Ohm's Law

The One-Sentence Version

Ohm's law says that how much current flows through something depends on how hard you push it (voltage) divided by how much it fights back (resistance).

That's it. Everything below is just that sentence, in more detail, applied to the parts sitting on your desk right now.

Written as a formula:

V = I × R

| Symbol | Unit | Name | What it actually means | |---|---|---|---| | V | Volts (V) | Voltage | How hard the electricity is being pushed | | I | Amps (A) | Current | How much electricity is actually flowing | | R | Ohms (Ω) | Resistance | How much the component fights that flow |

Why It Feels Abstract (and the fix)

You can't see any of these three things. So use water — the analogy is not a simplification, it's genuinely how the math behaves.

Picture a water tank feeding a pipe:

| Electricity | Water | If you increase it… | |---|---|---| | Voltage | Pressure in the tank | More pressure → more flow | | Current | Litres per second through the pipe | This is the thing you're changing | | Resistance | How narrow the pipe is | Narrower pipe → less flow |

Now Ohm's law reads like common sense: more pressure means more flow; a narrower pipe means less flow. Double the pressure and you double the flow. Double the narrowness and you halve it.

The one trap: voltage is not "electricity that gets used up." Voltage is a difference between two points — like the height difference in a waterfall. That's why you always measure voltage across something, and current through it.

The Three Forms

The same equation, rearranged. You'll use all three:

| You want to find… | Use | Typical question | |---|---|---| | Voltage | V = I × R | "How many volts will this drop?" | | Current | I = V / R | "Am I about to fry this LED?" | | Resistance | R = V / I | "Which resistor do I need?" |

If algebra isn't your thing: cover the letter you want with your thumb, and what's left is the formula.

Your Kit Has Three Resistor Values

Your kit ships with 30 resistors in three values. Each one exists because of an Ohm's law calculation someone already did for you:

| Value | Color bands | Why it's in the kit | |---|---|---| | 220 Ω | Red · Red · Brown · Gold | Current-limiting a single LED (Day 2 onward) | | 330 Ω | Orange · Orange · Brown · Gold | Current-limiting the RGB LED's three legs | | 10 kΩ | Brown · Black · Orange · Gold | Pull-downs for the DIP switch, and the photoresistor divider |

Reading the bands: first two bands are digits, the third is how many zeros to add, gold is the ±5% tolerance. Hold the resistor with the gold band on the right and read left to right. Red-Red-Brown = 2, 2, add one zero = 220. Brown-Black-Orange = 1, 0, add three zeros = 10,000.

Resistors are not polarized — there's no backwards. Either leg can face either way.

Worked Example: The Circuit On Your Breadboard Right Now

Day 2's cabin light is pin 12 → 220 Ω resistor → LED → ground. Let's prove the 220 Ω is the right choice.

What we know:

• The HERO board drives a pin at 5 V
• A red LED "drops" about 2 V across itself (this is fixed by physics, not by you)
• The resistor is 220 Ω

Step 1 — how many volts are left for the resistor?

5 V (supply) − 2 V (the LED takes this) = 3 V across the resistor

Step 2 — use I = V / R:

I = 3 V / 220 Ω = 0.0136 A = 13.6 mA

Step 3 — is that safe? A standard LED is happiest at 10–20 mA, and the board's pins are rated for 20 mA continuous (40 mA absolute maximum). 13.6 mA is comfortably inside both limits — bright light, no damage. That's the whole reason the resistor is in the kit.

What Happens If You Get It Wrong

This is the real test of whether you understand it. Same circuit, different resistor:

| Resistor | Current (I = 3 V / R) | Result | |---|---|---| | No resistor | 3 V / ~0 Ω → hundreds of mA | LED burns out, possibly the pin with it | | 220 Ω | 13.6 mA | Correct. Bright and safe | | 1 kΩ | 3 mA | Works, noticeably dimmer | | 10 kΩ | 0.3 mA | Barely visible glow |

Notice the pattern: resistance up, current down — every time, proportionally. That's Ohm's law doing its job.

This is also the rule of thumb worth memorising: when in doubt, go bigger. A too-large resistor costs you brightness. A too-small one costs you the component.

The Same Math, Two More Missions

The DIP switch pull-down (Day 3). A 10 kΩ resistor sits between the input pin and ground. When the switch closes, 5 V appears across it:

I = 5 V / 10,000 Ω = 0.0005 A = 0.5 mA

Half a milliamp — essentially free. That's why it's 10 kΩ: big enough to waste almost no current, small enough to hold the pin at a solid LOW when the switch is open. A plain wire (0 Ω) would short 5 V straight to ground.

The photoresistor divider (Days 6–7). Wired 5V — photoresistor — A0 — 10kΩ — GND, the two resistances split the 5 V between them in proportion to their size. As light changes the photoresistor's resistance, the voltage at A0 moves with it:

| Light level | Photoresistor ≈ | Voltage at A0 ≈ | |---|---|---| | Bright | 1 kΩ | 5 × 10/(10+1) = 4.5 V | | Dim | 10 kΩ | 5 × 10/(10+10) = 2.5 V | | Dark | 100 kΩ | 5 × 10/(10+100) = 0.45 V |

Nothing new is happening here — it's Ohm's law twice, in series. This is how a sensor that only changes resistance becomes a number your code can read.

Power: The Other Formula

Voltage times current gives you power — how fast energy turns into heat and light:

P = V × I (watts)

For a resistor specifically: P = I² × R

Your kit's resistors are rated ¼ watt (0.25 W). The hardest-working one in the whole course is that 220 Ω on Day 2:

P = (0.0136)² × 220 = 0.04 W

About a sixth of its rating. Nothing in this course will ever cook a resistor — but on bigger projects, this is the calculation that tells you whether a part needs to be physically larger.

Check Yourself

You've got this if you can answer these without scrolling up:

1. Say Ohm's law in your own words, without using the letters V, I, or R.
2. Your LED circuit has 3 V across a 220 Ω resistor. What's the current?
3. You swap in a 1 kΩ resistor. Does the LED get brighter or dimmer — and roughly by how much?
4. Why does the 220 Ω resistor need to be there at all?

(Answers: 1 — flow equals push divided by opposition. 2 — 3/220 = 13.6 mA. 3 — dimmer, about 4½× less current. 4 — without it, nothing limits the current and the LED draws far more than the ~20 mA it can survive.)