Voltage, current and resistance

Objectives

By the end of this lesson you will:

Compute an unknown voltage, current or resistance in series and parallel circuits with Ohm’s law
Compute the power a resistor dissipates, and choose a suitable power rating

Before you start

A calculator, paper and a pencil. The concepts and practice do not need a board.
For the lab: a TESAIoT Dev Kit or Eva Kit board with a USB cable, a breadboard, jumper wires, one each of 1 kΩ, 2.2 kΩ and 4.7 kΩ resistors (standard 1/4 W), and a digital multimeter
Never used a multimeter before? That is fine — this lab uses only voltage mode and resistance mode. Full detail is in Module 3.

Safety. Everything in this lesson uses only the board’s 3.3 V supply. Never let a jumper wire connect the 3V3 pin straight to GND (a short circuit), and always unplug the USB cable before moving wires on the breadboard.

See it work first

Pick up the board and look at it. Find the USB connector (where power comes in), then find the pins labelled 3V3 (or 3.3V) and GND on the header. If the silkscreen labels are too small to read, open the board’s pinout diagram instead.

GND is the whole board’s reference point. Every time we say “this pin is 3.3 V,” we always mean “3.3 V above GND” — the same way a building’s height is measured from the ground, not from sea level.

Now guess, before reading on: if you connected one 1 kΩ resistor between 3V3 and GND, how much current would flow, and would the resistor get hot? Write your answer down, then compare it against what we calculate in the next section.

Concepts

1. Ohm’s law: three quantities, one equation

Voltage (V), in volts (V), is the push that makes charge move. It is always measured between two points.
Current (I), in amperes (A), is the amount of charge flowing past one point per second.
Resistance (R), in ohms (Ω), is whatever opposes the current.

All three are tied together by Ohm’s law:

V = I × R        I = V / R        R = V / I

On a microcontroller board, current is usually in milliamps (mA) and resistance is usually in kilohms (kΩ). Using the pair mA and kΩ gives you volts out directly, with no zeros to count.

Example: a 1 kΩ resistor across 3.3 V

I = V / R = 3.3 V / 1 kΩ = 3.3 mA

Change it to 10 kΩ and the current drops by a factor of ten, to 0.33 mA, or 330 µA.

What about a wire with only 0.1 Ω of resistance? The same law says the current wants to flow at 3.3 / 0.1 = 33 A. This is exactly why shorting the 3V3 pin to GND is dangerous: the board’s voltage regulator gets extremely hot or shuts itself off, and thin traces on the board can burn.

2. Series and parallel

Left: series — the same current flows through every part, and the voltage divides. Right: parallel — every part sees the same voltage, and the current splits between them.

Series: parts connected one after another in a single path.

The current is the same through every part
Total resistance R = R1 + R2 + …
Each part’s voltage adds up to the supply voltage (Kirchhoff’s voltage law, KVL)

Example: 1 kΩ in series with 2.2 kΩ from 3.3 V

R total = 1 kΩ + 2.2 kΩ = 3.2 kΩ
I       = 3.3 V / 3.2 kΩ = 1.03 mA
V_R1    = 1.03 mA × 1 kΩ   = 1.03 V
V_R2    = 1.03 mA × 2.2 kΩ = 2.27 V      check: 1.03 + 2.27 = 3.30 V

Parallel: parts connected across the same two points.

Every part sees the same voltage
Currents split apart, then add back together (Kirchhoff’s current law, KCL)
1/R = 1/R1 + 1/R2 + … ; for exactly two, the shortcut is R = (R1 × R2) / (R1 + R2)
The total resistance is always smaller than the smallest part — use this to sanity-check an answer quickly

Example: 1 kΩ in parallel with 2.2 kΩ at 3.3 V

R total = (1 × 2.2) / (1 + 2.2) kΩ = 0.6875 kΩ = 687.5 Ω
I1 = 3.3 V / 1 kΩ   = 3.3 mA
I2 = 3.3 V / 2.2 kΩ = 1.5 mA
I total = 4.8 mA                 check: 3.3 V / 687.5 Ω = 4.8 mA

3. Power and power rating

A resistor turns electrical energy into heat. The rate it heats up is power (P), in watts (W).

P = V × I = I² × R = V² / R

Pick the formula that matches what you know: know the voltage across it and the resistance, use V² / R; know the current and the resistance, use I² × R.

Every resistor has a power rating. The leaded type used on most breadboards is typically 1/4 W. The small SMD resistors on a board usually have a lower rating — always check the manufacturer’s datasheet. A common rule of thumb is to choose a rating at least twice the power you calculated, because a resistor running at its full rating gets very hot, and its rating drops further as the surrounding air gets hotter.

Example 1: the question from “See it work first” — 1 kΩ across 3.3 V

P = V² / R = (3.3 V)² / 1000 Ω = 0.01089 W ≈ 10.9 mW

Far below 1/4 W (250 mW). The resistor barely warms up at all.

Example 2: 100 Ω across 5 V

P = (5 V)² / 100 Ω = 0.25 W

Exactly at the 1/4 W rating. It works, but it will be too hot to touch, and its life will be short. Following the two-times rule, you would need 0.5 W, so pick a 1/2 W part.

