2.1 Ohm's Law, Joule's Law & DC Circuit Calculations
Key Takeaways
- Ohm's Law defines the proportional relationship between electromotive force (E in volts), current (I in amperes), and resistance (R in ohms): E = I * R, I = E / R, and R = E / I.
- Joule's Law defines electrical power (P in watts) as the rate of energy dissipation: P = E * I, P = I² * R, and P = E² / R. Derating a resistive heating load from 240V to 208V reduces power output by 24.9% (operating at 75.1% rated wattage).
- In a DC series circuit, current is identical through all components (I_total = I1 = I2 = ...), total resistance is the algebraic sum of all resistances (R_total = R1 + R2 + ...), and the sum of component voltage drops equals source voltage (Kirchhoff's Voltage Law).
- In a DC parallel circuit, voltage is identical across all parallel branches (E_total = E1 = E2 = ...), total current is the sum of branch currents (Kirchhoff's Current Law), and total equivalent resistance is always lower than the smallest branch resistance.
- Solving complex series-parallel combination circuits requires a systematic reduction method: simplify parallel banks to equivalent series resistances (using product-over-sum or reciprocal formulas), calculate total circuit current, and expand back out to solve branch-specific currents and voltage drops.
2.1 Ohm's Law, Joule's Law & DC Circuit Calculations
Electrical theory forms the foundation of all National Electrical Code (NEC) rules, overcurrent protection sizing, equipment ratings, and branch-circuit calculations. On the Idaho Journeyman Electrician examination, candidates are expected to manipulate electrical equations quickly, solve multi-resistor network problems, and evaluate electrical power transformations without relying on guesswork.
1. Fundamental Electrical Quantities & Units
All electrical circuits govern the interaction of four fundamental physical quantities:
+-----------------------------------------------------------------------------+
| CORE ELECTRICAL QUANTITIES & SYMBOLS |
| |
| QUANTITY SYMBOL (FORMULA) UNIT (SYMBOL) PHYSICAL ANALOGY |
| ----------------------------------------------------------------------- |
| Electromotive E (or V) Volt (V) Water Pressure |
| Force (Voltage) (PSI) |
| |
| Current I Ampere (A) Water Flow Rate |
| (Intensity) (Gallons / Minute) |
| |
| Resistance R Ohm (Ω) Pipe Friction / |
| Restriction |
| |
| Electrical P Watt (W) Mechanical Work |
| Power Rate (Horsepower) |
+-----------------------------------------------------------------------------+
Detailed Parameter Definitions
- Electromotive Force / Voltage (E or V): The potential difference or electrical pressure required to force one ampere of current through a resistance of one ohm. In electrical calculations, E stands for Electromotive Force (often denoted as V for Voltage).
- Current (I): The rate of electron flow past a given point in a conductor. One ampere represents the movement of one coulomb of electrical charge (6.242 x 10^18 electrons) per second.
- Resistance (R): The opposition offered by a material to the flow of electrical current, measured in Ohms (Ω). Conductor resistance depends directly on material resistivity (ρ), length (L), cross-sectional area (A), and operating temperature.
- Power (P): The rate at which electrical energy is converted into another form of energy (such as heat, light, or mechanical motion), measured in Watts (W) or Joules per second. In mechanical power conversions, 1 Horsepower (HP) = 746 Watts.
2. Ohm's Law & The Ohm's Circle
Formulated by Georg Simon Ohm in 1827, Ohm's Law states that 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.
+-------------------+
| E |
| (Voltage) |
+---------+---------+
| I | R |
| (Amps) | (Ohms) |
+---------+---------+
E = I x R | I = E / R | R = E / I
Step-by-Step Ohm's Law Applications
Example 1.1: Determining Conductor Operating Current
A 240 V single-phase electric water heater element has an internal cold resistance of 12 Ω. What is the continuous current draw when energized at rated voltage?
