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TRN-AUTO19-05 · Lesson 2 · 16 clock hours · theory

Electrical Theory: Voltage, Current, Resistance, and Ohm's Law

Objective. Students will apply Ohm's Law and Watt's Law to calculate unknown values in series, parallel, and series-parallel circuits and predict how faults change those values.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a certified examiner or master technician in the trade or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

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Why Electrical Theory Matters in Automotive Work

Every system on a modern vehicle—starting, charging, lighting, HVAC, fuel injection, and advanced driver-assistance—runs on electricity. Before you touch a meter or a wiring diagram, you need a solid mental model of how voltage, current, and resistance relate to one another. Ohm's Law and Watt's Law give you that model. They let you calculate what a circuit should read before you measure anything, so a wrong meter reading instantly tells you something is wrong—and points you toward the fault.

The Three Fundamental Quantities

Voltage (V) is electrical pressure—the force that pushes electrons through a conductor. It is measured in volts (V) between two points. A fully charged 12-volt automotive battery actually measures closer to 12.6 V at rest. Voltage does not flow; it exists across two points.

Current (I) is the rate of electron flow past a point, measured in amperes (A). Current flows through a component. A headlight bulb, a blower motor, and a fuel injector each draw a specific amount of current when operating normally. Knowing that value lets you choose the right fuse and wire gauge.

Resistance (R) is the opposition to current flow, measured in ohms (Ω). Every conductor, winding, and load has resistance. Unwanted resistance—caused by corrosion, loose connections, or damaged wire—is one of the most common causes of electrical complaints in the shop.

Ohm's Law

Ohm's Law states that voltage equals current multiplied by resistance:

V = I × R

Rearranged to solve for the unknown:

Use the triangle method if it helps: write V on top, I and R on the bottom. Cover the quantity you want, and the remaining two show the operation.

Watt's Law (Power)

Power (P), measured in watts (W), describes how much work a circuit does per unit of time:

P = V × I

Combined with Ohm's Law, two more useful forms emerge:

Automotive fuses and wire gauges are rated for both current and the heat (power) that current produces. Exceeding those ratings causes insulation damage and fire risk.

Series Circuits

In a series circuit, current has only one path. The same current flows through every component. Key rules:

Fault prediction: an open (broken wire or burned-out component) in a series circuit stops all current—the entire circuit goes dead. A short across one resistor removes that resistance, raising current through the rest of the circuit.

Parallel Circuits

In a parallel circuit, each branch has the same voltage across it, but current divides among the branches. Key rules:

Fault prediction: an open in one branch kills only that branch; the others keep working. A short across one branch collapses voltage to near zero for the whole circuit and blows the fuse.

Series-Parallel Circuits

Most automotive circuits combine both types. A blower motor speed-control circuit, for example, places resistors in series with a parallel motor-and-switch network. Solve these by simplifying: reduce the parallel section to a single equivalent resistance first, then treat the result as a series circuit.

Predicting Faults with Math

Calculate expected values before you measure. If a front parking lamp circuit should draw 2.1 A but your ammeter reads 0.8 A, excess resistance is present—look for a bad ground or corroded connector. If current is higher than calculated, look for an unintended short path. Voltage-drop testing across individual connectors and wires is how you locate that hidden resistance.

Safety in the Electrical Lab

Always observe these precautions when working with DC circuit trainers and automotive electrical systems:

Key terms10

Voltage (V)Electrical pressure or potential difference between two points, measured in volts; the force that drives current through a circuit.
Current (I)The rate of electron flow through a conductor, measured in amperes (A); the same throughout a series circuit, divided among branches in a parallel circuit.
Resistance (R)Opposition to current flow in a circuit, measured in ohms (Ω); includes both intentional loads and unwanted resistance from corrosion or loose connections.
Ohm's LawThe relationship V = I × R, which allows any one of the three quantities to be calculated when the other two are known.
Watt's LawP = V × I; expresses electrical power in watts and is used to determine heat load in wires, fuses, and components.
Series CircuitA circuit with only one current path; total resistance equals the sum of all individual resistances, and an open anywhere stops all current.
Parallel CircuitA circuit with two or more current paths sharing the same voltage; total current equals the sum of branch currents, and an open in one branch does not affect others.
Voltage DropThe portion of source voltage consumed across a specific component or resistance; excessive voltage drop across a wire or connector indicates unwanted resistance.
Open CircuitA break in the current path that causes current to drop to zero in that path; in a series circuit it kills the entire circuit.
Short CircuitAn unintended low-resistance path that allows excessive current to flow, typically causing fuse or breaker operation and potential wiring damage.