Worked example

This is the actual circuit we will build in the lab. Work through the calculation with us, line by line.

A mixed circuit: R1 in series with the R2-parallel-R3 group. Point A is where we will calculate and measure the voltage relative to GND.

Problem: a 3.3 V supply feeds R1 = 1 kΩ to point A; from point A, R2 = 2.2 kΩ is in parallel with R3 = 4.7 kΩ down to GND. Find the voltage at point A, the current in every branch, and the power in every part.

Step 1: collapse the parallel part first

R2 ∥ R3 = (2.2 × 4.7) / (2.2 + 4.7) kΩ = 10.34 / 6.9 kΩ = 1.499 kΩ ≈ 1.50 kΩ

Step 2: now the circuit is just two parts in series

R total = 1 kΩ + 1.499 kΩ = 2.499 kΩ ≈ 2.50 kΩ
I       = 3.3 V / 2.499 kΩ = 1.32 mA          (the current through R1)
V_R1    = 1.32 mA × 1 kΩ = 1.32 V
V_A     = 3.3 V − 1.32 V = 1.98 V

Step 3: split the current across the two branches using the voltage at A

I2 = 1.98 V / 2.2 kΩ = 0.90 mA
I3 = 1.98 V / 4.7 kΩ = 0.42 mA
KCL check: 0.90 + 0.42 = 1.32 mA  matches the current through R1

Step 4: power

Part	Formula used	Power
R1	I² × R = (1.32 mA)² × 1 kΩ	1.74 mW
R2	V² / R = (1.98 V)² / 2.2 kΩ	1.78 mW
R3	V² / R = (1.98 V)² / 4.7 kΩ	0.83 mW
Total	check with V × I = 3.3 V × 1.32 mA	4.36 mW

The sum of the three parts (1.74 + 1.78 + 0.83 = 4.35 mW, differing from 4.36 due to rounding) equals the power the supply delivers. Every part is dozens of times below 1/4 W, so ordinary leaded resistors are comfortably fine here.

Notice the habit used throughout this example: once you finish calculating, always check with a second law (KVL, KCL, or total power). If the two numbers do not agree, some step is wrong.

Practice

Try these yourself first, then scroll down to check the solutions.

A 4.7 kΩ resistor sits across 3.3 V. What is the current, in mA and in µA?
An indicator LED circuit needs 2 mA from 3.3 V. What must the circuit’s total resistance be?
The total resistance of (a) two 10 kΩ resistors in parallel, (b) three 3.3 kΩ resistors in series
A 100 Ω resistor sits across 5 V. What power does it dissipate, and what power rating should you choose?
You measure 1.2 V across a 330 Ω resistor. What current flows through it?
A 3.3 V supply feeds 2.2 kΩ in series with a group of two 1 kΩ resistors in parallel. Find the total current, the voltage across the parallel group, and the current in each 1 kΩ resistor.

Solution

I = 3.3 V / 4.7 kΩ = 0.702 mA = 702 µA
R = V / I = 3.3 V / 2 mA = 1.65 kΩ
(a) 10 × 10 / (10 + 10) = 5 kΩ (b) 3 × 3.3 kΩ = 9.9 kΩ
P = 5² / 100 = 0.25 W. A 1/4 W part would run exactly at its rating; following the two-times rule, choose 1/2 W.
I = 1.2 V / 330 Ω = 3.64 mA. This is the safest way to measure current indirectly — we will use it again in Module 3.
1 kΩ ∥ 1 kΩ = 500 Ω. Total resistance = 2.2 kΩ + 0.5 kΩ = 2.7 kΩ. Total current = 3.3 V / 2.7 kΩ = 1.22 mA. Voltage across the parallel group = 1.22 mA × 0.5 kΩ = 0.611 V. Current in each part = 0.611 V / 1 kΩ = 0.611 mA (the two together give 1.22 mA, matching the total current).

Check your understanding

Answer at least 4 of the 5 questions in quiz.yaml correctly. For any you get wrong, go back and read the relevant section, then try the calculation again.

Lab

Lab goal: build the circuit from the worked example, measure the real voltages, and explain why they do not match the calculation exactly.

Measure the resistors before wiring anything. Set the multimeter to Ω mode and measure all three resistors one at a time, disconnected from anything else. Record their real values. Do not hold both leads with your fingers — your body’s resistance would end up in parallel with the reading.
Wire the circuit as shown, on the breadboard, with the board’s USB cable still unplugged. Five holes in the same breadboard column are connected together (details in the lesson Building a circuit on a breadboard). Connect the GND wire from the board first, then the 3V3 wire.
Guess before applying power: write the values you expect to measure into the “calculated” column of the table.
Plug in USB, then measure the voltages. Set the meter to DC voltage mode (V⎓); black lead in the COM jack touching GND, red lead in the VΩ jack touching the point being measured. Measure the real 3V3 voltage first, then the voltage at point A, then the voltage across R1 (red lead at 3V3, black lead at point A).
Compute the currents from the measured voltages. I1 = V_R1 / R1 (using the resistance you measured in step 1), I2 = V_A / R2, and I3 = V_A / R3, then check that I1 ≈ I2 + I3.
Unplug USB before taking the circuit apart.