Given: E = 240 V, R = 12 Ω I = E / R = (240 V) / (12 Ω) = 20.0 A
Example 1.2: Calculating Unknown Component Resistance
A 120 V industrial solenoid valve draws 2.5 A when energized. What is the DC resistance of the solenoid coil?
Given: E = 120 V, I = 2.5 A R = E / I = (120 V) / (2.5 A) = 48.0 Ω
3. Joule's Law & The 12-Formula Power Wheel
Joule's Law (The Power Law) establishes that the electrical power dissipated in a circuit is directly proportional to the product of voltage and current. By substituting Ohm's Law expressions (E = I x R and I = E / R) into the base power equation P = E x I, electricians derive the complete 12-formula wheel.
+-----------------------------------------------------------------------------+
| THE 12 ELECTRICAL FORMULA MATRIX |
| |
| TO FIND: FORMULA 1 FORMULA 2 FORMULA 3 |
| ----------------------------------------------------------------------- |
| Voltage (E) E = I * R E = P / I E = √(P * R) |
| |
| Current (I) I = E / R I = P / E I = √(P / R) |
| |
| Resistance (R) R = E / I R = E² / P R = P / I² |
| |
| Power (P) P = E * I P = I² * R P = E² / R |
+-----------------------------------------------------------------------------+
[!IMPORTANT] The Voltage Derating Trap on Resistive Loads (High-Yield Exam Concept): When a purely resistive heating appliance rated for 240 V is connected to a 208 V supply (common in commercial facilities fed by 208Y/120V systems), the resistance of the heating element remains constant, but the power output drops with the square of the voltage:
P_actual = P_rated x (V_applied / V_rated)^2 = P_rated x (208 / 240)^2 = P_rated x (0.8667)^2 = P_rated x 0.7511
A 4,800 W, 240 V heater operated on 208 V will only produce 4,800 x 0.7511 = 3,605.3 W (a 24.9% loss of heating capacity!).
Worked Example: Heat Dissipation in a Bad Connection
An improper, loose wire-nut splice has developed a contact resistance of 0.75 Ω. If a continuous commercial branch-circuit load of 16 A passes through this high-resistance terminal, calculate the thermal power dissipated at the splice.
Given: I = 16 A, R = 0.75 Ω P = I^2 x R = (16 A)^2 x 0.75 Ω = 256 x 0.75 = 192.0 Watts (Note: 192 Watts concentrated inside an enclosed junction box is equivalent to two 100W incandescent bulbs, rapidly melting conductor insulation and causing electrical fires).
4. DC Series Circuits
A series circuit provides only one continuous path for electric current to flow through all connected loads.
+-----------------------------------------------------------------------------+
| DC SERIES CIRCUIT |
| |
| +--------------[ R1 = 10 Ω ]--------------+ |
| | | |
| | v |
| +----+----+ [ R2 = 15 Ω ] |
| | E_source| | |
| | = 120 V | v |
| +----+----+ [ R3 = 25 Ω ] |
| | | |
| +-----------------------------------------+ |
+-----------------------------------------------------------------------------+
Fundamental Series Rules:
- Current is Constant: The current is identical at every point in the circuit: I_total = I_1 = I_2 = I_3 = ... = I_n
- Total Resistance is Additive: The total resistance is the direct algebraic sum of individual load resistances: R_total = R_1 + R_2 + R_3 + ... + R_n
- Kirchhoff's Voltage Law (KVL): The sum of all individual voltage drops across series components equals the total applied source voltage: E_total = V_1 + V_2 + V_3 + ... + V_n
- Individual Voltage Drops: Calculated via Ohm's Law: V_n = I_total x R_n.
