Ohm's Law in Practice: V = IR, Power & Safety
Five beginner-friendly lessons that take you from the idea behind voltage, current and resistance to sizing real resistors, checking power ratings and solving practical DC problems. Every lesson has worked examples, an interactive tool and a quiz.
Lesson 1 · Understanding Ohm's Law
Level: Beginner
- Name the three quantities — voltage, current, resistance — and their units.
- State what V = IR means physically and why it works.
- Rearrange the formula to find V, I or R, using the Ohm triangle.
Every DC circuit is a story about three characters. Voltage (symbol V, unit the volt) is the electrical "push" that makes charge move — think of it as water pressure. Current (symbol I, unit the ampere or amp) is the actual flow of charge, like the flow rate of water through a pipe. Resistance (symbol R, unit the ohm, Ω) is how much a component opposes that flow, like a narrow section of pipe. Change any one of these three and the other two respond in a predictable way — that predictability is Ohm's Law.
Ohm's Law is the relationship between those three characters, written as V = I × R: the voltage across a resistor equals the current through it multiplied by its resistance. Because it is a simple proportionality, you can rearrange it to find whichever quantity you don't yet know. The triangle below is the classic memory aid — cover the value you're solving for and read off the operation: I = V ÷ R and R = V ÷ I.
The water model makes the behavior obvious. Raise the water level (more voltage) and more water flows (more current). Squeeze the pipe (more resistance) and less flows. That is exactly what the formula says: doubling the voltage doubles the current, and doubling the resistance halves it. Hold onto that intuition — it turns every calculation in this module from memorized arithmetic into something you can reason about and sanity-check.
| Quantity | Symbol | Unit | What it is |
|---|---|---|---|
| Voltage | V | volt (V) | The push that drives current |
| Current | I | ampere (A) | The flow of charge |
| Resistance | R | ohm (Ω) | Opposition to current |
| Power | P | watt (W) | Rate of energy transfer (Lesson 3) |
Check A 9 V battery is connected to a 3 Ω resistor. What current flows?
I = V ÷ R = 9 ÷ 3 = 3 A.
Use the "cover the I" form of the triangle: current is voltage divided by resistance.
If you double the voltage across a fixed resistor, the current…
With R fixed, I = V/R, so current is directly proportional to voltage. Double V → double I.
Lesson 2 · Applying V = IR
Level: Beginner
- Solve Ohm's Law problems in a clear, repeatable order.
- Use units and SI prefixes (mA, kΩ, MΩ) correctly.
- Spot unrealistic answers by checking units and magnitude.
Now we solve problems properly, in the same order every time. First, write down the knowns with their units. Second, choose the right form of the formula — the one that has the unknown on its own. Third, substitute the numbers. Fourth, calculate. Fifth, check: does the unit make sense, and is the size of the answer reasonable? That last step catches most mistakes — a "3,000,000 A" through a tiny resistor is a red flag, not an answer.
Units matter as much as the math. Voltage is in volts, current in amperes, resistance in ohms. In real circuits you'll rarely see plain amps or ohms — small currents are in milliamps (mA, 1 mA = 0.001 A) and large resistances in kilo-ohms (kΩ, 1 kΩ = 1,000 Ω) or mega-ohms (MΩ, 1 MΩ = 1,000,000 Ω). Before you divide or multiply, convert everything to base units (V, A, Ω), do the math, then convert the answer back to a friendly form.
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| milli | m | 10⁻³ | 1 mA = 0.001 A |
| kilo | k | 10³ | 1 kΩ = 1,000 Ω |
| mega | M | 10⁶ | 1 MΩ = 1,000,000 Ω |
- Knowns: V = 120 V, I = 600 mA. Find R.
- Convert & choose: 600 mA = 0.6 A. Use R = V ÷ I.
- Substitute & calculate: R = 120 ÷ 0.6 = 200 Ω.
- Check: unit is ohms ✓, and 200 Ω is a sensible size for a 120 V, 0.6 A load ✓.
