2.1 Ohm's Law, Joule's Law & DC Circuit Analysis

2.1 Ohm's Law, Joule's Law & DC Circuit Analysis

Quick Answer: Ohm's law defines the foundational mathematical relationship governing all electrical circuits: electromotive force in volts ($E$ or $V$) equals the product of current in amperes ($I$) and resistance in ohms ($R$), or $V = I \times R$. Combined with Joule's law of electrical power ($P = V \times I$), electricians derive the 12 classic power wheel formulas. In series circuits, current remains uniform while voltage drops are additive ($R_T = R_1 + R_2 + \dots$). In parallel circuits, voltage across branches is identical while branch currents add to total line current ($1/R_T = 1/R_1 + 1/R_2 + \dots$). Mastering these principles is indispensable for solving complex load calculations, understanding conductor voltage drop, and diagnosing dangerous high-resistance faults on the Colorado Journeyman exam.

Every electrical installation governed by the National Electrical Code (NEC) operates according to unchanging physical laws. Whether you are sizing feeders for a commercial warehouse in Denver, troubleshooting a control panel in an industrial facility in Pueblo, or sitting for the Colorado Journeyman Electrician examination administered by PSI, electrical theory forms the bedrock of safe and code-compliant installations. Direct current (DC) theory serves as the primary stepping stone for all subsequent circuit analysis, transformer calculations, and three-phase power mathematics.

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The Fundamental Triad: Voltage, Current, and Resistance

All electrical circuits contain three primary interdependent physical quantities:

1. Electromotive Force / Voltage ($E$ or $V$): The electrical potential difference or pressure that forces free electrons to drift through a conductor. Measured in volts (V), named after Alessandro Volta. One volt drives one ampere of current through one ohm of resistance.
2. Current ($I$): The rate of electron charge flow past a given point in an electrical conductor. Measured in amperes (A), named after André-Marie Ampère. One ampere represents the flow of one coulomb of electrical charge ($6.242 \times 10^{18}$ electrons) per second ($1\text{ A} = 1\text{ C/s}$).
3. Resistance ($R$): The physical opposition a material offers to the movement of free electrons. Measured in ohms ($\Omega$), named after Georg Simon Ohm. Resistance converts electrical potential energy into thermal energy (heat).

The Direct Proportionality of Ohm's Law

Ohm's law states that current in an electrical circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance:

$$I = \frac{E}{R}, \quad E = I \times R, \quad R = \frac{E}{I}$$

• If resistance remains constant and voltage doubles, circuit current doubles.
• If voltage remains constant and resistance doubles, circuit current is cut in half.
• If resistance drops to zero (a dead short circuit), current surges toward infinity, constrained only by source impedance and upstream conductor resistance until overcurrent protective devices (OCPDs) trip.

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The Ohm's Law and Joule's Law Wheel (12 Fundamental Formulas)

James Prescott Joule established that the rate at which electrical energy is transformed into thermal heat—defined as power ($P$) and measured in watts (W)—is directly proportional to the product of electromotive force and electron flow:

$$P = E \times I$$

By substituting the basic Ohm's law equalities ($E = I \times R$ and $I = E / R$) into Joule's core power equation, electricians obtain the complete 12-formula Ohm's Law / Power Wheel:

Electrical Quantities, Units, Symbols, and Measurement Tools

Exam Tip: On licensing exams, pay close attention to whether the question provides voltage, current, resistance, or power. If a problem states an electric baseboard heater is rated for $240\text{ V}$ and $2{,}000\text{ W}$, its internal resistance is fixed ($R = E^2 / P = 240^2 / 2000 = 28.8,\Omega$). If that same heater is connected to a $120\text{ V}$ circuit, its resistance does not change, but its power dissipation drops to one-fourth ($P = E^2 / R = 120^2 / 28.8 = 500\text{ W}$).

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Series Circuit Analysis & Kirchhoff's Voltage Law (KVL)

A series circuit provides only a single continuous conductive loop for current to flow from the power supply, through each connected load, and back to the source.