- Total Power is Additive: Total power equals the sum of power consumed by individual components: P_total = P_1 + P_2 + P_3 + ... + P_n = E_total x I_total
Step-by-Step Series Problem Solution
Using the circuit illustrated above (E_source = 120 V, R_1 = 10 Ω, R_2 = 15 Ω, R_3 = 25 Ω):
- Step 1: Calculate Total Circuit Resistance (R_T): R_T = 10 Ω + 15 Ω + 25 Ω = 50.0 Ω
- Step 2: Calculate Total Circuit Current (I_T): I_T = E_T / R_T = (120 V) / (50 Ω) = 2.40 A
- Step 3: Calculate Voltage Drop Across Each Resistor: V_1 = I_T x R_1 = 2.40 A x 10 Ω = 24.0 V V_2 = I_T x R_2 = 2.40 A x 15 Ω = 36.0 V V_3 = I_T x R_3 = 2.40 A x 25 Ω = 60.0 V Check KVL: 24 V + 36 V + 60 V = 120.0 V (Matches Source)
- Step 4: Calculate Individual and Total Power Dissipated: P_1 = (2.4 A)^2 x 10 Ω = 57.6 W P_2 = (2.4 A)^2 x 15 Ω = 86.4 W P_3 = (2.4 A)^2 x 25 Ω = 144.0 W P_T = 57.6 + 86.4 + 144.0 = 288.0 W (Check: P_T = 120 V x 2.4 A = 288.0 W)
5. DC Parallel Circuits
A parallel circuit provides two or more independent branches for current to travel across a common voltage supply.
+-----------------------------------------------------------------------------+
| DC PARALLEL CIRCUIT |
| |
| +----------------+----------------+ |
| | | | |
| v (I1) v (I2) v (I3) |
| [ R1 = 20 Ω ] [ R2 = 30 Ω ] [ R3 = 60 Ω ] |
| | | | |
| +--------+----------------+----------------+ |
| | |
| +----+----+ |
| | E_source| = 120 V |
| +----+----+ |
| | |
| +------------------------------------------+ |
+-----------------------------------------------------------------------------+
Fundamental Parallel Rules:
- Voltage is Constant Across All Branches: E_total = E_1 = E_2 = E_3 = ... = E_n
- Kirchhoff's Current Law (KCL): The total current entering a junction must equal the sum of currents leaving that junction: I_total = I_1 + I_2 + I_3 + ... + I_n
- Equivalent Resistance Calculation: The total equivalent resistance is always strictly LESS than the resistance of the smallest individual branch.
- General Reciprocal Formula (Any Number of Resistors): 1 / R_total = 1 / R_1 + 1 / R_2 + 1 / R_3 + ... + 1 / R_n => R_total = (1) / (1 / R_1 + 1 / R_2 + ... + 1 / R_n)
- Product-Over-Sum Formula (Exactly Two Parallel Resistors): R_total = (R_1 x R_2) / (R_1 + R_2)
- Equal-Value Resistor Shortcut: If N identical resistors of resistance R are connected in parallel: R_total = R / N
Step-by-Step Parallel Problem Solution
Using the parallel circuit above (E_source = 120 V, R_1 = 20 Ω, R_2 = 30 Ω, R_3 = 60 Ω):
- Step 1: Calculate Total Equivalent Resistance (R_T): 1 / R_T = 1 / 20 + 1 / 30 + 1 / 60 = 3 / 60 + 2 / 60 + 1 / 60 = 6 / 60 = (1) / (10 Ω) R_T = 10.0 Ω
- Step 2: Calculate Individual Branch Currents (I_1, I_2, I_3): I_1 = E / R_1 = (120 V) / (20 Ω) = 6.0 A I_2 = E / R_2 = (120 V) / (30 Ω) = 4.0 A I_3 = E / R_3 = (120 V) / (60 Ω) = 2.0 A
- Step 3: Calculate Total Circuit Current (I_T): I_T = I_1 + I_2 + I_3 = 6.0 A + 4.0 A + 2.0 A = 12.0 A Verification via Ohm's Law: I_T = E_T / R_T = (120 V) / (10 Ω) = 12.0 A
- Step 4: Total Power Dissipated: P_T = E_T x I_T = 120 V x 12 A = 1,440.0 W
6. Series-Parallel Combination Circuits
Combination circuits contain both series and parallel elements connected together. Solving these requires systematic block reduction.
