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7.1 Electrical Principles: Voltage, Current, Resistance & Ohm's Law (V=IR, P=VI)

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1 Chapter 1: Guide Scope, AFQT & Service-Specific Policy

1.1 Editorial Scope and Service-Specific AFCT Policy 1.2 AFQT Meaning and Study Priorities 1.3 Service Composites and Targeted Skill Planning 1.4 Format-Agnostic Test Preparation and Pacing

2 Chapter 2: General Science (GS)

2.1 Life Science: Cell Biology, Human Anatomy & Genetics 2.2 Physical Science: Matter, Energy, Forces & Motion 2.3 Earth & Space Science: Geology, Meteorology & Astronomy 2.4 Chemistry Fundamentals: Atomic Structure, Elements & Chemical Reactions

3 Chapter 3: Arithmetic Reasoning (AR)

3.1 Word Problems: Rates, Ratios, Proportions & Percentages 3.2 Financial & Distance Math: Interest, Profit/Loss, Speed & Time 3.3 Work & Mixture Problems: Joint Rates & Percentage Concentrations 3.4 Multi-Step Problem Solving & Algebraic Word Problems

4 Chapter 4: Word Knowledge (WK)

4.1 General Academic & Technical Vocabulary 4.2 Context Clues, Synonyms & Antonyms 4.3 Word Roots, Prefixes & Suffixes (Etymology) 4.4 Advanced Technical & Operational Terminology

5 Chapter 5: Paragraph Comprehension (PC)

5.1 Identifying Main Ideas, Central Themes & Topic Sentences 5.2 Extracting Supporting Details & Factual Information 5.3 Drawing Logical Inferences & Author's Purpose 5.4 Analyzing Text Structure & Drawing Conclusions

6 Chapter 6: Mathematics Knowledge (MK)

6.1 Pre-Algebra & Number Operations: Exponents, Radicals & Order of Operations 6.2 High School Algebra: Linear Equations, Inequalities & Polynomials 6.3 Geometry Fundamentals: Angles, Triangles, Quadrilaterals & Circles 6.4 Coordinate Geometry, Perimeter, Area & Volume Formulas

7 Chapter 7: Electronics Information (EI)

7.1 Electrical Principles: Voltage, Current, Resistance & Ohm's Law (V=IR, P=VI) 7.2 Electric Circuits: Series, Parallel & Combination Circuit Analysis 7.3 Electrical Components: Resistors, Capacitors, Inductors & Transformers 7.4 Electronic Semiconductors, Transistors, Diodes & Logic Gates

8 Chapter 8: Auto & Shop Information (AS)

8.1 Automotive Systems: Internal Combustion Engines, Fuel & Ignition Systems 8.2 Automotive Mechanical Systems: Transmission, Suspension, Brakes & Cooling 8.3 Hand & Power Tools: Identification, Uses & Fasteners 8.4 Shop Practice, Woodworking, Metalworking & Safety Rules

9 Chapter 9: Mechanical Comprehension (MC)

9.1 Simple Machines: Levers, Pulleys, Inclined Planes & Mechanical Advantage 9.2 Force, Motion, Work & Power: Newton's Laws & Energy Transformations 9.3 Fluid Dynamics & Pressure: Pascal's Principle, Hydraulics & Pneumatics 9.4 Mechanical Components: Gears, Belts, Springs & Friction

10 Chapter 10: Spatial Assembly Practice (AO Skill)

10.1 Spatial Connections and Assembly Patterns

7.1 Electrical Principles: Voltage, Current, Resistance & Ohm's Law (V=IR, P=VI)

Key Takeaways Explain This Topic AI

Voltage (V or E) is the electromotive force or electrical potential difference measured in Volts (V), representing work per unit charge.

Current (I) is the rate of electron flow through a conductor measured in Amperes (A or Amps), where 1 Amp equals 1 Coulomb per second.