Rules Governing Series Circuits

1. Current is Constant: Because there is only one pathway for electron drift, the current is identical at every point in the circuit: $I_T = I_1 = I_2 = I_3 = \dots = I_n$
2. Resistance is Additive: Total equivalent circuit resistance equals the direct algebraic sum of all individual load resistances: $R_T = R_1 + R_2 + R_3 + \dots + R_n$
3. Kirchhoff's Voltage Law (KVL): The algebraic sum of all voltage drops across individual loads in a closed series loop equals the total applied source voltage: $E_T = V_1 + V_2 + V_3 + \dots + V_n$
4. Power is Additive: Total power consumed by the series circuit equals the sum of the powers dissipated by each individual component: $P_T = P_1 + P_2 + P_3 + \dots + P_n$

Worked Example: Step-by-Step Series Circuit Calculations

Scenario: A $120\text{ V}$ DC supply feeds a circuit with three resistive heating elements connected in series: $R_1 = 10,\Omega$, $R_2 = 20,\Omega$, and $R_3 = 30,\Omega$.

• Step 1: Calculate Total Resistance ($R_T$): $R_T = R_1 + R_2 + R_3 = 10\,\Omega + 20\,\Omega + 30\,\Omega = 60\,\Omega$
• Step 2: Calculate Total Circuit Current ($I_T$): $I_T = \frac{E_T}{R_T} = \frac{120\text{ V}}{60\,\Omega} = 2.0\text{ A}$
• Step 3: Calculate Individual Voltage Drops ($V_1, V_2, V_3$):
  • $V_1 = I_T \times R_1 = 2.0\text{ A} \times 10,\Omega = 20\text{ V}$
  • $V_2 = I_T \times R_2 = 2.0\text{ A} \times 20,\Omega = 40\text{ V}$
  • $V_3 = I_T \times R_3 = 2.0\text{ A} \times 30,\Omega = 60\text{ V}$
  • Verification (KVL): $20\text{ V} + 40\text{ V} + 60\text{ V} = 120\text{ V}$ (Checks out perfectly).
• Step 4: Calculate Power Dissipation for Each Resistor:
  • $P_1 = I_T^2 \times R_1 = (2.0)^2 \times 10 = 40\text{ W}$
  • $P_2 = I_T^2 \times R_2 = (2.0)^2 \times 20 = 80\text{ W}$
  • $P_3 = I_T^2 \times R_3 = (2.0)^2 \times 30 = 120\text{ W}$
  • $P_T = P_1 + P_2 + P_3 = 40 + 80 + 120 = 240\text{ W}$
  • Verification: $P_T = E_T \times I_T = 120\text{ V} \times 2.0\text{ A} = 240\text{ W}$.

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Parallel Circuit Analysis & Kirchhoff's Current Law (KCL)

A parallel circuit connects two or more components across the same two common electrical nodes, providing multiple independent paths for electron flow. Standard commercial, residential, and industrial building wiring is configured in parallel so that each branch circuit receives full nominal line voltage, and the switching off or failure of one appliance does not interrupt power to the remaining loads.

Rules Governing Parallel Circuits

1. Voltage is Constant: The electromotive force across each individual branch equals the total system supply voltage: $E_T = V_1 = V_2 = V_3 = \dots = V_n$
2. Kirchhoff's Current Law (KCL): The total current entering any circuit junction or node must equal the total current exiting that junction. In a parallel circuit, total current delivered by the source is the algebraic sum of the individual branch currents: $I_T = I_1 + I_2 + I_3 + \dots + I_n$
3. Equivalent Resistance is Reciprocal: The reciprocal of the total equivalent resistance equals the sum of the reciprocals of each individual branch resistance: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$ Critical Axiom: Total equivalent resistance ($R_T$) in a parallel circuit is always strictly less than the lowest value branch resistor.
4. Total Power is Additive: Exactly as in a series circuit, total power consumed is the sum of the power dissipated in all branches: $P_T = P_1 + P_2 + P_3 + \dots + P_n$

The Reciprocal Method and Shortcut Formulas

For any number of parallel resistors, calculate $R_T$ using the general reciprocal equation:

$R_T = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}}$

When dealing with specific configurations on timed exams, use these algebraic shortcuts:

• Product Over Sum (for exactly two parallel resistors): $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$
• Equal Value Resistors in Parallel: If $n$ resistors of identical resistance value $R$ are connected in parallel, the total resistance is simply the individual value divided by the number of resistors: $R_T = \frac{R}{n}$ (Example: Four $100,\Omega$ resistors in parallel yield $R_T = 100 / 4 = 25,\Omega$.)