+-----------------------------------------------------------------------------+
| SERIES-PARALLEL COMBINATION |
| |
| +--------[ R1 = 8 Ω ]---------+ |
| | | |
| | Node A | |
| | +-------+ | |
| | | | | |
| | v v | |
| +----+----+ [R2=12Ω] [R3=24Ω] | |
| | E=120 V | | | | |
| +----+----+ +-------+ | |
| | Node B | |
| | | | |
| +-------[ R4 = 4 Ω ]----------+ |
+-----------------------------------------------------------------------------+
Systematic 5-Step Reduction Method:
- Identify Parallel Sub-Banks: Locate resistors connected between identical node pairs.
- Calculate Equivalent Parallel Resistance (R_p): Replace the parallel branch with a single equivalent resistor.
- Combine Series Elements (R_T): Add all remaining series resistors and equivalent parallel blocks.
- Calculate Total Circuit Current (I_T): I_T = E_source / R_T.
- Work Backwards for Local Values: Determine voltage drops across series elements, subtract to find the remaining voltage across the parallel bank, and calculate individual branch currents.
Complete Worked Solution:
- Step 1: Simplify the Parallel Bank (R_2 and R_3): R_p(2-3) = (R_2 x R_3) / (R_2 + R_3) = (12 x 24) / (12 + 24) = 288 / 36 = 8.0 Ω
- Step 2: Calculate Total Series Resistance (R_T): R_T = R_1 + R_p(2-3) + R_4 = 8 Ω + 8 Ω + 4 Ω = 20.0 Ω
- Step 3: Calculate Total Circuit Current (I_T): I_T = E_source / R_T = (120 V) / (20 Ω) = 6.0 A
- Step 4: Calculate Series Voltage Drops: V_R1 = I_T x R_1 = 6.0 A x 8 Ω = 48.0 V V_R4 = I_T x R_4 = 6.0 A x 4 Ω = 24.0 V
- Step 5: Calculate Voltage Across Parallel Bank (Node A to Node B): V_NodeAB = E_source - (V_R1 + V_R4) = 120 V - (48 V + 24 V) = 120 - 72 = 48.0 V (Or: V_NodeAB = I_T x R_p(2-3) = 6.0 A x 8.0 Ω = 48.0 V)
- Step 6: Calculate Individual Branch Currents through R_2 and R_3: I_R2 = V_NodeAB / R_2 = (48 V) / (12 Ω) = 4.0 A I_R3 = V_NodeAB / R_3 = (48 V) / (24 Ω) = 2.0 A Check KCL: I_R2 + I_R3 = 4.0 A + 2.0 A = 6.0 A = I_T
7. Quick Circuit Comparison Reference
| Electrical Property | Series Circuit | Parallel Circuit | Combination Circuit |
|---|---|---|---|
| Current (I) | Same through all loads (I_T = I_1 = I_2) | Splits across branches (I_T = Sum I_n) | Main line carries I_T; splits in parallel banks |
| Voltage (E) | Divides across loads (E_T = Sum V_n) | Same across all branches (E_T = E_1 = E_2) | Drops occur across series parts; same across shared nodes |
| Total Resistance (R_T) | R_T = R_1 + R_2 + ... (Increases) | 1/R_T = 1/R_1 + 1/R_2 + ... (Decreases) | Combine step-by-step using reduction rules |
| Component Failure | Open circuit stops all current flow | Open branch leaves other branches active | Depends on whether open occurs in main trunk or sub-branch |
A commercial electric duct heater is rated at 4,800 Watts at 240 Volts single-phase. If this heater is connected to a 208 Volt single-phase supply in a commercial building, what is its actual operating power output?
Two resistors are connected in parallel across a 120V source: Resistor A is 30 Ω and Resistor B is 60 Ω. A third resistor of 10 Ω is connected in series with this parallel combination. What is the total current flowing from the 120V source?
In a series circuit consisting of three resistive loads of 15 Ω, 25 Ω, and 40 Ω connected across a 240V DC supply, what is the voltage drop across the 25 Ω resistor?