Resistance (R) is the opposition to current flow measured in Ohms (Ω), determined by conductor material resistivity, length, cross-sectional area, and temperature.

Ohm's Law defines the fundamental mathematical relationship between voltage, current, and resistance: V = I × R, I = V / R, and R = V / I.

Electrical Power (P) measured in Watts (W) represents the rate of electrical energy consumption: P = V × I = I²R = V² / R.

Last updated: July 2026

7.1 Electrical Principles: Voltage, Current, Resistance & Ohm's Law

Quick Answer: Electrical circuits are governed by three core variables: Voltage ( V V V ) (electrical pressure measured in Volts), Current ( I I I ) (rate of charge flow measured in Amperes), and Resistance ( R R R ) (opposition to flow measured in Ohms). They are linked by Ohm's Law ( V = I × R V = I \times R V = I × R ) and Power ( P = V × I = I 2 R = V 2 / R P = V \times I = I^2 R = V^2 / R P = V × I = I 2 R = V 2 / R ) measured in Watts. Mastery of these formulas and step-by-step calculations supports Electronics Information problem solving.

1. Atomic Theory & Fundamentals of Electrical Charge

To understand electricity, you must start at the atomic level. All matter is composed of atoms containing three subatomic particles:

Protons: Positively charged particles located in the atom's nucleus.

Neutrons: Electrically neutral particles located in the nucleus.

Electrons: Negatively charged particles orbiting the nucleus in discrete energy shells. [ Atomic Structure & Electron Movement ]

  • Nucleus (Protons + Neutrons) ---> Outer Valence Shell
  • [ + + + ] ( ( ( - ) ) )
  • |
  • [ Valence Electron ]
  • |
  • (Energy Applied)
  • v
  • [ FREE ELECTRON ]
  • (Current Flow)

The outermost shell of an atom is called the valence shell , and the electrons in this shell are valence electrons . The electrical properties of any material depend directly on how loosely or tightly these valence electrons are bound to the nucleus:

Conductors: Materials with 1 to 3 valence electrons (e.g., Copper, Silver, Aluminum, Gold). These electrons easily break free to become free electrons , allowing electric current to flow with minimal resistance.

Insulators: Materials with 5 to 8 valence electrons (e.g., Rubber, Glass, Plastic, Ceramic, Air). Their valence shells are nearly full, tightly holding electrons and preventing current flow.

Semiconductors: Materials with exactly 4 valence electrons (e.g., Silicon, Germanium). Under different conditions (temperature or impurity doping), they can act as either conductors or insulators.

Electrical Charge ( Q Q Q )

Electrical charge is measured in Coulombs ( C C C ) . One Coulomb represents the combined negative charge of approximately 6.242 × 10 18 6.242 \times 10^{18} 6.242 × 1 0 18 electrons .

2. Fundamental Electrical Quantities: V, I, R, and G

Every electrical circuit involves four fundamental parameters that define how energy moves:

  • Quantity Symbol Unit Unit Abbreviation Water Analogy
  • Voltage (Electromotive Force) V V V or E E E Volt V \text{V} V Water pressure from a pump or elevated tank
  • Current I I I Ampere A \text{A} A Volume rate of water flow (gallons per minute)
  • Resistance R R R Ohm Ω \Omega Ω Pipe diameter or restriction resisting flow
  • Conductance G G G Siemens (or Mho) S \text{S} S Ease with which water flows through the pipe
  • Voltage ( V V V or E E E )

Voltage—also called Electromotive Force (EMF) or Potential Difference —is the force or pressure that pushes free electrons through a conductor. One Volt is defined as the potential difference required to move 1 Coulomb of charge through a circuit while expending 1 Joule of energy ( 1 V = 1 J/C 1\text{ V} = 1\text{ J/C} 1 V = 1 J/C ).

