Ohm's law and the power formulas, with examples
Ohm's law and the power formulas in every form, worked examples for bulbs, heaters and LEDs, what current a home appliance draws, and how to choose a fuse or wire with safety in mind.
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In short
- Ohm's law is V = I × R. Voltage equals current times resistance.
- Power is P = V × I, which also gives P = I² × R and P = V² ÷ R.
- Any two of voltage, current, resistance and power give the other two.
- A 60 W bulb on 230 V draws about 0.26 A, and a 1,500 W heater draws about 6.5 A.
- These formulas fit simple resistive loads. Motors and AC power factors need more, and mains work is for a qualified electrician.
Ohm's law is the first relationship every electronics student learns, and it is one that working electricians use daily. It links voltage, current and resistance, and together with the power formula it lets you find any two of the four main quantities from the other two. With it, you can choose a resistor for an LED, find what current a heater draws, or work out why a long thin wire gets warm. This page gives the formulas, explains them in plain terms and works through common examples. It finishes with the limits of the law, as real circuits do not always behave so simply.
The four quantities
Voltage (V), measured in volts, is the electrical push that drives charge round a circuit.
Current (I), measured in amperes or amps, is the flow of charge.
Resistance (R), measured in ohms (Ω), is how strongly a material opposes that flow.
Power (P), measured in watts, is the rate at which energy is used.
A water pipe is the usual comparison. Pressure is like voltage, the flow rate is like current, and a narrow section of pipe is like resistance. More pressure means more flow, and a narrower pipe means less.
The formulas
Ohm's law, in its three forms:
V = I × R
I = V ÷ R
R = V ÷ I
The power formula:
P = V × I
Putting the two together gives the other forms:
P = I² × R
P = V² ÷ R
I = P ÷ V
R = V² ÷ P
| You know | Find I | Find R | Find P |
|---|---|---|---|
| V and R | V ÷ R | given | V² ÷ R |
| V and I | given | V ÷ I | V × I |
| V and P | P ÷ V | V² ÷ P | given |
| I and R | given | given | I² × R |
| I and P | given | P ÷ I² | given |
| R and P | √(P ÷ R) | given | given |
With the calculator, pick the pair you know, enter the values, and it finds the rest.
Worked examples
A bulb. A 60 W filament bulb runs on 230 V. Its current is 60 ÷ 230 = 0.2609 A, or about 261 mA. Its hot resistance is 230² ÷ 60 = 881.7 Ω.
A resistor across the mains. 230 V across 100 Ω gives 2.3 A and 529 W. A small resistor would fail at once, so this is a sum for a heating element, not a bench experiment.
A room heater. A 1,500 W heater on 230 V draws 1,500 ÷ 230 = 6.52 A, and its element has a resistance of about 35.3 Ω.
An electric geyser. A 2,000 W water heater draws 8.7 A on a 230 V supply. That is why it has its own circuit and a socket rated for the load.
A phone charger. A charger rated at 5 V and 2 A delivers 5 × 2 = 10 W. A fast charger at 9 V and 2 A delivers 18 W.
Choosing a resistor for an LED
This is one of the most common uses of the law at a hobby bench. An LED has a nearly fixed voltage drop, so you must limit its current with a resistor.
Take a red LED with a 2 V drop, wanted at 20 mA on a 9 V battery. The resistor sees the remaining 9 − 2 = 7 V:
R = 7 ÷ 0.02 = 350 Ω
Resistors come in standard values, so choose the next one up, 390 Ω, which keeps the current slightly lower. The power in it is 7² ÷ 390 = 0.126 W, so a common 0.25 W resistor is suitable.
Resistors in series and parallel
Before applying Ohm's law to a circuit, combine the resistors.
In series, the resistances simply add. 100 Ω and 220 Ω in series make 320 Ω.
In parallel, the combined resistance is smaller than either:
1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂
Two 100 Ω resistors in parallel make 50 Ω. Household appliances are connected in parallel, so that each gets the full 230 V, and their currents add up on the supply line. That is why too many heaters on one extension cord can overload it.
Current draw and fuses at home
To see what a home circuit carries, add the wattage of everything on it and divide by the voltage.
| Appliance | Power | Current at 230 V |
|---|---|---|
| LED bulb | 9 W | 0.04 A |
| Ceiling fan | 75 W | 0.33 A |
| Laptop charger | 65 W | 0.28 A |
| Room heater | 1,500 W | 6.52 A |
| Geyser | 2,000 W | 8.70 A |
| Induction cooktop | 2,000 W | 8.70 A |
A kettle and an induction cooktop on the same 16 A socket come close to the limit. Protection devices such as MCBs are chosen to protect the wiring, so never fit a larger one to stop a breaker tripping. If it trips, the wiring or the load needs checking.
Energy use on your bill is power multiplied by time. A 2,000 W geyser used for half an hour takes 1 kWh, which is one unit. Multiply the units by your rate to get the cost, and remember that state tariffs rise in slabs, so the last units in a month often cost the most.
Why thin and long wires get hot
Every wire has a little resistance, and I² × R tells you how much heat it makes. Because the current is squared, doubling the current gives four times the heat. So a thin extension cord feeding a heater can become warm, even dangerously so. The voltage drop along a long cable follows V = I × R too, which explains why a motor at the end of a long run may start slowly.
This is also why power lines use very high voltages. For the same power, higher voltage means lower current, and so far less heat lost on the way.
Where Ohm's law stops working
The law applies to materials with constant resistance, such as ordinary resistors at a steady temperature. Other cases behave differently.
Filament bulbs. Resistance rises sharply as the filament heats up, so a cold bulb measures far less than its working resistance.
Diodes and LEDs. Current does not grow in proportion to voltage.
Motors and transformers. On AC, these have inductance, so current and voltage do not peak together. Power then needs a power factor, as P = V × I × power factor. That is why a real motor's current can be higher than P ÷ V suggests.
Three-phase supply. Three-phase loads use a separate formula with √3.
A quick way to check your answer
Before trusting a result, try a rough sum in your head. If a load is about 2,300 W on a 230 V supply, the current must be about 10 A, because 230 × 10 = 2,300. If your calculation shows 100 A or 1 A, a decimal point has slipped. Watch for milliamps in particular, as 20 mA is 0.02 A, not 0.2 A.
Staying safe
Formulas help you to understand a circuit, but mains voltage can kill. Switch off at the board before touching anything, use an insulated tester, and call a licensed electrician for new circuits, sockets for heavy loads and anything that smells of burning or trips repeatedly. Work on low-voltage batteries and bench power supplies is the right place to practise.
What this calculator assumes
- The load is a plain resistance, with voltage and current in step.
- You enter a pair of known values, and all four quantities are then shown.
- Units are volts, amps, ohms and watts. For milliamps, divide by 1,000 before entering.
- Results are rounded for display, and small currents are also shown in mA.
- It does not allow for power factor, temperature change or three-phase supply.