Lecture 4 - Resistors (ECE 1004)

Potentiometers and Variable Resistors

• Potentiometers (variable resistors) typically feature three terminals and an adjustable mechanical control:

  • The outer two pins provide the total, fixed resistance across the entire resistive element.

  • The center pin connects to an internal wiper or lever that moves back and forth across the resistive material when the control knob (made of white plastic, brass, or other materials) or slide mechanism is adjusted.

  • Example: Across a potentiometer rated at $10\,\text{k}\Omega$, measuring from one outer pin to the other outer pin yields exactly $10\,\text{k}\Omega$. Measuring from either outer pin to the center pin yields a variable resistance value dependent on the knob position.

Dual Subscript Notation and Current Direction Assumptions

• Dual Subscript Notation for Current:

  • Current variables are expressed using two subscript letters indicating start and end points.

  • $I_{ab}$ represents the electric current flowing from node $a$ to node $b$.

  • $I_{ba}$ represents the electric current flowing in the opposite direction, from node $b$ to node $a$.

  • Mathematically, $I_{ba} = -I_{ab}$.

• Current Direction Assumptions in Circuit Analysis:

  • Circuit analysis permits arbitrarily assigning a variable name and reference direction arrow to current in any branch.

  • Physical intuition helps establish likely directions: DC voltage sources (batteries) drive conventional current out of their positive ($+$) higher-voltage terminal, through the external circuit network, and back into their negative ($-$) lower-voltage terminal.

  • Splitting and Recombining Current:

  • When current $I_1$ leaves a battery positive terminal and reaches a junction, it splits into parallel path currents (e.g., $I_2$ and $I_3$).

  • At the returning bottom junction point, these branch currents sum back together ($I_2 + I_3 = I_1$) before returning to the negative terminal.

  • Calculating Assumed Currents:

  • Performing nodal or loop calculations based on assigned reference arrows will yield positive scalar values if the assumed directions match physical flow.

  • A negative calculated scalar indicates the true physical current flows opposite to the assumed arrow direction.

Topologies: Series and Parallel Connections

• Series Connection Definition:

  • Two circuit elements are connected in series if they share exactly one connection point (node), and that shared point is exclusive (no other current-carrying element or wire is connected to that same node).

  • Analogy: A continuous chain where each link connects exclusively to the next link.

  • Relationship Analogy: An exclusive relationship between two entities where no third party is allowed to join the node.

  • If a third element joins the junction point, exclusivity is broken, and the original two elements are no longer in series because current can split into the third path.

• Parallel Connection Definition:

  • Two circuit elements are connected in parallel if they share two common connection points (nodes).

  • Analogy: Two people holding both of their hands together simultaneously.

• Identifying Series vs. Parallel in Multi-Element Circuits:

  • If elements share one exclusive node, they are in series.

  • If elements share two distinct common nodes, they are in parallel.

  • Complex topologies may contain distinct series pairs and parallel pairs simultaneously (e.g., elements $A$ and $B$ in series, $C$ and $D$ in parallel, $E$ and $F$ in series, $F$ and $G$ in parallel).

• Electrical Properties of Parallel Elements:

  • Elements connected in parallel share the exact same voltage drop across them ($V_a = V_b = V_c$) because their terminals are tied to the exact same pair of electrical nodes.

  • According to Ohm's Law ($V = I \times R$), currents through parallel elements are identical only if their resistance values are identical. If resistances differ, the branch currents will differ despite having identical terminal voltages.

• Electrical Properties of Series Elements:

  • Elements connected in series carry the exact same electric current ($I_a = I_b = I_c$) because charge entering the exclusive node has no alternative path to exit.

  • If the series elements have different resistance values, the voltage drops across each individual element will differ.

Circuit Nodes and Kirchhoff's Current Law (KCL)

• Circuit Nodes:

  • Real-world circuit wiring features arbitrary bends and arrangements, unlike idealized rectangular textbook schematics.

  • A node is defined as an entire continuous connection point or region joining two or more circuit elements.

  • Because ideal wires have zero resistance, shrinking, bending, or expanding a wire segment between element terminals does not alter the electrical node.