Item	Calculated (nominal values)	Recalculated (measured values)
3V3 voltage	3.30 V	use your measured value
V_A	1.98 V
V_R1	1.32 V
I1 = V_R1 / R1	1.32 mA
I2 + I3	1.32 mA

Interpreting the results: gold-band resistors carry a ±5% tolerance, and the real 3V3 rail may not be exactly 3.30 V. If your measured values differ from the “recalculated” column by no more than about 1–2%, that counts as a match. If the difference is larger, find the cause before moving on. Common causes are plugging into the wrong row, mixing up which resistor is which, or a broken jumper wire.

Going further

The next lesson, Voltage dividers and analogue sensors, picks up right where point A leaves off — it is itself a kind of voltage divider. We will use the same principle to explain a knob on the board, and to convert the number the ADC reads back into volts.

Reflect

When you guessed in “See it work first,” did you think the 1 kΩ resistor would get hot? How did the real answer change how you think about “3.3 V power” now?

References

Review questions

Answer on your own first, then open the answer.

A 2.2 kΩ resistor is connected across 3.3 V. What current flows? (Objective 1)
1. 1.5 mA
2. 7.26 mA
3. 0.67 mA
4. 1.5 A
Show answer

Answer: A. 1.5 mA

I = V / R = 3.3 V / 2.2 kΩ = 1.5 mA ใช้คู่หน่วย V กับ kΩ จะได้ mA ออกมาตรง ๆ ข้อ 7.26 mA มาจากการเอาแรงดันคูณความต้านทาน ซึ่งผิดสูตร
A 1 kΩ resistor in parallel with 3.3 kΩ. What is the combined resistance? (Objective 1)
1. 4.3 kΩ
2. 767 Ω
3. 2.15 kΩ
4. 1.15 kΩ
Show answer

Answer: B. 767 Ω

(1 × 3.3) / (1 + 3.3) kΩ = 0.767 kΩ ความต้านทานขนานต้องน้อยกว่าตัวที่น้อยที่สุด (1 kΩ) เสมอ จึงตัดตัวเลือกอื่นได้ทันที
Which statements are true for two resistors in series with a supply? (choose all that apply) (Objective 1)
1. กระแสที่ไหลผ่านทั้งสองตัวเท่ากัน
2. แรงดันคร่อมทั้งสองตัวบวกกันได้แรงดันแหล่งจ่าย
3. ความต้านทานรวมน้อยกว่าตัวที่น้อยที่สุด
4. แรงดันคร่อมทั้งสองตัวเท่ากันเสมอ
Show answer

Answer: A. กระแสที่ไหลผ่านทั้งสองตัวเท่ากัน · B. แรงดันคร่อมทั้งสองตัวบวกกันได้แรงดันแหล่งจ่าย

อนุกรมมีทางเดินกระแสทางเดียว กระแสจึงเท่ากัน และแรงดันแบ่งกันตามสัดส่วนความต้านทาน (KVL) ความต้านทานรวมน้อยกว่าตัวที่น้อยที่สุดเป็นสมบัติของวงจรขนาน
A 47 Ω resistor sits across 3.3 V. Which power rating should you choose with at least a 2× margin? (Objective 2)
1. 1/8 W
2. 1/4 W
3. 1/2 W
4. เลือกอะไรก็ได้ ไฟ 3.3 V ไม่ทำให้ตัวต้านทานร้อน
Show answer

Answer: C. 1/2 W

P = V² / R = 3.3² / 47 = 0.232 W เผื่อสองเท่าได้ 0.46 W จึงต้องใช้ 1/2 W ตัว 1/4 W (0.25 W) ทำงานเกือบเต็มพิกัดและร้อนมาก แรงดันต่ำไม่ได้แปลว่ากำลังต่ำเสมอ ถ้าความต้านทานต่ำพอ
You know the current through a resistor and its resistance. Which formula gives the power directly? (Objective 2)
1. P = V / I
2. P = I² × R
3. P = R / I
4. P = V² × R
Show answer

Answer: B. P = I² × R

P = V × I และแทน V = I × R จะได้ P = I² × R ถ้ารู้แรงดันคร่อมกับความต้านทานให้ใช้ P = V² / R แทน

Cite this lesson

If you teach from this lesson or reuse it in slides or documents, credit it with the text below. If you changed it, add (adapted) after the title.

"Voltage, current and resistance" from TESA Open Knowledge by the Thai Embedded Systems Association (TESA), https://github.com/tesaiot/tesa-qualification-program, licensed under CC BY-NC 4.0

Thai attribution: "แรงดัน กระแส และความต้านทาน" จาก TESA Open Knowledge โดยสมาคมสมองกลฝังตัวไทย (Thai Embedded Systems Association: TESA) https://github.com/tesaiot/tesa-qualification-program สัญญาอนุญาต CC BY-NC 4.0

Lesson link: https://tesaiot.github.io/tesa-qualification-program/en/courses/electronics-and-instruments/m01-circuits/l01-voltage-current-resistance/

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