2.1 Ohm's Law, Watt's Law & DC Series/Parallel Circuits

Ohm's Law, Watt's Law & DC Series/Parallel Circuits

Mastery of direct current (DC) theory and fundamental electrical mathematics is the foundation upon which all trade calculations, circuit diagnostics, and National Electrical Code (NEC) load sizing requirements rest. Whether determining the voltage drop across branch circuit conductors, calculating heater element wattage, or analyzing control circuits in industrial panels, journeyman electricians must be able to fluidly apply Ohm's Law and Watt's Law across series, parallel, and combination circuit topologies.

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1. Fundamentals of Ohm's Law

In 1827, German physicist Georg Simon Ohm formulated the relationship governing electrical conduction: the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

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|                                 OHM'S LAW FORMULA MATRIX                                |
|                                                                                         |
|               [ VOLTAGE (E or V) ]  =  [ CURRENT (I) ]  ×  [ RESISTANCE (R) ]          |
|                                                                                         |
|          To Find Voltage (E):          To Find Current (I):      To Find Resistance (R):|
|               E = I × R                     I = E / R                 R = E / I         |
|            (Volts = A × Ω)               (Amps = V / Ω)            (Ohms = V / A)       |
+-----------------------------------------------------------------------------------------+

Primary Electrical Quantities and Physical Units:

1. Electromotive Force / Voltage ($E$ or $V$): The electrical pressure or potential difference that drives charge carriers through a closed circuit. Measured in Volts (V). One volt equals one Joule of work performed per Coulomb of charge moved ($1\text{ V} = 1\text{ J/C}$).
2. Current ($I$): The rate of electrical charge flow past a given point in a conductor. Measured in Amperes (A). One ampere equals one Coulomb of charge moving per second ($1\text{ A} = 1\text{ C/s} = 6.242 \times 10^{18}\text{ electrons/second}$).
3. Resistance ($R$): The physical opposition to the flow of electric current, converting electrical energy into heat. Measured in Ohms ($\Omega$). One ohm allows one ampere of current to flow when an electromotive force of one volt is applied ($1\ \Omega = 1\text{ V/A}$).

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2. Watt's Law and Electrical Power Relationships

Electrical power represents the rate at which electrical energy is converted into another form of energy (such as heat, light, or mechanical work). James Watt's mathematical relationships, combined with Ohm's Law, form the standard twelve-formula electrical calculation wheel.

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|                                THE 12 OHM'S / WATT'S FORMULAS                           |
|                                                                                         |
|     [ POWER (P) in Watts ]          [ VOLTAGE (E) in Volts ]    [ CURRENT (I) in Amps ] |
|       • P = E × I                     • E = I × R                 • I = E / R           |
|       • P = I² × R                    • E = P / I                 • I = P / E           |
|       • P = E² / R                    • E = √(P × R)              • I = √(P / R)        |
|                                                                                         |
|                             [ RESISTANCE (R) in Ohms ]                                  |
|                               • R = E / I                                               |
|                               • R = E² / P                                              |
|                               • R = P / I²                                              |
+-----------------------------------------------------------------------------------------+

Power Relationships and $I^2R$ Heat Losses:

• $P = E \times I$: Direct product of applied potential and total current.
• $P = I^2 \times R$ (Joule's Law / Copper Loss): Demonstrates that power dissipation through resistance increases with the square of the current. If circuit current doubles, heat loss in conductors and connections quadruples ($2^2 = 4$).
• $P = \frac{E^2}{R}$: Demonstrates the impact of voltage fluctuations on resistive loads. If voltage supplied to a fixed resistance heating element drops by $10%$ (to $90%$ nominal), heat output drops to $0.90^2 = 81%$ of rated output.

Trade Application Example: Industrial Strip Heater Sizing

An industrial drying oven operates two $240\text{V}$ resistive heating elements rated at $2,400\text{W}$ each.

• Resistance of each element: $R = \frac{E^2}{P} = \frac{240^2}{2,400} = \frac{57,600}{2,400} = 24\ \Omega$.
• Current drawn by each element at $240\text{V}$: $I = \frac{P}{E} = \frac{2,400}{240} = 10\text{ A}$.
• If voltage drops to $208\text{V}$, the power output of each element becomes: $P_{\text{new}} = \frac{E^2}{R} = \frac{208^2}{24} = \frac{43,264}{24} = 1,802.67\text{ W}$ (a $24.9%$ reduction in heating capacity).

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3. Series DC Circuits & Kirchhoff's Voltage Law (KVL)

A series circuit provides only one continuous path for current to flow. All components are connected end-to-end.