Worked examples4

1. A single headlight bulb is connected to a 12.6 V source. The bulb has a resistance of 6 Ω. Calculate current draw and power consumption.
  1. Identify knowns: V = 12.6 V, R = 6 Ω.
  2. Apply Ohm's Law to find current: I = V ÷ R = 12.6 ÷ 6 = 2.1 A.
  3. Apply Watt's Law to find power: P = V × I = 12.6 × 2.1 = 26.46 W.
Answer: Current = 2.1 A; Power = 26.46 W (approximately 26.5 W).
2. A blower motor circuit has the blower motor (R1 = 3 Ω) in series with a speed resistor (R2 = 1.5 Ω). Source voltage is 12 V. Find total resistance, circuit current, and voltage drop across each component.
  1. Identify knowns: V_source = 12 V, R1 = 3 Ω, R2 = 1.5 Ω, series circuit.
  2. Total resistance: R_total = R1 + R2 = 3 + 1.5 = 4.5 Ω.
  3. Circuit current: I = V ÷ R_total = 12 ÷ 4.5 = 2.667 A (round to 2.67 A).
  4. Voltage drop across R1 (motor): V1 = I × R1 = 2.67 × 3 = 8.01 V.
  5. Voltage drop across R2 (speed resistor): V2 = I × R2 = 2.67 × 1.5 = 4.005 V ≈ 4.00 V.
  6. Check: V1 + V2 = 8.01 + 4.00 = 12.01 V ≈ 12 V. ✓ (small difference due to rounding).
Answer: R_total = 4.5 Ω; I = 2.67 A; V_motor = 8.01 V; V_resistor = 4.00 V.
3. Two taillamp bulbs are wired in parallel across 12 V. Bulb A has a resistance of 12 Ω and Bulb B has a resistance of 6 Ω. Find total resistance, total current, and current through each bulb.
  1. Identify knowns: V = 12 V, R_A = 12 Ω, R_B = 6 Ω, parallel circuit.
  2. Total parallel resistance: R_total = (R_A × R_B) ÷ (R_A + R_B) = (12 × 6) ÷ (12 + 6) = 72 ÷ 18 = 4 Ω.
  3. Total current: I_total = V ÷ R_total = 12 ÷ 4 = 3 A.
  4. Current through Bulb A: I_A = V ÷ R_A = 12 ÷ 12 = 1 A.
  5. Current through Bulb B: I_B = V ÷ R_B = 12 ÷ 6 = 2 A.
  6. Check: I_A + I_B = 1 + 2 = 3 A = I_total. ✓
Answer: R_total = 4 Ω; I_total = 3 A; I_A = 1 A; I_B = 2 A.
4. Bulb B in the circuit above burns out (open circuit). Predict what happens to total current and to Bulb A's brightness.
  1. An open in Bulb B removes that branch. The circuit now has only Bulb A (12 Ω) connected across 12 V.
  2. New total current: I_total = V ÷ R_A = 12 ÷ 12 = 1 A.
  3. Bulb A voltage is still 12 V (parallel branches share source voltage), so Bulb A brightness is unchanged.
  4. Total current drawn from the source drops from 3 A to 1 A because one branch is gone.
Answer: Total current drops from 3 A to 1 A. Bulb A continues to glow at the same brightness because it still receives 12 V.

The math in this lesson2

Learners compute total resistance of series and parallel bulb circuits.
A-SSE.1A-CED.4
Brush up in the math halls at newSchoolWorks.com:
In class: Instructor-led lecture with whiteboard worked examples: solve for V, I, R, and P in 10 progressively complex automotive circuits (headlight circuit, blower motor circuit, multiple-lamp arrangement); students solve matching problems in pairs. · Students complete a 15-problem circuit-math worksheet using automotive wiring diagrams as the source data, calculating expected readings before any meter is connected. · Hands-on trainer-board exercise using Elenco or equivalent DC circuit trainers: students build series, parallel, and combination circuits to match calculated values, then intentionally open and short one branch and record how meter readings change.
Materials: Whiteboard, circuit calculation worksheet, DC circuit trainer boards, digital multimeters (Fluke 115 or equivalent), resistor kits, wire jumpers, automotive wiring diagram excerpts (OEM or Mitchell 1 prints)
Check: Students solve an in-class 10-problem circuit calculation quiz with 80% accuracy; trainer-board readings must match calculated values within ±5%.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

ASE A6 – Electrical/Electronic Systems
1. A circuit has a source voltage of 14 V and a total resistance of 7 Ω. What is the circuit current?
ASE A6 – Electrical/Electronic Systems
2. In a series circuit with three resistors (4 Ω, 6 Ω, and 2 Ω) connected to 12 V, what is the voltage drop across the 6 Ω resistor?
ASE A6 – Electrical/Electronic Systems
3. A blower motor draws 8 A from a 12 V source. What is the power consumed by the motor?
ASE A6 – Electrical/Electronic Systems
4. Two resistors, 10 Ω and 10 Ω, are connected in parallel. What is their combined resistance?
ASE A6 – Electrical/Electronic Systems
5. A technician measures 0 V across a lamp in a series circuit that should be lit. The most likely cause is:
ASE A6 – Electrical/Electronic Systems
6. A technician finds that a parallel lighting circuit blows its fuse immediately after one bulb socket is repaired. The most likely cause of the repeated fuse failure is:
ASE A6 – Electrical/Electronic Systems
7. Which tool setting error can damage a digital multimeter and create an unsafe condition when measuring current in an automotive circuit?
ASE A6 – Electrical/Electronic Systems
8. A series-parallel circuit has a 2 Ω resistor in series with two parallel branches of 6 Ω each. The source is 12 V. What is the total circuit current?
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