  • In heavy power systems, physical nodes consist of large, thick copper bars called busses (or bus bars) designed to distribute high currents.

  • Node Identification Example:

  • Node $A$: Connects the positive terminal of a $10\,\text{V}$ voltage source to one end of a $5\,\Omega$ resistor.

  • Node $B$: Connects the opposite side of the $5\,\Omega$ resistor, the top terminal of a $2\,\Omega$ resistor, a $3\,\Omega$ resistor, and a $2\,\text{A}$ source.

  • Node $C$: Connects the bottom terminals of the $10\,\text{V}$ source, $2\,\Omega$ resistor, $3\,\Omega$ resistor, and $2\,\text{A}$ source.

  • Total nodes in this network = 3.

• Kirchhoff's Current Law (KCL):

  • Established by Gustav Kirchhoff based on the principle of conservation of electric charge.

  • Formal Statement: The algebraic sum of all electric currents at any circuit node is equal to zero:     $\sum I = 0$

  • Practical Formulation: Total current entering a node must equal total current leaving that node:     $\text{Current}_{IN} = \text{Current}_{OUT}$

  • Sign Convention: Assigning positive signs ($+$) to currents entering a node and negative signs ($-$) to currents exiting a node yields:     $I_{A} + I_{C} - I_{B} - I_{D} = 0$

  • Application to Series Nodes:

  • At a two-element node where current $I_A$ enters and $I_B$ leaves, KCL dictates $I_A - I_B = 0 \implies I_A = I_B$, proving series elements share identical current.

  • Numerical Node Example:

  • If currents of $1\,\text{A}$ and $3\,\text{A}$ enter a single node, and an unknown current $I_{\text{out}}$ leaves the node into a series branch, KCL dictates:       $I_{\text{out}} = 1\,\text{A} + 3\,\text{A} = 4\,\text{A}$

Kirchhoff's Voltage Law (KVL) and Loop Analysis

• Kirchhoff's Voltage Law (KVL):

  • Formal Statement: The algebraic sum of all voltage rises and voltage drops around any closed loop in a circuit is equal to zero:     $\sum V = 0$

  • Conservation Principle: Electrical energy supplied by sources (voltage rises) around a closed loop is completely dissipated or absorbed by the remaining elements (voltage drops) in that loop:     $\sum V_{\text{rises}} = \sum V_{\text{drops}}$

• Loop Traversal Rules and Polarity Signs:

  • KVL equations are written by traveling continuously around a closed circuit loop in either a clockwise or counter-clockwise direction.

  • The mathematical sign assigned to each element's voltage in the KVL sum is determined by the polarity sign encountered first upon entering the element:

  • Entering a negative terminal ($-$) and exiting a positive terminal ($+$) represents a voltage rise (written as $-V$       if tracking drops, or $+V$ if tracking rises).

  • Entering a positive terminal ($+$) and exiting a negative terminal ($-$) represents a voltage drop.

  • Direction Invariance: KVL holds true regardless of whether the loop traversal is evaluated clockwise or counter-clockwise.

• Energy Behavior of Elements:

  • Supplying Energy: An element supplies energy when current leaves its positive terminal (voltage rise in the direction of current flow).

  • Dissipating Energy: An element dissipates energy (e.g., a resistor converting electrical energy to heat) when current enters its positive terminal and exits its negative terminal (voltage drop).

  • Opposing Sources (Battery Charging):

  • If two unequal voltage sources oppose each other in a single loop, current is driven by the larger source.

  • Current enters the positive terminal of the smaller voltage source, causing it to absorb energy (charging the battery) rather than supply energy.

• KVL Equation Formulation Examples:

  • Clockwise traversal through a loop containing source $V_a$ (rise) and passive components $V_b$, $V_c$ (drops):     $V_b + V_c - V_a = 0 \implies V_a = V_b + V_c$

  • Traversal with multi-loop shared branches:     $-V_d + V_e - V_{cd} = 0$

• Detailed KVL Numerical Example 1 (Single Source Series Loop):

  • Circuit Parameters: A $10\,\text{V}$ DC source connected in series with five resistors ($R_1 = 10\,\Omega$, $R_2 = 15\,\Omega$, $R_3 = 20\,\Omega$, $R_4 = 30\,\Omega$, $R_5 = 25\,\Omega$) with a measured loop current of $I = 0.1\,\text{A}$.