+-----------------------------------------------------------------------------------------+
|                                 SERIES DC CIRCUIT RULES                                 |
|                                                                                         |
|   1. Current is CONSTANT:           I_Total = I₁ = I₂ = I₃ = ... = I_n                  |
|   2. Resistance is ADDITIVE:        R_Total = R₁ + R₂ + R₃ + ... + R_n                  |
|   3. Voltage Drops are ADDITIVE:    E_Total = V₁ + V₂ + V₃ + ... + V_n                  |
|   4. Total Power is ADDITIVE:       P_Total = P₁ + P₂ + P₃ + ... + P_n                  |
+-----------------------------------------------------------------------------------------+

Kirchhoff's Voltage Law (KVL):

Kirchhoff's Voltage Law states that the algebraic sum of all voltages around any closed loop in an electrical network is equal to zero: $\sum V = 0$. In practical terms, the source voltage applied to a series circuit equals the sum of the individual voltage drops across each series component ($E_{\text{source}} = V_1 + V_2 + \dots + V_n$).

Voltage Divider Rule:

The voltage drop across any individual resistor ($R_x$) in a series circuit is directly proportional to its share of the total circuit resistance: $V_x = E_{\text{Total}} \times \left(\frac{R_x}{R_{\text{Total}}}\right)$

Step-by-Step Series Calculation:

A $120\text{V}$ DC control circuit connects three field resistors in series: $R_1 = 15\ \Omega$, $R_2 = 25\ \Omega$, and $R_3 = 20\ \Omega$.

1. Total Resistance ($R_T$): $R_T = 15 + 25 + 20 = 60\ \Omega$.
2. Total Current ($I_T$): $I_T = \frac{E_T}{R_T} = \frac{120\text{V}}{60\ \Omega} = 2.0\text{ A}$.
3. Individual Voltage Drops:
   • $V_1 = I_T \times R_1 = 2.0\text{ A} \times 15\ \Omega = 30\text{V}$
   • $V_2 = I_T \times R_2 = 2.0\text{ A} \times 25\ \Omega = 50\text{V}$
   • $V_3 = I_T \times R_3 = 2.0\text{ A} \times 20\ \Omega = 40\text{V}$
   • Verification (KVL): $30\text{V} + 50\text{V} + 40\text{V} = 120\text{V}$.
4. Power Dissipation:
   • $P_1 = 30\text{V} \times 2\text{ A} = 60\text{W}$; $P_2 = 50\text{V} \times 2\text{ A} = 100\text{W}$; $P_3 = 40\text{V} \times 2\text{ A} = 80\text{W}$.
   • $P_T = 60 + 100 + 80 = 240\text{W}$ ($P_T = 120\text{V} \times 2\text{ A} = 240\text{W}$).

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4. Parallel DC Circuits & Kirchhoff's Current Law (KCL)

A parallel circuit connects components across common electrical nodes, creating multiple independent branches for current flow.

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|                                PARALLEL DC CIRCUIT RULES                                |
|                                                                                         |
|   1. Voltage is CONSTANT:           E_Total = V₁ = V₂ = V₃ = ... = V_n                  |
|   2. Current is ADDITIVE:           I_Total = I₁ + I₂ + I₃ + ... + I_n                  |
|   3. Resistance is RECIPROCAL:      1/R_Total = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/R_n        |
|   4. Total Power is ADDITIVE:       P_Total = P₁ + P₂ + P₃ + ... + P_n                  |
+-----------------------------------------------------------------------------------------+

Kirchhoff's Current Law (KCL):

Kirchhoff's Current Law states that the algebraic sum of currents entering and leaving any node (junction) is zero: $\sum I_{\text{enter}} = \sum I_{\text{leave}}$. In a parallel circuit, the total supply current entering the main junction equals the sum of the currents flowing through each parallel branch.

Methods for Calculating Equivalent Parallel Resistance ($R_T$):

1. General Reciprocal Formula (Any number of branches): $R_T = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}}$
2. Product-Over-Sum Formula (Strictly for Two Parallel Branches): $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$
3. Equal Resistors Formula ($N$ identical resistors of value $R$): $R_T = \frac{R}{N}$

Critical Rule: The total equivalent resistance ($R_T$) of any parallel circuit is always less than the resistance of the smallest branch resistor.

Current Divider Rule (Two Parallel Branches):

$I_1 = I_T \times \left(\frac{R_2}{R_1 + R_2}\right), \quad I_2 = I_T \times \left(\frac{R_1}{R_1 + R_2}\right)$

Step-by-Step Parallel Calculation:

A $24\text{V}$ DC industrial control power supply feeds three parallel branches: $R_1 = 12\ \Omega$, $R_2 = 20\ \Omega$, and $R_3 = 30\ \Omega$.

