Lesson 5.5 — Ohm's Law in Practice

§00

Why Green Risk?

🟢 Green Risk — No Hazard

This is the only Green-risk lesson in Module 5. It is pure calculation — no equipment, no energized systems, no PPE required. Every other lesson in this module involves hazard exposure. Here, the only thing you are working with is math.

Ohm's Law is a math skill. This lesson is about making that math second nature so you can apply it in the field without reaching for a calculator. A technician who can run these numbers mentally can diagnose a high-resistance fault, size a fuse, or verify a power supply load before the next technician has finished powering up their meter.

No lockout required. No gloves required. Just pencil, paper, and the six equations you will know cold by the end of this lesson.

§01

Overview

Ohm's Law is the most-used equation in electrical work. V = IR. Everything else — power, voltage drop, fuse sizing, wire selection — flows from that single relationship. A field technician who can apply it quickly and accurately can diagnose problems faster than any other method, because most electrical faults show up as unexpected voltage, unexpected current, or unexpected resistance.

This lesson is entirely practical. There are no abstract circuits drawn on a chalkboard. Every problem is drawn from real field situations: a solenoid that will not actuate, a fuse that keeps blowing, a motor running rough because one winding is bad. By the end of this lesson you will have solved all ten of them yourself.

What's in This Lesson

10 real-world field scenarios — solve each one yourself, then check your answer
Timed drill tool — build speed with random V/I/R problems
Voltage divider extension — a critical application for sensing circuits
5-question assessment — verify your understanding before moving on

§02

Learning Objectives

L5-05-01
Solve for V, I, or R given any two quantities in a DC circuit
L5-05-02
Apply Ohm's Law to diagnose high-resistance faults (voltage drop)
L5-05-03
Calculate expected current draw to detect overload conditions
L5-05-04
Use Ohm's Law to verify fuse and breaker sizing
L5-05-05
Calculate wire voltage drop for a given load and run length
L5-05-06
Work through a 5-step troubleshooting scenario using only V = IR and P = VI

§03

Quick Reference — The Six Equations

Cover the unknown to read the formula

Cover the quantity you want to find. What remains is the formula to use.

        Cover V → V = I × R

        Cover I → I = V / R

        Cover R → R = V / I

All Six Working Equations

V = I × R

Voltage

I = V / R

Current

R = V / I

Resistance

P = V × I

Power (basic)

P = I² × R

Power (current form)

P = V² / R

Power (voltage form)

Which Formula Do You Use?

Use the form that contains your two known values. If you know V and R but not I, use I = V/R. If you know P and V but need R, rearrange P = V²/R to get R = V²/P. Always identify your two knowns first before reaching for the formula.

§04

Unit Prefixes and Conversions

The most common source of wrong answers in the field is a bad unit conversion. Before you plug numbers into any formula, confirm they are all in base units (Volts, Amps, Ohms, Watts).

Prefix	Symbol	Multiplier	Example	Convert to Base
Mega	M	× 1,000,000	1 MΩ	1,000,000 Ω
Kilo	k	× 1,000	4.7 kΩ	4,700 Ω
(base)	—	× 1	12 Ω	12 Ω
Milli	m	× 0.001	250 mA	0.25 A
Micro	µ	× 0.000001	47 µF	0.000047 F

⚠ Critical Field Conversions — Technicians Get These Wrong

Input: 2.2 kΩ
Base: 2,200 Ω

Input: 500 mA
Base: 0.5 A

Input: 1.5 MΩ
Base: 1,500,000 Ω

§05

Field Scenarios — Solve It Yourself

Each card describes a real situation a LEO technician might encounter. Read the given values, work the problem on paper or in your head, then enter your answer and check it. Answers within ±5% are accepted as correct.

Scenario 1 of 10 — Solenoid Current Check

A 24 VDC solenoid has a coil resistance stamped at 48 Ω. The solenoid is not actuating. Before testing with a meter, calculate the expected current draw.

V = 24 V | R = 48 Ω | Find: I (Amps)

I = V / R = 24 / 48 = 0.5 A

If your meter reads significantly less than 0.5 A, the coil is open (a wire break inside the solenoid body). If it reads more than expected, the resistance is lower than the stamped value — possible shorted turns in the coil. Either condition indicates a failed solenoid. This expected-current value becomes your diagnostic baseline before you ever touch a meter to the circuit.

Scenario 2 of 10 — Fuse Sizing Check

A 480 VAC heater draws 15 A. A 10 A fuse is installed. Calculate the power the heater is trying to draw and explain why the fuse keeps blowing.

