5.1 Ohm's Law

5.1 Ohm's Law

Quick Answer: For a resistive DC element, V = I × R. Solve for current with I = V / R and for resistance with R = V / I. Convert every quantity to volts, amperes, and ohms before you multiply or divide, then sense-check the magnitude against typical aircraft values such as a 28 V DC bus.

Ohm’s law is the calculation backbone of CAAS SAR-66 Module 3 topic 3.6 DC Circuits. Almost every later network problem—series strings, parallel branches, voltage dividers, and internal-resistance loading—starts by applying V = IR to one element or to an equivalent resistance. If this relationship is automatic, Kirchhoff and combination circuits become bookkeeping. If it is shaky, the rest of Module 3 collapses under time pressure.

The Three Forms

These three forms are algebraically identical. On the exam, choose the form that isolates the unknown in one step. Do not rearrange under stress if the printed form already matches the stem.

Physical meaning: For a metallic conductor at constant temperature, current is proportional to the applied potential difference. Doubling V at fixed R doubles I. Doubling R at fixed V halves I. Resistance is the constant of proportionality: R = V/I.

Unit Discipline (Where Marks Are Lost)

Module 3 stems freely mix prefixes:

Rule: Convert first, calculate second, then optionally convert the answer back to the unit the options use. Mixing 28 V with 100 mA without converting yields a “resistance” of 0.28 Ω when the true value is 280 Ω.

Worked Example 1 — Find Current

A landing-light control resistor of 14 Ω is connected across a 28 V DC bus (ideal source for this item).

I = V / R = 28 / 14 = 2 A.

Sanity check: 28 V and a low double-digit ohm value should produce amperes, not milliamperes. If your calculator shows 0.002 A, you likely treated 28 as millivolts or 14 as kilohms.

Worked Example 2 — Find Voltage Drop

A feeder carries 15 A through a total cable-plus-joint resistance of 0.08 Ω.

V = I × R = 15 × 0.08 = 1.2 V.

The bus may still be near 28 V at the source, but the load end is 1.2 V lower because of that feeder drop. Ohm’s law applies to any resistive segment, not only to the named load.

Worked Example 3 — Find Resistance

A heater draws 3.5 A from a 115 V DC shop supply (DC for this practice item).

R = V / I = 115 / 3.5 ≈ 32.86 Ω (about 33 Ω if options are rounded).

If options include 32.9 Ω, 0.030 Ω, and 402.5 Ω, reject inverted V/I or I×V mistakes immediately.

Worked Example 4 — Mixed Prefixes

A sensor sees 24 V across 4.7 kΩ.

Convert: R = 4700 Ω.

I = 24 / 4700 ≈ 0.00511 A = 5.11 mA.

Exam options often list milliamperes. Leaving the answer as 0.00511 A is correct but may not match the printed choice format—read the units on each option.

Worked Example 5 — Two-Step with Partial Data

A series string has total R_T = 120 Ω on a 24 V supply. What is the current, and what voltage appears across a 40 Ω portion of that string?

1. I = V / R_T = 24 / 120 = 0.2 A (same current through every series element).
2. V_40 = I × 40 = 0.2 × 40 = 8 V.

You did not need a separate law for the second step—Ohm’s law on the segment after finding the series current.

Proportional Reasoning (Exam Traps)

Trap 1 — Doubling V and R

If V doubles and R doubles, I is unchanged because I = V/R. Candidates who “feel” that more voltage means more current, while forgetting the matching resistance change, pick “I doubles.”

Trap 2 — Inverse vs Direct

At fixed V, if R is reduced to one-third, I becomes three times larger—not one-third. Inverse proportion means the product I × R stays equal to V.

Trap 3 — Confusing Ohm with Power

Power P = VI = I²R = V²/R is related but not a substitute for Ohm’s law. A stem that gives only P and V still needs I = P/V or R = V²/P before you talk about “Ohm.” Do not invent V = P × R.

Trap 4 — Parallel intuition on a single resistor

Ohm’s law on one resistor does not care about other branches until you have found the voltage across that resistor. Applying supply voltage to a branch that is not actually at supply potential (for example, a resistor mid-string) produces wrong current.

Aircraft Context as a Sanity Check

Many large-aircraft DC systems centre on roughly 28 V distribution. If a simple Ohm’s-law item about a bus-powered lamp returns 2800 V or 0.0028 V, you misplaced a decimal or a kilo prefix. Use context to catch errors; do not replace the formula with “it must be 28 V” when the stem gives different numbers.

Method Checklist for Module 3

1. Underline the unknown (V, I, or R).
2. Convert all given values to V, A, Ω.
3. Select V = IR, I = V/R, or R = V/I.
4. Substitute and calculate.
5. Convert the result to the option’s unit if needed.
6. Sense-check magnitude and reject inverted formulae.

Master these six steps until they take under fifteen seconds. Topic 3.6 then becomes Kirchhoff and network topology on top of a reliable Ohm engine—not three separate subjects fighting for working memory.