1. Equivalent Total Resistance ($R_T$): $\frac{1}{R_T} = \frac{1}{12} + \frac{1}{20} + \frac{1}{30} = \frac{5}{60} + \frac{3}{60} + \frac{2}{60} = \frac{10}{60} = \frac{1}{6} \implies R_T = 6\ \Omega$
2. Branch Currents:
   • $I_1 = \frac{24\text{V}}{12\ \Omega} = 2.0\text{ A}$
   • $I_2 = \frac{24\text{V}}{20\ \Omega} = 1.2\text{ A}$
   • $I_3 = \frac{24\text{V}}{30\ \Omega} = 0.8\text{ A}$
3. Total Current Verification (KCL):
   • $I_T = I_1 + I_2 + I_3 = 2.0 + 1.2 + 0.8 = 4.0\text{ A}$.
   • Using Ohm's Law: $I_T = \frac{E_T}{R_T} = \frac{24\text{V}}{6\ \Omega} = 4.0\text{ A}$.

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5. Combination (Series-Parallel) DC Circuits

Real-world electrical installations—such as conductor runs with line resistance feeding parallel equipment banks—are combination circuits containing both series and parallel elements.

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|                    STEP-BY-STEP CIRCUIT REDUCTION METHODOLOGY                           |
|                                                                                         |
|   [STEP 1: Identify Sub-Networks]                                                       |
|   Locate purely series or purely parallel resistor groups furthest from source.         |
|                                     |                                                   |
|                                     v                                                   |
|   [STEP 2: Calculate Equivalent Resistance of Clusters]                                 |
|   Solve parallel groups with reciprocal / product-over-sum: R_parallel = (R_a × R_b) / (R_a + R_b)|
|                                     |                                                   |
|                                     v                                                   |
|   [STEP 3: Redraw the Simplified Equivalent Circuit]                                    |
|   Replace each cluster with a single equivalent resistor symbol.                        |
|                                     |                                                   |
|                                     v                                                   |
|   [STEP 4: Calculate Total Circuit Resistance and Source Current]                       |
|   Sum remaining series elements: R_Total = R_series + R_parallel. Then I_Total = E / R_Total|
|                                     |                                                   |
|                                     v                                                   |
|   [STEP 5: Step Forward to Solve Individual Drops and Branch Currents]                  |
|   Use V = I × R for series drops, then apply KVL and KCL across internal branches.     |
+-----------------------------------------------------------------------------------------+

Comprehensive Worked Reduction Problem:

Consider a $120\text{V}$ DC circuit where resistor $R_1 = 10\ \Omega$ is connected in series with a parallel branch consisting of $R_2 = 30\ \Omega$ and $R_3 = 60\ \Omega$, followed by another series resistor $R_4 = 10\ \Omega$.

1. Simplify Parallel Cluster ($R_{2,3}$): $R_{2,3} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega$
2. Calculate Total Circuit Resistance ($R_T$): $R_T = R_1 + R_{2,3} + R_4 = 10\ \Omega + 20\ \Omega + 10\ \Omega = 40\ \Omega$
3. Calculate Total Circuit Current ($I_T$): $I_T = \frac{E_T}{R_T} = \frac{120\text{V}}{40\ \Omega} = 3.0\text{ A}$
4. Calculate Series Voltage Drops ($V_{R1}$ and $V_{R4}$):
   • $V_{R1} = I_T \times R_1 = 3.0\text{ A} \times 10\ \Omega = 30\text{V}$
   • $V_{R4} = I_T \times R_4 = 3.0\text{ A} \times 10\ \Omega = 30\text{V}$
5. Calculate Parallel Bank Voltage Drop ($V_{\text{parallel}}$):
   • Applying KVL: $V_{\text{parallel}} = E_T - V_{R1} - V_{R4} = 120\text{V} - 30\text{V} - 30\text{V} = 60\text{V}$.
   • Or via Ohm's Law: $V_{\text{parallel}} = I_T \times R_{2,3} = 3.0\text{ A} \times 20\ \Omega = 60\text{V}$.
6. Calculate Individual Branch Currents ($I_{R2}$ and $I_{R3}$):
   • $I_{R2} = \frac{V_{\text{parallel}}}{R_2} = \frac{60\text{V}}{30\ \Omega} = 2.0\text{ A}$
   • $I_{R3} = \frac{V_{\text{parallel}}}{R_3} = \frac{60\text{V}}{60\ \Omega} = 1.0\text{ A}$
   • KCL Check: $I_T = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A}$.
7. Total Power Summary:
   • $P_{R1} = 30\text{V} \times 3.0\text{ A} = 90\text{W}$
   • $P_{R2} = 60\text{V} \times 2.0\text{ A} = 120\text{W}$
   • $P_{R3} = 60\text{V} \times 1.0\text{ A} = 60\text{W}$
   • $P_{R4} = 30\text{V} \times 3.0\text{ A} = 90\text{W}$
   • $P_{\text{Total}} = 90 + 120 + 60 + 90 = 360\text{W}$ ($P_T = 120\text{V} \times 3.0\text{ A} = 360\text{W}$).