V = 480 V | I = 15 A | Find: P (Watts)

P = V × I = 480 × 15 = 7,200 W (7.2 kW)

The 10 A fuse is undersized for a 15 A continuous load. Per NEC 210.20(A), overcurrent protection for continuous loads must be sized at 125% of the load current: 15 A × 1.25 = 18.75 A — round up to the next standard size, which is 20 A. The 10 A fuse was never going to survive this circuit. Replace with a 20 A fuse rated for 480 VAC.

Scenario 3 of 10 — Voltage Drop Diagnosis

A 120 V motor starter is wired with #14 AWG wire over a 150 ft run (300 ft total circuit). The motor terminal voltage measures 108 V while drawing 10 A. Calculate the resistance of the wiring that is causing this voltage drop.

V_source = 120 V | V_load = 108 V | I = 10 A | Find: R of wire (Ω)

V_drop = 120 − 108 = 12 V → R = V_drop / I = 12 / 10 = 1.2 Ω

Theoretical resistance of 300 ft of #14 AWG copper is approximately 0.98 Ω. Your calculated 1.2 Ω is 0.22 Ω higher than the wire alone — that excess resistance is a bad connection somewhere on the run. A 12 V drop on a 120 V circuit is a 10% drop, far exceeding the NEC 3% recommendation for branch circuits. Inspect every termination point on the run.

Scenario 4 of 10 — Unknown Resistance from Field Measurements

During a panel inspection, you measure 480 V at the supply terminals and 467 V at the load terminals. The load draws 25 A. What is the resistance of the conductors and connections in this circuit?

V_supply = 480 V | V_load = 467 V | I = 25 A | Find: R (Ω)

V_drop = 480 − 467 = 13 V → R = 13 / 25 = 0.52 Ω

0.52 Ω is significant for a high-current power circuit. At 25 A, that resistance dissipates P = I²R = 625 × 0.52 = 325 W as heat in the conductors and connections — wasted power and a potential fire hazard over time. Investigate corroded terminations, undersized conductors, a loose lug, or a connection made without proper torque.

Scenario 5 of 10 — Motor Winding Resistance Check

A 480 V, 3-phase motor has three winding resistances measured with an ohmmeter: Phase A = 2.1 Ω, Phase B = 2.1 Ω, Phase C = 3.8 Ω. Is Phase C normal?

Phase A: 2.1 Ω | Phase B: 2.1 Ω | Phase C: 3.8 Ω | Select the correct assessment:

Yes — 3.8 Ω is within normal manufacturing tolerance. The motor is fine. No — 3.8 Ω vs 2.1 Ω is an 81% deviation. Phase C is abnormal and the motor needs inspection before further use. Cannot determine without running the motor under load first.

Answer: No — Phase C is abnormal and requires immediate investigation.

Phase C resistance is 81% higher than Phases A and B. Applying Ohm's Law: since all three phases see the same line voltage, Phase C draws less current (I = V/R — higher R, lower I). Less current through that winding means less magnetic flux and less torque contribution. An 81% resistance asymmetry indicates a partial open winding, damaged insulation, or an incorrect rewind. Do not continue operating this motor. Tag it out, pull it, and send it for inspection and possible rewind.

Scenario 6 of 10 — Power Supply Load Check

A 24 VDC power supply is rated 5 A max (120 W). Six PLC input cards are connected, each drawing 350 mA. What is the total current draw in mA? Is the supply overloaded?

6 cards × 350 mA each | Supply rated: 5 A / 120 W | Find: Total current (mA)

6 × 350 mA = 2,100 mA = 2.1 A

P = V × I = 24 × 2.1 = 50.4 W. The supply is rated 120 W. Load is 50.4 / 120 = 42% — well within rating. NOT overloaded. Best practice for PLC panel power supplies: keep loading below 75% of rating to allow headroom for startup surges and future card additions. At current loading you could add three more identical cards (total 9 × 350 mA = 3,150 mA = 75.6 W = 63% loaded) before approaching the recommended limit.

Scenario 7 of 10 — Wire Gauge Selection

You are running a new 120 VAC circuit for a 1,200 W exhaust fan 75 feet from the panel (150 ft total circuit length). The maximum acceptable voltage drop is 3%. What is the maximum allowable resistance of the conductors?

V = 120 V | P = 1,200 W | Max V_drop = 3% | Find: R_max of wire (Ω)

Hint: Step 1 — find I from P = VI. Step 2 — find V_drop = 3% of 120 V. Step 3 — find R = V_drop / I.

Step 1: I = P / V = 1,200 / 120 = 10 A
Step 2: V_drop(max) = 120 × 0.03 = 3.6 V
Step 3: R_max = 3.6 / 10 = 0.36 Ω

Total circuit = 150 ft. #12 AWG copper has approximately 0.159 Ω per 100 ft → 150 ft = 0.24 Ω — within the 0.36 Ω limit. #14 AWG is 0.253 Ω per 100 ft → 150 ft = 0.38 Ω — slightly over the limit. Use #12 AWG minimum for this run. In practice, go one size larger for fan circuits because motor inrush can cause additional drop during startup.