  • Individual Voltage Drops ($V_i = I \times R_i$):

  • $V_1 = 0.1\,\text{A} \times 10\,\Omega = 1.0\,\text{V}$

  • $V_2 = 0.1\,\text{A} \times 15\,\Omega = 1.5\,\text{V}$

  • $V_3 = 0.1\,\text{A} \times 20\,\Omega = 2.0\,\text{V}$

  • $V_4 = 0.1\,\text{A} \times 30\,\Omega = 3.0\,\text{V}$

  • $V_5 = 0.1\,\text{A} \times 25\,\Omega = 2.5\,\text{V}$

  • KVL Verification:     $V_T = V_1 + V_2 + V_3 + V_4 + V_5$     $10\,\text{V} = 1.0\,\text{V} + 1.5\,\text{V} + 2.0\,\text{V} + 3.0\,\text{V} + 2.5\,\text{V} = 10.0\,\text{V}$

• Detailed KVL Numerical Example 2 (Opposing Sources Loop):

  • Circuit Configuration: A $3\,\text{V}$ voltage source, a $4\,\Omega$ resistor, a $5\,\text{V}$ voltage source connected in opposing orientation, and a $6\,\Omega$ resistor in a single closed loop.

  • Assumed Clockwise Current $I$:

  • Voltage terms encountered clockwise: $-3\,\text{V}$ (rise across $3\,\text{V}$ source), $+4I$ (drop across $4\,\Omega$), $+5\,\text{V}$ (drop across opposing $5\,\text{V}$ source), $+6I$ (drop across $6\,\Omega$).

  • Equation Derivation:     $-3 + 4I + 5 + 6I = 0$     $2 + 10I = 0$     $10I = -2$     $I = -0.2\,\text{A}$

  • Physical Interpretation: The negative result ($-0.2\,\text{A}$) confirms that the actual conventional current flows counter-clockwise at $0.2\,\text{A}$, driven by the dominant $5\,\text{V}$ source.

Equivalent Resistance Calculations

• Resistor Passive Sign Convention:

  • The terminal where conventional electric current enters a resistor is always designated as the positive ($+$) polarity end for voltage drop calculations.

• Series Resistors Equivalent Resistance ($R_{eq}$):

  • For $n$ resistors in series carrying identical current $I$:     $V_{\text{total}} = V_1 + V_2 + \dots + V_n = I R_1 + I R_2 + \dots + I R_n = I(R_1 + R_2 + \dots + R_n)$

  • General Formula:     $R_{eq} = R_1 + R_2 + R_3 + \dots + R_n$

• Parallel Resistors Equivalent Resistance ($R_{eq}$):

  • For $n$ resistors in parallel sharing identical voltage $V$:     $I_{\text{total}} = I_1 + I_2 + \dots + I_n = \frac{V}{R_1} + \frac{V}{R_2} + \dots + \frac{V}{R_n} = V\left(\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\right)$

  • Reciprocal Formula:     $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n} \implies R_{eq} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}}$

• Two-Resistor Parallel Shortcut Formula:

  • For exactly two resistors in parallel:     $R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}$

  • Rule of Thumb: Calculate the product of the two resistances in the numerator and divide by their sum in the denominator.

• Properties and Mathematical Behavior of Parallel Networks:

  • Identical Parallel Resistors: Connecting two identical resistors $R$ in parallel yields an equivalent resistance of exactly half ($0.5R$).

  • Example: Two $1\,\text{k}\Omega$ resistors in parallel yield:       $R_{eq} = \frac{1\,\text{k}\Omega \times 1\,\text{k}\Omega}{1\,\text{k}\Omega + 1\,\text{k}\Omega} = \frac{1}{2}\,\text{k}\Omega = 0.5\,\text{k}\Omega = 500\,\Omega$

  • Smallest Resistance Rule: The equivalent resistance of any parallel network is always strictly smaller than the single smallest individual resistor in that network.