Ohm's Law solver
Enter any two of V, I, R — the third is solved and the power is shown. Change values to watch the current and power bars respond.
Check A device draws 150 mA at 5 V. What is its effective resistance?
Convert: 150 mA = 0.15 A. R = V ÷ I = 5 ÷ 0.15 ≈ 33.3 Ω.
Always convert mA to A before dividing, then the result is in ohms.
What is 4.7 kΩ expressed in ohms?
kilo = 1,000, so 4.7 kΩ = 4.7 × 1,000 = 4,700 Ω.
Lesson 3 · Electrical Power
Level: Beginner → Intermediate
- Define electrical power and read P = VI.
- Derive P = I²R and P = V²/R by substituting Ohm's Law.
- Choose the power formula that matches the values you have.
Electrical power refers to the rate at which electrical energy is transferred and is measured in watts (W). The basic equation to remember is P = V × I, which tells us that the power either delivered to or absorbed by a component is equal to the voltage across it multiplied by the current flowing through it. For example, if you have a 12-volt supply providing 2 amps, the power output is 24 watts. This equation is particularly useful when you know both the voltage and the current values.
Ohm's Law connects voltage (V), current (I), and resistance (R), allowing us to express power in two additional ways that are useful when not all three values are available. If you substitute I = V/R into P = V·I, you arrive at P = V² ÷ R, which is handy when the voltage and resistance are known. Alternatively, if you substitute V = I·R into P = V·I, you get P = I² × R, which is useful when you know the current and resistance. All three expressions represent the same power; simply choose the formula that corresponds to the two quantities you already have, making your calculations simpler.
| Formula | Best used when you know… | Quick note |
|---|---|---|
| P = V·I | voltage and current | The defining form |
| P = I²·R | current and resistance | Power grows with the square of current |
| P = V²/R | voltage and resistance | More voltage or less resistance → more power |
A 9 V battery drives a 3 Ω resistor.
- First find the current: I = V ÷ R = 9 ÷ 3 = 3 A.
- By P = V·I: P = 9 × 3 = 27 W.
- Check with P = I²R: 3² × 3 = 9 × 3 = 27 W ✓.
- Check with P = V²/R: 9² ÷ 3 = 81 ÷ 3 = 27 W ✓. All agree.
Power calculator
Enter any two of V, I, R. Every valid formula is highlighted with its result — use the one that matches your knowns.
Check A 5 V supply drives 100 mA through a resistor. What is the power, using P = V·I?
100 mA = 0.1 A. P = V·I = 5 × 0.1 = 0.5 W (500 mW).
You can confirm with the other forms: R = 5/0.1 = 50 Ω, so P = I²R = 0.01 × 50 = 0.5 W, and P = V²/R = 25/50 = 0.5 W.
You know the current and the resistance. Which formula is the most direct for power?
With I and R known, P = I²R needs no extra step. (P = V·R is not a valid power formula.)
Lesson 4 · Power Dissipation & Circuit Safety
Level: Beginner → Intermediate
- Explain what "dissipation" means and why resistors get warm.
- Read a resistor's power rating and apply a safe margin.
- Decide whether a part is safe for a given circuit.
A resistor doesn't store the energy that flows through it — it converts it into heat. That conversion is power dissipation, and its rate is exactly the P you calculated in Lesson 3 (P = I²R). A little warmth is normal and harmless; too much is not. Every resistor is built to safely dissipate a maximum power, printed or implied by its size: common ratings are ¼ W, ½ W, 1 W, 2 W and larger. If you force more power through it than it's rated for, the resistor overheats — its value drifts, it can burn, smoke, or even start a fire.