Current ( I I I )

Current is the rate at which electrical charge flows past a given point in a circuit per second ( I = Q / t I = Q / t I = Q / t ). One Ampere is equal to 1 Coulomb of charge passing a point in 1 second ( 1 A = 1 C/s 1\text{ A} = 1\text{ C/s} 1 A = 1 C/s ).

Important classification practice Distinction: Conventional Current vs. Electron Flow

Conventional Current Flow: Assumes charge flows from Positive (+) to Negative (-) . Standard circuit schematics and engineering analysis use conventional flow.

Electron Flow Theory: Describes actual physical electron movement from Negative (-) to Positive (+) (since electrons are negatively charged and repelled by negative terminals).

Resistance ( R R R )

Resistance is the natural opposition a material offers to the flow of electric current, converting electrical energy into thermal energy (heat). The resistance of a conductor is determined by four physical factors ( R = ρ L A R = \rho \frac{L}{A} R = ρ A L ​ ):

Material (Resistivity ρ \rho ρ ): Copper has low resistivity; Nichrome has high resistivity.

Length ( L L L ): Resistance is directly proportional to length. Doubling wire length doubles its resistance.

Cross-Sectional Area ( A A A ): Resistance is inversely proportional to cross-sectional area (wire thickness). A thicker wire has less resistance than a thin wire.

Temperature: For most metals (Positive Temperature Coefficient or PTC), resistance increases as temperature rises.

Conductance ( G G G )

Conductance is the mathematical reciprocal of resistance ( G = 1 R G = \frac{1}{R} G = R 1 ​ ). It measures how easily current flows through a material and is expressed in Siemens ( S S S ).

3. Ohm's Law ( V = I × R V = I \times R V = I × R )

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. [ Ohm's Law Memory Circle ]

  • +-------+
  • | V |
  • +-------+
  • | I | R |
  • +-------+

V = I × R | I = V / R | R = V / I

Equations: Voltage: V = I × R \text{Voltage: } V = I \times R Voltage: V = I × R Current: I = V R \text{Current: } I = \frac{V}{R} Current: I = R V ​ Resistance: R = V I \text{Resistance: } R = \frac{V}{I} Resistance: R = I V ​

  • Step-by-Step Calculation Examples
  • Example 1: Solving for Current ( I I I )

A 120 V 120\text{V} 120 V household outlet powers an electric heater with an internal heating element resistance of 24 Ω 24\Omega 24Ω . Calculate the current drawn by the heater.

  • Given: V = 120 V V = 120\text{V} V = 120 V , R = 24 Ω R = 24\Omega R = 24Ω
  • Formula: I = V R I = \frac{V}{R} I = R V ​

Calculation: I = 120 V 24 Ω = 5.0 Amperes I = \frac{120\text{V}}{24\Omega} = 5.0\text{ Amperes} I = 24Ω 120 V ​ = 5.0 Amperes

Example 2: Solving for Voltage ( V V V )

An automotive headlight draws 2.5 A 2.5\text{A} 2.5 A of current and has an internal resistance of 4.8 Ω 4.8\Omega 4.8Ω . What supply voltage is required?

  • Given: I = 2.5 A I = 2.5\text{A} I = 2.5 A , R = 4.8 Ω R = 4.8\Omega R = 4.8Ω
  • Formula: V = I × R V = I \times R V = I × R

Calculation: V = 2.5 A × 4.8 Ω = 12.0 Volts V = 2.5\text{A} \times 4.8\Omega = 12.0\text{ Volts} V = 2.5 A × 4.8Ω = 12.0 Volts

Example 3: Solving for Resistance ( R R R )

When a 24 V 24\text{V} 24 V power supply is connected across an unknown resistor, a multimeter measures a current of 0.5 A 0.5\text{A} 0.5 A ( 500 mA 500\text{mA} 500 mA ). Find the resistance value.

