $R_{eq}$:

$I = \frac{V}{R_{eq}} = V \times \left(\frac{1}{R_{eq}}\right)$

• Division by $V$: Dividing both sides of both equations by $V$ gives:

$\frac{I}{V} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$

$\frac{I}{V} = \frac{1}{R_{eq}}$

• Transitive Equivalence: By the transitive property of equality:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$

• Solving for $R_{eq}$: Taking the reciprocal (raising both sides to the $-1$ power):

$R_{eq} = \left(\sum_{i=1}^{n} \frac{1}{R_i}\right)^{-1}$

$QED$

Current Division Rule

• Origin: Derived directly from the parallel resistor proof, analogous to how the series resistor proof produces the voltage division rule.

• Purpose: Allows direct calculation of an individual branch current ($I_a$ or $I_b$) from total current ($I$) in a single computation.

• Strict Operational Constraint: Current division only works for two resistors in parallel. It cannot be applied directly to networks with three or more parallel resistors without preliminary reduction.

Fundamentals of Resistor Combinations

Series Identification and Rules

• Definition: Two circuit elements are connected in series if and only if exactly two elements are connected to a single shared node, with no other branching paths or elements attached to that node.

• Combination Rule: Resistances in series add directly:

$R_{eq} = R_1 + R_2$

• Example Calculation: Two resistors ($2\,\Omega$ and $3\,\Omega$) connected end-to-end share a single middle node connected exclusively to them.

$R_{eq} = 2\,\Omega + 3\,\Omega = 5\,\Omega$

• Mandatory Unit Requirement: Numerical answers without physical units (e.g., writing "$5$" instead of "$5\,\Omega$") are strictly incorrect.

Parallel Identification and Rules

• Definition: Two or more circuit elements are in parallel if they share identical node connections on both ends.

• Combination Rule: Combined reciprocal sum inverted:

$R_{eq} = \left(\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\right)^{-1}$

• Example Calculation: Three $10\,\Omega$ resistors connected between the exact same top and bottom nodes:

$\frac{1}{R_{eq}} = \frac{1}{10\,\Omega} + \frac{1}{10\,\Omega} + \frac{1}{10\,\Omega} = \frac{3}{10\,\Omega}$

$R_{eq} = \left(\frac{3}{10\,\Omega}\right)^{-1} = \frac{10}{3}\,\Omega \approx 3.33\,\Omega$

Circuit Analysis by Reduction Method

Engineering Philosophy and Methodology

• Cookbook Process Approach: Methodical, standardized procedural steps ensure reproducible, error-free analysis.

• Efficiency Standard: Engineering efficiency demands performing circuit reduction only as far as necessary to extract required values, minimizing unnecessary manual and computational labor.

Step-by-Step Circuit Reduction Walkthrough

Circuit Specifications

• Independent DC Voltage Source: $100\,\text{V}$

• Resistor $R_1 = 10\,\Omega$

• Resistor $R_2 = 50\,\Omega$

• Resistor $R_3 = 20\,\Omega$

Step 1: Label All Nodes

• Node Definition: A node is any point of connection between two or more circuit elements.

• Circuit Audit: Inspection reveals exactly 3 nodes in this circuit:

  • Node 1: Connection between the positive terminal of the $100\,\text{V}$ source and the input of the $10\,\Omega$ resistor.

  • Node 2: Junction connecting the output of the $10\,\Omega$ resistor, the input of the $50\,\Omega$ resistor, and the input of the $20\,\Omega$ resistor.

  • Node 3: Reference junction connecting the bottom terminals of the $50\,\Omega$ resistor, $20\,\Omega$ resistor, and the negative terminal of the $100\,\text{V}$ source.

Step 2: Reduce the Circuit Systematically

• Pairwise Logical Testing:

  • Test Pair ($10\,\Omega$ and $50\,\Omega$):

    • Are they in series? No. Node 2 connects three elements ($10\,\Omega$, $50\,\Omega$, $20\,\Omega$), violating the series definition.

    • Are they in parallel? No. They share Node 2, but do not share a second node connection.

  • Test Pair ($10\,\Omega$ and $20\,\Omega$):

    • Are they in series? No. Node 2 connects more than two elements.

    • Are they in parallel? No. They do not share identical node connections on both ends.

  • Test Pair ($50\,\Omega$ and $20\,\Omega$):

    • Are they in series? No. Node 2 connects three elements.

    • Are they in parallel? Yes. Both resistors are connected between Node 2 and Node 3.

• First Reduction (Parallel Combination):

  • Combine $50\,\Omega$ and $20\,\Omega$ in parallel:

$R_{eq1} = \left(\frac{1}{50\,\Omega} + \frac{1}{20\,\Omega}\right)^{-1} = \left(\frac{2 + 5}{100\,\Omega}\right)^{-1} = \frac{100}{7}\,\Omega \approx 14.2857\,\Omega$

*   *Node Behavior in Parallel Reduction:* Combining resistors in parallel removes individual branches, but **nodes do not disappear** (Node 2 and Node 3 both remain intact).

• Redrawn Diagram 1 (Blue Version Verification):

  • $100\,\text{V}$ source between Node 1 and Node 3.

  • $10\,\Omega$ resistor between Node 1 and Node 2.

  • $R_{eq1} = 14.2857\,\Omega$ between Node 2 and Node 3.

• Second Reduction (Series Combination):

  • Inspect $10\,\Omega$ and $R_{eq1} = 14.2857\,\Omega$ at Node 2. Only two elements connect at Node 2; therefore, they are in series.

$R_{total} = 10\,\Omega + 14.2857\,\Omega = 24.2857\,\Omega$

*   *Node Behavior in Series Reduction:* Combining elements in series causes the intermediate node (Node 2) to disappear from the reduced diagram.