The practical rule is to leave a margin: don't run a resistor near its limit. A common guideline is to keep the actual dissipation to about half (or at most ~70%) of the rating, so the part stays cool, accurate and long-lived. This matters in real designs — a ¼ W resistor is fine for a tiny LED circuit but far too small for a resistor dropping several volts at high current. Always compute P, compare it to the rating, and when in doubt, choose the next size up.
| Rating | Typical part | Good for |
|---|---|---|
| ¼ W (0.25) | Small through-hole / most SMD | LEDs, low-current signals |
| ½ W (0.5) | Medium through-hole | Light loads, prototypes |
| 1 W | Larger metal-film / axial | General power dropping |
| 2 W | Big axial | Higher current / voltage drops |
| 5 W + | Wirewound / ceramic | Heavily loaded circuits |
Picking a resistor by its value alone. A 100 Ω resistor that drops 9 V carries 90 mA and dissipates 0.81 W — a ¼ W 100 Ω part would overheat even though the value is correct. Check both value and power.
Resistor power-rating checker
Enter the voltage across a resistor, its resistance and its rating. The checker computes the dissipation and tells you if the rating is safe.
Check A 100 Ω resistor has 9 V across it. What power does it dissipate, and is a ¼ W rating safe?
I = 9/100 = 0.09 A; P = V²/R = 81/100 = 0.81 W. A ¼ W (0.25 W) rating is not safe — 0.81 W is more than three times it. Use at least 1 W (2 W for a comfortable margin).
This is the classic "right value, wrong power" trap — always check the wattage.
What happens if you push more power through a resistor than its rating?
Exceeding the rating causes overheating: the value drifts and the part can fail catastrophically. Ratings are hard limits, not suggestions.
Lesson 5 · Real-World Problem Sets
Level: Beginner → Intermediate — challenge
- Apply V = IR, power and dissipation to realistic scenarios.
- Solve progressively harder problems and check your answers.
- Choose resistor values and verify power ratings in context.
Time to put it all together. The problems below grow in difficulty: find a current, size a resistor for an LED, compute a power draw, and judge a power rating. Type your answer and hit Check for instant feedback; if you're stuck, reveal the worked solution. Treat each one as you would a real bench problem — identify the knowns, pick the formula, convert units, calculate, then sanity-check.
Problem 1 · Current draw A 12 V source powers a 6 Ω heater. What current does it draw?
I = V/R = 12/6 = 2 A.
Problem 2 · Choosing a resistor A 5 V supply drives an LED (forward voltage 2 V) at 20 mA. What series resistor do you need?
The resistor drops 5 − 2 = 3 V at 20 mA (0.02 A): R = 3/0.02 = 150 Ω. Its power is P = I²R = 0.0004 × 150 = 0.06 W, so a ¼ W part is comfortable.
Problem 3 · Power consumption A 12 V motor draws 1.5 A. What is its power consumption, and what is its effective resistance?
P = V·I = 12 × 1.5 = 18 W. Effective R = V/I = 12/1.5 = 8 Ω.
Problem 4 · Rating adequacy A 220 Ω resistor has 11 V across it and is rated ½ W. First find the power it dissipates.
P = V²/R = 11²/220 = 121/220 = 0.55 W. That exceeds the ½ W (0.5 W) rating, so it is not adequate — the part would overheat. Use a 1 W resistor.
| # | Scenario | Given | Answer |
|---|---|---|---|
| 1 | Current draw | 12 V, 6 Ω | 2 A |
| 2 | LED series resistor | 5 V, Vf 2 V, 20 mA | 150 Ω (0.06 W) |
| 3 | Power draw | 12 V, 1.5 A | 18 W, 8 Ω |
| 4 | Rating check | 11 V, 220 Ω, ½ W | 0.55 W → not adequate |
A 9 V battery is connected directly across a 100 Ω resistor. Roughly what power is dissipated?
P = V²/R = 81/100 = 0.81 W. (I = 90 mA; P = I²R = 0.0081 × 100 = 0.81 W.)
You now have a practical toolkit for V, I, R and power. Continue to DC Circuit Analysis for KCL, KVL and Thevenin, or revisit First Circuits for hands-on breadboarding and LED colour codes.