Ohm's Law and Electrical Power: Formulas With Worked Examples
Almost every calculation in basic electronics starts with two relationships: Ohm's law, which links voltage, current and resistance, and the power law, which tells you how much energy a part turns into heat every second. If you can use these two confidently, you can size resistors, pick wire, check whether a component will overheat and estimate your electricity bill.
This article explains both laws, shows how to rearrange them, and then works through several numeric examples of the kind you will meet in lab records, university exams and real hobby projects.
The Formulas
Ohm's law says that the current through a resistor is proportional to the voltage across it. The constant of proportionality is the resistance, measured in ohms (Ω). Voltage is in volts (V) and current in amperes (A).
Ohm's law V = I x R I = V / R R = V / I Power (watts) P = V x I P = I^2 x R (substitute V = I x R) P = V^2 / R (substitute I = V / R) Energy E (joules) = P (watts) x t (seconds) E (kWh) = P (kW) x t (hours)
The three power forms give the same answer; you simply choose the one that uses the quantities you already know. When current through a part is fixed (for example in a series string), P = I^2 x R is handy. When the voltage across a part is fixed (for example a heater on the mains), P = V^2 / R is quicker.
| Quantity | Symbol | Unit | Common prefixes |
|---|---|---|---|
| Voltage | V | volt (V) | mV, kV |
| Current | I | ampere (A) | µA, mA |
| Resistance | R | ohm (Ω) | kΩ, MΩ |
| Power | P | watt (W) | mW, kW |
| Energy | E | joule (J) or kWh | 1 kWh = 3.6 MJ |
Worked Examples
Example 1: Current and power in a resistor
A 470 Ω resistor is connected across a 12 V supply. Find the current and the power dissipated, and choose a suitable wattage rating.
I = V / R = 12 / 470 = 0.02553 A = 25.5 mA P = V^2 / R = 144 / 470 = 0.306 W Check: P = V x I = 12 x 0.02553 = 0.306 W
A common 1/4 W (0.25 W) resistor would be overloaded. A good rule is to keep dissipation at or below half the rating, so a 1 W resistor is the sensible choice here, or at minimum a 1/2 W part, which will run warm.
Example 2: Mains heater
A room heater is rated 1000 W at 230 V. Find the current it draws and its hot resistance.
I = P / V = 1000 / 230 = 4.35 A R = V^2 / P = 230^2 / 1000 = 52900 / 1000 = 52.9 Ω
If you measure the element cold with a multimeter you will usually read a lower value, because the resistance of the heating wire rises with temperature. Ohm's law holds at each instant; the resistance itself changes.
Example 3: Electricity bill
The same heater runs 3 hours a day for 30 days. The tariff is ₹8 per unit (1 unit = 1 kWh).
Energy per day = 1 kW x 3 h = 3 kWh Energy per month = 3 x 30 = 90 kWh Cost = 90 x 8 = ₹720
Example 4: Voltage drop in a cable
A 12 V LED strip draws 5 A through a cable whose two conductors together measure 0.2 Ω. How much voltage reaches the strip, and how much power is lost in the cable?
Drop in cable = I x R = 5 x 0.2 = 1.0 V Voltage at strip = 12 - 1.0 = 11.0 V Power lost = I^2 x R = 25 x 0.2 = 5 W
Notice that power loss grows with the square of current. Doubling the current to 10 A in the same cable would quadruple the loss to 20 W. This is why long low-voltage runs need thicker wire, and why power companies transmit at high voltage and low current.
Example 5: Finding an unknown resistance
A small DC motor draws 180 mA from a 6 V supply while running. Its effective resistance at that operating point is R = 6 / 0.18 = 33.3 Ω and it consumes P = 6 x 0.18 = 1.08 W. Remember that a motor is not a pure resistor; at stall it draws much more current, so this figure only describes one operating condition.
Where Ohm's Law Applies and Where It Does Not
Ohm's law describes ohmic devices, whose resistance stays constant over the operating range: carbon-film and metal-film resistors, copper wire at a steady temperature, and resistive heating elements at a fixed temperature. A plot of current against voltage for these is a straight line through the origin.
Diodes, LEDs, transistors and filament lamps are non-ohmic. A silicon diode conducts almost nothing until about 0.6 V and then current rises steeply. You can still calculate an instantaneous ratio V/I, but it is not a constant, so you cannot use a single resistance value to predict behaviour at other voltages. In LED and transistor circuits we apply Ohm's law to the resistor in series, not to the semiconductor itself.
| Device | Ohmic? | How to calculate |
|---|---|---|
| Fixed resistor | Yes | V = IR directly |
| Filament lamp | No, R rises when hot | Use rated V and P |
| Diode or LED | No | Subtract forward voltage, apply Ohm's law to series resistor |
| Thermistor (NTC) | No, R falls with heat | Use datasheet R-T table |
Common Mistakes
- Mixing units: 5 V across 2.2 kΩ is 2.27 mA, not 2.27 A. Write the prefixes in every step.
- Choosing resistor wattage equal to the calculated power. Allow at least a 2x margin so the part does not run hot.
- Using the cold resistance of a lamp or heater to calculate its running current.
- Forgetting that P = V^2 / R uses the voltage across the component, not the supply voltage, when other parts share the supply.
- Confusing power (W) with energy (Wh or kWh). A 10 W device running for 10 hours uses 100 Wh = 0.1 kWh.
Summary
Ohm's law, V = IR, and the power equations P = VI = I^2 R = V^2 / R are enough to solve a large share of everyday circuit problems. Always convert units, pick the power formula that uses known values, give resistors a generous wattage margin, and remember that semiconductors and lamps do not follow a fixed resistance. Practise with a multimeter: measure a resistor, apply a known voltage, and confirm that the current you measure matches your calculation.