Scenario 8 of 10 — 480 V Motor Expected Current

A 480 V, 3-phase, 25 HP motor has a nameplate efficiency of 90% and power factor of 0.87. Calculate the expected line current at full load.

HP = 25 | V_L = 480 V (3-phase) | η = 90% | PF = 0.87 | Find: I_L (A)

Hint: 1 HP = 746 W. Output power ÷ efficiency = input power. For 3-phase: I_L = P_in / (√3 × V_L × PF). Use √3 ≈ 1.732.

Step 1: P_out = 25 × 746 = 18,650 W
Step 2: P_in = 18,650 / 0.90 = 20,722 W
Step 3: I_L = 20,722 / (1.732 × 480 × 0.87) = 20,722 / 724.4 ≈ 28.6 A

This motor should draw approximately 28.6 A per phase under full load. The NEC table FLA for a 25 HP / 480 V motor is listed at 34 A — the nameplate FLA includes a built-in safety margin. If your clamp meter reads significantly above 34 A, the motor is overloaded or has a mechanical fault. If it reads well below 28 A, verify the driven equipment is properly engaged — the motor may be underloaded or the coupling may be slipping.

Scenario 9 of 10 — Heating Element Replacement

A process heater runs on 240 VAC and should produce 3,600 W of heat output. The heating element burns out and a replacement must be ordered. What resistance should the new element have?

V = 240 V | P = 3,600 W | Find: R (Ω)

Hint: Use P = V² / R, rearranged to R = V² / P.

R = V² / P = 240² / 3,600 = 57,600 / 3,600 = 16 Ω

Order a replacement element rated 16 Ω at operating temperature. Note that heating element resistance changes with temperature — measure cold resistance as a baseline, but understand the hot resistance will differ slightly. When ordering, specify resistance, wattage, voltage rating, and physical dimensions. Verify each new element with an ohmmeter before installation — an element reading 14 Ω or 18 Ω will produce significantly different heat output than specified.

Scenario 10 of 10 — The Five-Step Troubleshoot Challenge

A 24 VDC control relay is not energizing. Supply voltage at the panel = 24.1 V. Voltage measured at the relay coil terminals = 9.3 V. Relay coil resistance = 480 Ω. Walk through the diagnosis step by step.

Click each button in order to reveal the next diagnostic step. Complete all five steps to finish this scenario.

1

Step 1 — What Current Should the Relay Receive?

I_expected = V_supply / R_coil = 24.1 / 480 = 0.0502 A = 50.2 mA

If the circuit were healthy, the relay coil would see 50.2 mA — enough to develop the magnetic field required to pull in the armature. This is your baseline.

2

Step 2 — What Current Is Actually Flowing Through the Relay?

I_actual = V_relay / R_coil = 9.3 / 480 = 0.01938 A ≈ 19.4 mA

The relay is only receiving 19.4 mA — about 39% of the expected current. Most DC relay coils require at least 80% of rated voltage to guarantee pull-in. At 9.3 V (39% of 24 V), this relay simply cannot develop enough magnetic force to operate reliably.

3

Step 3 — How Much Voltage Is Being Lost in the Wiring?

V_drop = V_supply − V_relay = 24.1 − 9.3 = 14.8 V

14.8 volts — more than half the supply voltage — is being dropped somewhere between the panel terminal and the relay coil terminals. That voltage is being converted to heat at a bad connection.

4

Step 4 — What Is the Resistance of the Fault?

R_fault = V_drop / I_actual = 14.8 / 0.01938 = 763 Ω

There is approximately 763 Ω of fault resistance in series with the relay coil. For comparison, the relay coil itself is 480 Ω. The fault resistance is 1.6 times the load resistance — it is completely dominating the circuit and stealing the majority of the supply voltage.

5

Step 5 — Conclusion and Action

✓ Diagnosis: High-resistance fault (~763 Ω) in series with the relay coil

This is a classic high-resistance connection fault. The fault is absorbing 62% of the supply voltage (14.8 V out of 24.1 V). Possible causes: corroded terminal screw, broken wire strand making intermittent contact, oxidized connector pin, a splice that was not properly crimped, or a failed inline fuse holder. Action: use your meter on the millivolt range and probe across each connection from the panel output to the relay coil terminal. The connection with voltage across it is your fault. Fix the connection and retest — the relay should then receive full voltage and actuate normally.

Scenarios Completed

0 / 10 completed

§06

Timed Drill — Optional Practice

Build speed and accuracy with randomized V / I / R problems. Select difficulty and problem count, then work through each problem as fast as you can. Correct answers within ±5% score a point. Hard mode introduces kilo and milli prefixes.