• Redrawn Diagram 2 (Green Version Verification):

  • $100\,\text{V}$ source connected across a single equivalent resistance $R_{total} = 24.2857\,\Omega$ between Node 1 and Node 3.

Step 3: Solve for Unknown Variables (Working Backwards)

• Green Diagram Analysis (Source Current):

  • Calculate total current $I$ supplied by the $100\,\text{V}$ source using Ohm's Law:

$I = \frac{V}{R_{total}} = \frac{100\,\text{V}}{24.2857\,\Omega} \approx 4.1176\,\text{A} \approx 4.12\,\text{A}$

*   *Passive Sign Convention Check:* Current $I = 4.12\,\text{A}$ flows out of the positive terminal of the voltage source.

• Blue Diagram Analysis (Intermediate Voltages):

  • Total current $I = 4.12\,\text{A}$ flows through both series elements $10\,\Omega$ and $R_{eq1}$.

  • Voltage across $10\,\Omega$ resistor ($V_{12}$):

$V_{12} = I \times 10\,\Omega = 4.12\,\text{A} \times 10\,\Omega = 41.2\,\text{V}$

*   Voltage across $R_{eq1}$ ($V_{23}$):

$V_{23} = I \times R_{eq1} = 4.12\,\text{A} \times 14.2857\,\Omega = 58.85\,\text{V} \approx 58.9\,\text{V}$

• Original Diagram Analysis (Branch Currents):

  • Current $I = 4.12\,\text{A}$ arrives at Node 2 and splits into branch currents $I_1$ (through $50\,\Omega$) and $I_2$ (through $20\,\Omega$).

  • Branch Current $I_1$ (through $50\,\Omega$ resistor):

$I_1 = \frac{V_{23}}{50\,\Omega} = \frac{58.85\,\text{V}}{50\,\Omega} \approx 1.177\,\text{A} \approx 1.18\,\text{A}$

*   Branch Current $I_2$ (through $20\,\Omega$ resistor):

$I_2 = \frac{V_{23}}{20\,\Omega} = \frac{58.85\,\text{V}}{20\,\Omega} \approx 2.9425\,\text{A} \approx 2.94\,\text{A}$

Step 4: Power Verification and Balance

• Resistor Power Formulas:

$P = V \times I$

$P = \frac{V^2}{R}$

$P = I^2 \times R$

• Power Dissipated by Resistors (Positive by Passive Sign Convention):

  • $10\,\Omega$ Resistor:

$P_{10} = I^2 \times R = (4.12\,\text{A})^2 \times 10\,\Omega = 169.74\,\text{W} \approx 169.7\,\text{W}$

*   $50\,\Omega$ Resistor:

$P_{50} = I_1^2 \times R = (1.177\,\text{A})^2 \times 50\,\Omega = 69.27\,\text{W} \approx 69.4\,\text{W}$

*   $20\,\Omega$ Resistor:

$P_{20} = I_2^2 \times R = (2.9425\,\text{A})^2 \times 20\,\Omega = 173.17\,\text{W} \approx 172.9\,\text{W}$

• Power Associated with Independent Source:

  • Current leaves the positive terminal; therefore, power is supplied (negative sign under Passive Sign Convention):

$P_s = -V \times I = -100\,\text{V} \times 4.12\,\text{A} = -412\,\text{W}$

• Law of Conservation of Power Audit:

  • Sum of all powers in a closed circuit must equal zero:

$\sum P = P_{10} + P_{50} + P_{20} + P_s$

$\sum P = 169.7\,\text{W} + 69.4\,\text{W} + 172.9\,\text{W} - 412\,\text{W} = -0.1\,\text{W}$

Precision Standards and Error Management

• Source of Discrepancy: The non-zero power summation ($-0.1\,\text{W}$ to $-0.3\,\text{W}$ or $-300\,\text{mW}$) is an artifact of intermediate rounding errors (such as truncating $\frac{100}{7}\,\Omega \approx 14.2857\,\Omega$ to $14.3\,\Omega$).

• Error Significance Check: The error magnitude ($300\,\text{mW}$) is 3 to 4 orders of magnitude smaller than the circuit power values (tens to hundreds of watts). Thus, the error magnitude is acceptable and does not invalidate the solution.

• Mandatory Class Precision Rules:

  1. Zero Intermediate Rounding: Do not round numbers during intermediate calculations; preserve full precision until the final numerical answer is reached.

  2. Decimal Precision: Carry intermediate calculations to 3 or 4 decimal places (or keep full fraction representations).

  3. Final Answer Format: Convert all final answers to decimal form with proper units.

  4. Significant Figures: Significant figure rules are not enforced in this engineering analysis context.

  5. Node Labeling Synchronization: In-class examples require strict adherence to board node labels to maintain a shared technical language.

Questions & Discussion

• Question: How do you determine where to place node labels in a circuit diagram?

• Response: Nodes are defined by the physical connection of two or more elements. Junction points or dots in schematics represent these connections. Moving along an ideal wire without crossing an element does not change the node; all connected wires up to element terminals belong to the exact same node.

• Question: Can a node exist in the middle of a single line?

• Response: No. Placing a node in the middle of an unbroken line segment without a branching connection or element terminal is redundant; it remains the same single continuous node.

• Question: How was the fraction inversion calculated for the parallel combination of $50\,\Omega$ and $20\,\Omega$?

• Response: Step-by-step fraction manipulation:

$\frac{1}{R_{eq1}} = \frac{1}{50} + \frac{1}{20} = \frac{2}{100} + \frac{5}{100} = \frac{7}{100}$

$R_{eq1} = \left(\frac{7}{100}\right)^{-1} = \frac{100}{7}\,\Omega \approx 14.2857\,\Omega$