  • Given: V = 24 V V = 24\text{V} V = 24 V , I = 0.5 A I = 0.5\text{A} I = 0.5 A
  • Formula: R = V I R = \frac{V}{I} R = I V ​
  • Calculation: R = 24 V 0.5 A = 48.0 Ω R = \frac{24\text{V}}{0.5\text{A}} = 48.0\Omega R = 0.5 A 24 V ​ = 48.0Ω
  • 4. Electrical Power ( P P P ) and Energy ( E E E )

Electrical Power ( P P P ) is the rate at which electrical energy is transferred, converted, or dissipated in a circuit per unit time. Power is measured in Watts ( W W W ) , where 1 Watt equals 1 Joule per second ( 1 W = 1 J/s 1\text{ W} = 1\text{ J/s} 1 W = 1 J/s ).

Watt's Law & Power Equations

Power can be calculated using Voltage, Current, and Resistance by combining Watt's Law ( P = V × I P = V \times I P = V × I ) with Ohm's Law: Primary Power Formula: P = V × I \text{Primary Power Formula: } P = V \times I Primary Power Formula: P = V × I In terms of Current & Resistance: P = I 2 × R \text{In terms of Current \& Resistance: } P = I^2 \times R In terms of Current & Resistance: P = I 2 × R In terms of Voltage & Resistance: P = V 2 R \text{In terms of Voltage \& Resistance: } P = \frac{V^2}{R} In terms of Voltage & Resistance: P = R V 2 ​ [ The Electrical Power Wheel ]

  • Power (P) Voltage (V)
  • P = V × I V = I × R
  • P = I² × R V = P / I
  • P = V² / R V = √(P × R)
  • -----------------------------------
  • Current (I) Resistance (R)
  • I = V / R R = V / I
  • I = P / V R = P / I²
  • I = √(P / R) R = V² / P
  • Worked Power Example

A 100 Ω 100\Omega 100Ω resistor has 2.0 A 2.0\text{A} 2.0 A of current flowing through it. What is the power dissipated by the resistor?

  • Given: R = 100 Ω R = 100\Omega R = 100Ω , I = 2.0 A I = 2.0\text{A} I = 2.0 A
  • Formula: P = I 2 × R P = I^2 \times R P = I 2 × R

Calculation: P = ( 2.0 ) 2 × 100 = 4 × 100 = 400 Watts P = (2.0)^2 \times 100 = 4 \times 100 = 400\text{ Watts} P = ( 2.0 ) 2 × 100 = 4 × 100 = 400 Watts

Electrical Energy ( E E E )

Electrical energy is the total power consumed over a period of time: Energy ( E ) = P × t \text{Energy } (E) = P \times t Energy ( E ) = P × t

Standard international unit: Joule ( J J J ) ( 1 W × 1 s 1\text{ W} \times 1\text{ s} 1 W × 1 s ).

Commercial power unit: Kilowatt-hour (kWh) ( 1 kWh = 1 , 000 Watts × 1 hour = 3.6 × 10 6 Joules 1\text{ kWh} = 1,000\text{ Watts} \times 1\text{ hour} = 3.6 \times 10^6\text{ Joules} 1 kWh = 1 , 000 Watts × 1 hour = 3.6 × 1 0 6 Joules ).

Loading diagram... Ohm's Law and Electrical Power Relationships

  • 5. Direct Current (DC) vs. Alternating Current (AC)
  • Electric current exists in two distinct forms: Direct Current (DC) Alternating Current (AC)
  • Voltage Voltage
  • ^ ^ +Vpeak
  • | | / \ -- RMS = 0.707 Vpeak
  • +V |------------------ (Constant) | / \ /
  • | 0 +--+------+--+---> Time
  • +------------------> Time | / \
  • Vpeak| / \
  • Direct Current (DC)

In Direct Current , electric charge flows in one continuous direction with constant or steady magnitude. DC is produced by chemical batteries, solar photovoltaic cells, thermocouples, and DC power supplies.

Alternating Current (AC)

In Alternating Current , the movement of electric charge periodically reverses direction and constantly changes magnitude in a sinusoidal waveform. AC is generated by rotating alternators in commercial power plants.