Difficulty

Problems

0

Correct

1

Question

0:00

Time

10

Total

Drill Complete!

—

§07

The Voltage Divider — Ohm's Law Extended

Two resistors in series with a voltage source create a voltage divider. This is one of the most common sub-circuits in sensing and control work. Analog inputs, reference voltages, transducer interfaces, and level-shifting circuits all use this principle.

V_out = V_in × R₂ / (R₁ + R₂) Voltage Divider Formula

V_in = 12 V | R1 = 8 kΩ | R2 = 4 kΩ

Worked Example

Given: V_in = 12 V, R1 = 8 kΩ (8,000 Ω), R2 = 4 kΩ (4,000 Ω)

    // Step 1 — Total series resistance

    Rtotal = R1 + R2 = 8,000 + 4,000 = 12,000 Ω

    // Step 2 — Current through the series circuit

    I = Vin / Rtotal = 12 / 12,000 = 0.001 A (1 mA)

    // Step 3 — Voltage across R2 (this is V_out)

    Vout = I × R2 = 0.001 × 4,000 = 4.0 V

    // Or use the divider formula directly:

    Vout = 12 × 4,000 / (8,000 + 4,000) = 12 × 0.333 = 4.0 V

Voltage Across Each Resistor

In series circuits, the voltage is divided proportionally to resistance.

V across R1 = I × R1 = 0.001 × 8,000 = 8 V
V across R2 = I × R2 = 0.001 × 4,000 = 4 V
Check: 8 + 4 = 12 V ✓ (Kirchhoff's Voltage Law: voltages must sum to supply)

Field Application — Analog Signal Scaling

A PLC analog input expects 0–5 V. A sensor outputs 0–12 V. A voltage divider with R1 = 7 kΩ and R2 = 3 kΩ scales the signal: V_out = 12 × 3k / (7k + 3k) = 12 × 0.30 = 3.6 V at full scale. Adjust resistor values to match your target range. This is exactly how signal conditioning boards work inside control panels.

§08

Assessment

Answer all five questions. Each question tests a different objective from this lesson.

Q1 — A relay coil is rated 24 VDC with a coil resistance of 960 Ω. What is the current through the coil when energized?

A) 40 mA B) 400 mA C) 25 mA D) 4 A

Q2 — A 480 V, 30 A circuit has a measured voltage drop of 24 V across the conductors. What is the resistance of the conductors?

A) 144 Ω B) 16 Ω C) 0.8 Ω D) 1.25 Ω

Q3 — A 240 V baseboard heater is rated 2,000 W. What is its resistance?

A) 480 Ω B) 28.8 Ω C) 8.33 Ω D) 120 Ω

Q4 — #14 AWG conductors in a 120 VAC circuit have a combined resistance of 0.9 Ω. The circuit draws 8 A. What is the voltage drop across the conductors?

A) 1.35 V B) 7.2 V C) 133.3 V D) 0.11 V

Q5 — A 4.7 kΩ resistor and a 330 Ω resistor are connected in series. What is the total resistance?

A) 4.7 Ω B) 5,030 Ω C) 334.7 Ω D) 4,700 Ω

§09

Summary

The Core Insight

Ohm's Law is not just theory — it is a measurement interpretation tool. When voltage is lower than expected, resistance is higher than it should be. When current is lower than expected, something is open or resistance has increased. When a fuse blows, either the current exceeds its rating or the fuse was undersized from the start. Six equations tell you all of this, precisely, from your meter readings alone.

What You Covered in This Lesson

The six working equations derived from V = IR and P = VI
Unit prefix conversions — the most common source of calculation errors in the field
Solenoid current verification and how deviations indicate fault type (Scenario 1)
Fuse sizing and NEC 125% continuous load rule (Scenario 2)
Voltage drop diagnosis — converting a voltage measurement into a resistance fault value (Scenarios 3, 4)
Motor winding asymmetry detection using resistance comparison (Scenario 5)
Power supply load calculation and headroom analysis (Scenario 6)
Wire gauge selection from allowable voltage drop limits (Scenario 7)
3-phase motor current calculation including efficiency and power factor (Scenario 8)
Heating element resistance from power and voltage specifications (Scenario 9)
Five-step high-resistance fault diagnosis — a complete troubleshooting workflow (Scenario 10)
The voltage divider — Ohm's Law in sensing and signal conditioning applications

Key Numbers to Carry in Your Head

NEC 3% — max recommended voltage drop for branch circuits

NEC 125% — fuse/breaker sizing for continuous loads

746 W — 1 horsepower in electrical watts

√3 = 1.732 — 3-phase power constant

Up Next

Lesson 5.6 — Series and Parallel Circuits. You will apply Ohm's Law to multi-element circuits, calculate equivalent resistance for both configurations, and learn why parallel circuits behave counterintuitively when you add more loads.