  • Example A ($10\,\text{k}\Omega$in parallel with$1\,\text{k}\Omega):\n      R_{eq} = \frac{10 \times 1}{10 + 1} = \frac{10}{11} \approx 0.91\,\text{k}\Omega\n      (Note that 0.91\,\text{k}\Omega < 1\,\text{k}\Omega).\n - Example B ($100\,\text{k}\Omega in parallel with $1\,\text{k}\Omega$):       $R_{eq} = \frac{100 \times 1}{100 + 1} = \frac{100}{101} \approx 0.99\,\text{k}\Omega$       (Note that $0.99\,\text{k}\Omega < 1\,\text{k}\Omega$).

  • Real-World Utility Application: Utility power distribution lines frequently run two thick copper or aluminum cables side-by-side in parallel. Operating parallel conductors significantly reduces overall line resistance, expanding capacity like adding parallel water pipes.

• Numerical Practice Problem (Parallel Shortcut):

  • Calculate equivalent resistance for $6\,\text{k}\Omega$ in parallel with $3\,\text{k}\Omega$:     $R_{eq} = \frac{6\,\text{k}\Omega \times 3\,\text{k}\Omega}{6\,\text{k}\Omega + 3\,\text{k}\Omega} = \frac{18}{9} = 2\,\text{k}\Omega$

Circuit Reduction Methodology and Solved Examples

• Systematic Circuit Reduction Procedure:

  1. Identify pure series or pure parallel resistor combinations in the schematic.

  2. Calculate the equivalent resistance ($R_{eq}$) for each identified group.

  3. Redraw the simplified circuit schematic, replacing combined elements with single equivalent resistors.

  4. Repeat steps 1–3 sequentially, working progressively from the side of the circuit furthest away from the main power source back toward the source.

  5. Continue until the entire network is reduced to a single voltage source connected across a single equivalent resistance ($R_{\text{total}}$).

  6. Apply Ohm's law ($I_{\text{total}} = \frac{V_{\text{source}}}{R_{\text{total}}}$) to find total system current.

  7. Work backward through redrawn circuit intermediate steps to solve for specific branch voltages or branch currents as required.

• Complete Worked Reduction Example 1:

  • Network Components: A circuit containing parallel pair ($75\,\Omega$and$25\,\Omega) connected in series with parallel pair ($100\,\Omega and $50\,\Omega$).

  • Step 1: Reduce first parallel combination ($75\,\Omega \parallel 25\,\Omega$):     $R_{eq1} = \frac{75 \times 25}{75 + 25} = \frac{1875}{100} = 18.75\,\Omega$

  • Step 2: Reduce second parallel combination ($100\,\Omega \parallel 50\,\Omega$):     $R_{eq2} = \frac{100 \times 50}{100 + 50} = \frac{5000}{150} = \frac{100}{3} \approx 33.33\,\Omega$

  • Step 3: Combine $R_{eq1}$ and $R_{eq2}$ in series:     $R_{\text{total}} = R_{eq1} + R_{eq2} = 18.75\,\Omega + 33.33\,\Omega = 52.08\,\Omega \approx 52.1\,\Omega$

  • Verification: Measuring across input terminals with an ohmmeter yields $52.1\,\Omega$

• Complete Worked Reduction Example 2 (Network with Diagonal Wiring):

  • Circuit Description: Schematic contains a diagonal branch with a $3\,\Omega$ resistor and a $6\,\Omega$ resistor sharing two common node terminals, connected in series with an $8\,\Omega$ resistor, all in parallel with a $10\,\Omega$ resistor.

  • Step 1: Recognize diagonal $3\,\Omega$ and $6\,\Omega$ resistors share two common nodes, meaning they are in parallel:     $R_{eq1} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2\,\Omega$

  • Step 2: Combine $R_{eq1}$ ($2\,\Omega$) in series with the adjacent $8\,\Omega$ resistor:     $R_{\text{branch}} = 2\,\Omega + 8\,\Omega = 10\,\Omega$

  • Step 3: Combine $R_{\text{branch}}$ ($10\,\Omega$) in parallel with the remaining $10\,\Omega$ resistor:     $R_{\text{total}} = \frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5\,\Omega$