Ohm’s Law Calculation Audit

Ohm’s Law Calculation Audit

Follow voltage, current, resistance, and power from a real circuit to a safe component choice

foundations
math-foundations
calculation-audit
electronics
beginner
Ada Ada · calculation audit
Start with the unresolved circuit

Why can one resistor be the difference between a useful LED and a damaged output?

You connect a red indicator LED to a 3.3 V general-purpose input/output (GPIO) pin. The LED needs about 2.0 V, but the pin supplies 3.3 V. Where does the remaining voltage go, and how do you stop too much current flowing?

Ada’s answer is to make the physical path visible first, then use Ohm’s law. The same method scales from this 5 mA indicator to the chapter’s 12 V lamp, where a wrong assumption means 4 A and 48 W rather than a small change in brightness.

Companion to Ohm’s Law and Power. Every value below is retold from that chapter’s examples.

See the relationship before changing it

The figure reads from left to right. The blue card is lamp-path resistance. The middle card applies this page's rule. The green card is lamp current. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only lamp-path resistance, so the numeric fixture does not switch without explanation.

Lamp-path resistance changes lamp current An input card leads through the rule current = 12 V / resistance to the lamp current result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. More resistance reduces current for the same twelve-volt source.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 3 ohm.

  2. 2

    Name the relationship. current = 12 V / resistance

  3. 3

    Substitute with units. 12 V / 3 ohm = 4.00 A

  4. 4

    Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change lamp-path resistance

Try Predict the direction of current = 12 V / resistance. Test another lamp-path resistance, then compare lamp current.

3 ohm
Chapter baseline
Lamp current

Observe More resistance reduces current for the same twelve-volt source. Reset lamp-path resistance to 3 and compare lamp current.

Explain At 3 ohms, the 12 V source drives 4 A; increasing resistance lowers that current.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only lamp-path resistance moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only lamp-path resistance. It is not a field prediction.

Try

Trace the 3.3 V path, then move the resistor slider.

Observe

Above 260 ohms, a larger lamp-path resistance makes current and resistor heating fall together.

Explain

Lamp-path resistance uses the 1.3 V left after the LED’s drop and limits how quickly charge flows.

See the voltage before doing the algebra

The diagram follows the only closed path: the pin raises the electrical potential to 3.3 V, the resistor drops 1.3 V, and the red LED drops about 2.0 V before the path returns to ground. Those two drops add back to the supply: 1.3 V + 2.0 V = 3.3 V.

Voltage and current through a GPIO resistor and red LED A closed circuit path runs from a 3.3 volt GPIO through a resistor with a 1.3 volt drop, then through a red LED with a 2 volt drop, and back to ground. A teal arrow shows 5 milliamps around the loop. GPIO pin 3.3 V 260 Ω resistor drops 1.3 V red LED drops 2.0 V the same 5 mA flows around the whole loop
Read the loop clockwise. Current does not get “used up” between components. The same 5 mA flows through both; the supply voltage is shared as two physical drops.

Derive the resistor in four named moves

  1. 1

    Name the resistor’s voltage. Subtract the LED’s 2.0 V drop from the 3.3 V pin: VR = 3.3 V − 2.0 V = 1.3 V

  2. 2

    Put current in base units. “Milli” means one-thousandth, so 5 mA = 5/1000 A = 0.005 A.

  3. 3

    Make resistance the subject. Start with V = I × R, then divide both sides by I: R = V / I

  4. 4

    Substitute numbers with units. R = 1.3 V / 0.005 A = 260 V/A = 260 ΩOne volt per ampere is one ohm, so the units close as well as the number.

Now check heat. Power is voltage multiplied by current: P = 1.3 V × 0.005 A = 0.0065 W = 6.5 mW. That is the chapter’s hand-derived worked example, with the conversion shown rather than hidden.

Move one value: what does resistance buy?

Try Keep the chapter’s 1.3 V resistor drop fixed. Move resistance from 150 Ω to 680 Ω and predict whether current rises or falls.

260 Ω
Resistance
Current, I = V/R
Resistor power, P = VI
5 mA target

Ready: predict the current, then calculate.

Observe At 260 Ω the calculator returns 5.00 mA and 6.50 mW, exactly matching the derivation. Doubling resistance approximately halves current.

Explain With voltage fixed, R sits in the denominator of I = V/R. Making the denominator larger makes the current smaller.

Carry the same method into the chapter’s larger loads

Physical question Named move Worked result
How much of a 5 V sensor signal may reach a 3.3 V ADC? Form the required divider ratio. Vout/Vin = 3.3/5 = 0.66.
What does a 12 V source drive through a 3 Ω lamp path? Divide voltage by resistance. I = 12 V / 3 Ω = 4 A.
How quickly does that lamp convert energy? Multiply voltage by current. P = 12 V × 4 A = 48 W.

The equation is unchanged, but the physical decision changes. At 48 W, source capacity, switch rating, wire, fuse, connector, and heat margin all matter before power is applied.

Check yourself

Why is 3.3 V / 5 mA not the right LED-resistor calculation?
Answer: The LED already drops about 2.0 V. Only the remaining 1.3 V appears across the resistor.
If the resistor becomes 520 Ω while its voltage stays 1.3 V, what current should you expect?
Answer: I = 1.3 V / 520 Ω = 0.0025 A = 2.5 mA.
Why must the 12 V lamp example trigger more checks than the LED?
Answer: Its 4 A current and 48 W power make conductor, switch, fuse, connector, and heat limits part of the design.
Technical boundaries. These nominal direct-current examples do not model resistor tolerance, LED forward-voltage spread, GPIO output sag, self-heating, nonlinear loads, contact resistance, or startup transients. Calculate first, then verify the real circuit.

What the audit buys you: you can now read V = IR as a physical story—voltage pushes, resistance limits, current flows—and carry every unit into a component and safety decision.

Every number above is taken from the companion chapter and re-derived step by step.