Key AC Waveform Parameters:

Frequency ( f f f ): The number of complete cycles per second, measured in Hertz (Hz) . In North America, standard power frequency is 60 Hz 60\text{ Hz} 60 Hz ( 50 Hz 50\text{ Hz} 50 Hz in Europe).

Period ( T T T ): The time required to complete one full cycle ( T = 1 f T = \frac{1}{f} T = f 1 ​ ). At 60 Hz 60\text{ Hz} 60 Hz , T = 1 60 ≈ 16.67 milliseconds T = \frac{1}{60} \approx 16.67\text{ milliseconds} T = 60 1 ​ ≈ 16.67 milliseconds .

Peak Voltage ( V peak V_{\text{peak}} V peak ​ ): The maximum instantaneous voltage reached in either positive or negative polarity.

Root-Mean-Square Voltage ( V RMS V_{\text{RMS}} V RMS ​ ): The effective value of an AC voltage that delivers the equivalent thermal power to a heating element as a DC voltage of the same value. V RMS = 0.707 × V peak V_{\text{RMS}} = 0.707 \times V_{\text{peak}} V RMS ​ = 0.707 × V peak ​ V peak = 1.414 × V RMS V_{\text{peak}} = 1.414 \times V_{\text{RMS}} V peak ​ = 1.414 × V RMS ​

practice review Tip: Wall outlet voltage in the United States is rated at 120 V AC (RMS) 120\text{V AC (RMS)} 120 V AC (RMS) . Its actual peak voltage is 120 × 1.414 ≈ 170 Volts 120 \times 1.414 \approx 170\text{ Volts} 120 × 1.414 ≈ 170 Volts !

6. electronics practice referenceula & Unit Quick Reference

Electrical Quantity Symbol Base Unit Key Formula(s) practice review Shortcut

Voltage V V V or E E E Volt (V) V = I ⋅ R V = I \cdot R V = I ⋅ R , V = P I V = \frac{P}{I} V = I P ​ , V = P ⋅ R V = \sqrt{P \cdot R} V = P ⋅ R ​ 1 V = 1 Joule per Coulomb 1\text{ V} = 1\text{ Joule per Coulomb} 1 V = 1 Joule per Coulomb

Current I I I Ampere (A) I = V R I = \frac{V}{R} I = R V ​ , I = P V I = \frac{P}{V} I = V P ​ , I = P R I = \sqrt{\frac{P}{R}} I = R P ​ ​ 1 A = 1 Coulomb per second 1\text{ A} = 1\text{ Coulomb per second} 1 A = 1 Coulomb per second

Resistance R R R Ohm ( Ω \Omega Ω ) R = V I R = \frac{V}{I} R = I V ​ , R = P I 2 R = \frac{P}{I^2} R = I 2 P ​ , R = V 2 P R = \frac{V^2}{P} R = P V 2 ​ R = ρ L A R = \rho \frac{L}{A} R = ρ A L ​ (Longer = more R, Thicker = less R)

Power P P P Watt (W) P = V ⋅ I P = V \cdot I P = V ⋅ I , P = I 2 ⋅ R P = I^2 \cdot R P = I 2 ⋅ R , P = V 2 R P = \frac{V^2}{R} P = R V 2 ​ 1 Horsepower (HP) = 746 Watts 1\text{ Horsepower (HP)} = 746\text{ Watts} 1 Horsepower (HP) = 746 Watts

AC RMS Voltage V RMS V_{\text{RMS}} V RMS ​ Volt (V) V RMS = 0.707 ⋅ V peak V_{\text{RMS}} = 0.707 \cdot V_{\text{peak}} V RMS ​ = 0.707 ⋅ V peak ​ Peak to Peak ( V p-p ) = 2 ⋅ V peak V_{\text{p-p}}) = 2 \cdot V_{\text{peak}} V p-p ​ ) = 2 ⋅ V peak ​