Worked Example: Step-by-Step Parallel Circuit Calculations

Scenario: A $240\text{ V}$ DC supply feeds three parallel branch loads: $R_1 = 20,\Omega$, $R_2 = 30,\Omega$, and $R_3 = 60,\Omega$.

• Step 1: Calculate Total Equivalent Resistance ($R_T$): $\frac{1}{R_T} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3}{60} + \frac{2}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10}$ $R_T = 10\,\Omega$ (Notice that $10,\Omega$ is less than the smallest branch resistor of $20,\Omega$.)
• Step 2: Calculate Individual Branch Currents:
  • $I_1 = \frac{E}{R_1} = \frac{240\text{ V}}{20,\Omega} = 12\text{ A}$
  • $I_2 = \frac{E}{R_2} = \frac{240\text{ V}}{30,\Omega} = 8\text{ A}$
  • $I_3 = \frac{E}{R_3} = \frac{240\text{ V}}{60,\Omega} = 4\text{ A}$
• Step 3: Verify Kirchhoff's Current Law (KCL): $I_T = I_1 + I_2 + I_3 = 12\text{ A} + 8\text{ A} + 4\text{ A} = 24\text{ A}$ $I_T = \frac{E}{R_T} = \frac{240\text{ V}}{10\,\Omega} = 24\text{ A} \quad \text{(Matches perfectly)}$
• Step 4: Calculate Total Circuit Power ($P_T$): $P_T = E \times I_T = 240\text{ V} \times 24\text{ A} = 5{,}760\text{ W}$
  • By individual branch sum: $P_1 = 240 \times 12 = 2{,}880\text{ W}$; $P_2 = 240 \times 8 = 1{,}920\text{ W}$; $P_3 = 240 \times 4 = 960\text{ W}$.
  • $P_T = 2{,}880 + 1{,}920 + 960 = 5{,}760\text{ W}$.

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Combination (Series-Parallel) Circuits: Network Reduction

Most complex control circuits, transformer windings, and multi-wire branch circuits encounter combinations of series and parallel components. To solve a series-parallel network, apply the method of equivalent network reduction:

1. Identify Isolated Sub-Networks: Locate resistor groups that are purely in series with each other or purely in parallel.
2. Collapse Sub-Networks: Replace those clusters with a single equivalent resistor ($R_{\text{eq}}$).
3. Redraw the Simplified Circuit: Continue collapsing the network stage-by-stage until only a single equivalent resistance ($R_T$) remains across the source.
4. Calculate Total Current ($I_T$): Apply Ohm's law ($I_T = E / R_T$).
5. Expand Backward: Unfold the circuit in reverse order, applying Ohm's law, KVL, and KCL to find specific component voltage drops and branch currents.

Worked Example: Multi-Step Series-Parallel Combination Circuit

Problem: A $120\text{ V}$ DC source is connected to resistor $R_1 = 15,\Omega$ in series with a parallel bank consisting of $R_2 = 40,\Omega$ and $R_3 = 40,\Omega$.

• Stage 1: Solve the Parallel Cluster ($R_{2,3}$): $R_{2,3} = \frac{R_2}{n} = \frac{40\,\Omega}{2} = 20\,\Omega$
• Stage 2: Solve the Remaining Series Circuit ($R_T$): $R_T = R_1 + R_{2,3} = 15\,\Omega + 20\,\Omega = 35\,\Omega$
• Stage 3: Calculate Total Circuit Current ($I_T$): $I_T = \frac{E}{R_T} = \frac{120\text{ V}}{35\,\Omega} \approx 3.4286\text{ A}$
• Stage 4: Calculate Voltage Drops:
  • Voltage across $R_1$: $V_1 = I_T \times R_1 = 3.4286\text{ A} \times 15,\Omega = 51.43\text{ V}$
  • Voltage across parallel bank ($V_{2,3}$): $V_{2,3} = 120\text{ V} - 51.43\text{ V} = 68.57\text{ V}$ (or $3.4286\text{ A} \times 20,\Omega = 68.57\text{ V}$)
• Stage 5: Calculate Individual Currents in the Parallel Branches:
  • $I_2 = \frac{V_{2,3}}{R_2} = \frac{68.57\text{ V}}{40,\Omega} = 1.7143\text{ A}$
  • $I_3 = \frac{V_{2,3}}{R_3} = \frac{68.57\text{ V}}{40,\Omega} = 1.7143\text{ A}$
  • Check: $I_2 + I_3 = 1.7143 + 1.7143 = 3.4286\text{ A} = I_T$.

