Electronics · Math shelf · Optional — for the curious · build 2026.09.23-1644
V = IR
The three quantities
Symbol
Name
Unit
You've measured it
V
voltage
volts (V)
meter across two points, DC volts mode
I
current
amps (A) — ours are small, so usually milliamps (mA)
teacher demo only (the mA jack)
R
resistance
ohms (Ω)
meter on Ω, part out of the circuit
The law
V = I × R. The voltage drop across a resistance equals the current through it, times the resistance. Know any two, and the third has no choice.
Three ways to hold it
The same law rearranges to answer three different questions:
You know
You want
Use
current and resistance
the voltage drop
V = I × R
voltage and resistance
the current
I = V ÷ R
voltage and current
the resistance
R = V ÷ I
The triangle trick: cover the one you want. Cover V → I next to R means multiply. Cover I → V over R means divide. Cover R → V over I.
Get the units straight first
The law works when the units agree: volts with amps and ohms. Our currents are milliamps, so convert before plugging in: 14 mA = 0.014 A.
The mA · kΩ shortcut
Milli (÷1,000) and kilo (×1,000) cancel each other, so the law also works directly in the units our circuits speak: mA × kΩ = V, V ÷ kΩ = mA, V ÷ mA = kΩ. Example: 1 mA through 4.7 kΩ drops 1 × 4.7 = 4.7 V. (The Milli, Kilo, Mega sheet on this shelf covers the prefixes.)
Worked examples — your actual circuit
How much current lights the LED? The classic loop: 5 V supply, red LED, 220 Ω resistor. The LED drop is ≈ 2.0 V, so the resistor gets the rest: 5.0 − 2.0 = 3.0 V across 220 Ω. Current: I = V ÷ R = 3.0 ÷ 220 ≈ 0.014 A ≈ 14 mA. That's where the "≈ 14 mA" in lesson 1 comes from — now it's yours.
Pick a resistor. You want a gentler 10 mA through that red LED. The resistor still has to absorb 3.0 V: R = V ÷ I = 3.0 ÷ 0.010 = 300 Ω. No 300 Ω in the kit? Round up to the next size you have — more resistance means less current, and less current is the safe direction for an LED.
Predict a drop. 2 mA flows through a 1 kΩ resistor. Shortcut units: V = 2 mA × 1 kΩ = 2 V. If your meter reads something very different across that resistor, the circuit — or the prediction — has a bug worth finding.
Now you know why
Lesson 4's mystery: a resistor reading 0 V means I = 0 — because V = I × R, and anything times zero is zero. No drop, no flow.
Lesson 5's dimmer: turning the pot raises R, and I = V ÷ R says the current must fall. Bigger squeeze, smaller flow, dimmer LED.
Why 220 Ω: it lands a red LED near 14 mA — bright, and comfortably inside what the LED can take.
Practice
Answers at the bottom — predict before you peek, and check any you can with the meter.
A blue LED drops 3.2 V in the classic 5 V loop. How much voltage does the 220 Ω resistor get, and what current flows?
3 mA flows through a 2.2 kΩ resistor. What's the drop across it?
A mystery resistor has 4.0 V across it while 8 mA flows through. What is it?
You measure 5.0 V across a resistor and 0.0 mA through it. What does V = IR say the resistance is — and what's actually going on?
Check yourself with the meter
Problem 1 is buildable: blue LED, 220 Ω, 5 V. Measure the resistor's drop and compare to your math. Real components are ±5 % or so — if you're within a few tenths of a volt, you and Ohm agree.
Answers
1. 5.0 − 3.2 = 1.8 V across the resistor; I = 1.8 ÷ 220 ≈ 8 mA — real reason blue LEDs run dimmer in the same loop. 2. 3 mA × 2.2 kΩ = 6.6 V. 3. R = 4.0 V ÷ 8 mA = 0.5 kΩ = 500 Ω. 4. The formula says R = 5.0 ÷ 0 — division by zero, undefined. That's the math's way of saying the path is broken: an open circuit is effectively infinite resistance.