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Power Dissipation, Resistor Ratings & Thermal Management

Electrical resistance inevitably generates heat. When current passes through a conductor, terminal lug, or resistive load, power is dissipated according to Joule's formula $P = I^2 \times R$.

Resistor Wattage Ratings

Every physical resistor and electrical component has a maximum continuous power dissipation rating (e.g., $1/4\text{ W}$, $1/2\text{ W}$, $1\text{ W}$, $5\text{ W}$, $50\text{ W}$). If a component is forced to dissipate more watts than its physical surface area can radiate into the surrounding air, its temperature climbs until its insulation chars, its resistive element melts, or an open circuit occurs.

• Engineering Best Practice / Safety Factor: Electricians and technicians size power resistors to operate at no more than 50% to 60% of their continuous wattage rating in enclosed panels or high ambient temperatures. If a circuit dissipates $3.2\text{ W}$, a $5\text{ W}$ resistor will run unacceptably hot; a $10\text{ W}$ wire-wound resistor must be selected.
• Conductor I²R Losses: The same formula governs heating in electrical wiring. If loose terminal terminations introduce a contact resistance of only $0.5,\Omega$ in a circuit carrying a $30\text{ A}$ dryer load, the power dissipated at that loose terminal lug is: $P = I^2 \times R = (30\text{ A})^2 \times 0.5\,\Omega = 900 \times 0.5 = 450\text{ W}$ Dissipating $450\text{ W}$ of concentrated thermal energy inside a standard wall box or breaker terminal produces intense heat, melting conductor insulation and creating an imminent structural fire hazard.

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Field Troubleshooting & Voltage Drop Diagnostics

Mastery of Ohm's law enables electricians to diagnose field circuit faults quickly using standard multimeter measurements:

1. Open-Circuit Voltage vs. Loaded Voltage:
   • A broken or loose conductor connection might show nominal voltage ($120\text{ V}$) when measured with a high-impedance digital multimeter because modern meters draw negligible current ($< 1,\mu\text{A}$), meaning zero voltage drop occurs across the high-resistance connection ($V = I \times R = 0 \times R = 0\text{ V}$).
   • As soon as a real load (e.g., a $10\text{ A}$ motor) is switched on, current surges through the high-resistance connection. If the bad splice has a resistance of $4,\Omega$, the voltage drop across the splice becomes $V = 10\text{ A} \times 4,\Omega = 40\text{ V}$. The load voltage immediately collapses from $120\text{ V}$ down to $80\text{ V}$, causing motor stalls or lighting brownouts.
2. Direct Resistance Testing:
   • Always verify power is completely locked out and tagged out (LOTO) before connecting an ohmmeter. Measuring resistance across an energized circuit instantly blows the meter's internal fast-acting fuse and can destroy the instrument.

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Colorado Journeyman Exam Traps & Common Calculation Pitfalls

• Trap 1: The Inverse Power Effect: Remember that $P = E^2 / R$ indicates power is inversely proportional to resistance when voltage is constant. If you halve the circuit resistance across a constant $120\text{ V}$ line, power dissipation doubles. Conversely, under constant current ($P = I^2 \times R$), power is directly proportional to resistance: halving resistance halves power.
• Trap 2: Forgetting to Square Terms: When calculating power with $P = I^2 \times R$ or $P = E^2 / R$, candidates frequently forget to square the current or voltage, resulting in an answer off by orders of magnitude. Always write down the squared value explicitly ($240^2 = 57{,}600$; $120^2 = 14{,}400$).
• Trap 3: Misapplying Parallel Resistance Rules: If your calculated equivalent resistance for a parallel circuit is larger than any of the individual branch resistors, your math is incorrect. Total equivalent parallel resistance must always be smaller than the smallest branch resistor.