Power Formula With R and I: P = I²R Explained

The power formula using resistance (R) and current (I) is P = I²R: multiply the current in amperes by itself, then multiply by resistance in ohms to get power in watts. This calculator takes current and resistance as inputs and returns electrical power; it also shows the voltage implied by those values.

Formula

The power formula in terms of current and resistance is:

P = I²R

• P is electrical power in watts (W).
• I is current in amperes (A), also called amps.
• R is resistance in ohms (Ω).

The calculator’s second result uses Ohm’s law, V = IR, to calculate voltage in volts (V) from the same current and resistance. The two formulas work together for a resistive circuit: substituting V = IR into P = VI gives P = I²R.

Worked Example

Using the calculator’s default values, current is 2 A and resistance is 10 Ω.

1. Square the current: 2² = 4.
2. Multiply by resistance: 4 × 10 = 40 W.
3. Find the corresponding voltage: V = IR = 2 × 10 = 20 V.

The calculator therefore reports power of 40.00 W and voltage of 20.00 V. The power result follows directly from P = I²R.

How the Formula Works

Electrical power describes how quickly electrical energy is transferred or converted. In a simple circuit, power can also be calculated as P = VI, where voltage is in volts and current is in amperes. Ohm’s law relates voltage, current, and resistance: V = IR. Combining these relations gives two useful alternatives:

• Starting with P = VI and replacing V with IR gives P = I²R.
• Replacing I with V/R gives P = V²/R.

Use the version whose two known quantities match the information you have. If current and resistance are known, use P = I²R. If voltage and resistance are known, use P = V²/R. If voltage and current are known, use P = VI.

The square on current matters. Doubling current while keeping resistance constant makes the calculated power four times as large, not twice as large: (2I)²R = 4I²R. Doubling resistance at the same current doubles the result. This relationship is especially useful when considering the power converted to heat in a resistor, although actual operating limits depend on the component and circuit.

When to Use P = I²R

This form is useful when a circuit’s current and resistance are known or can be measured. It is commonly used to calculate the power dissipated in a resistor or another component that can be treated as a resistance under the conditions being considered.

For example, if 0.5 A flows through a 24 Ω resistor, then P = 0.5² × 24 = 6 W. The current is squared first, giving 0.25, and 0.25 × 24 gives 6. This result is the calculated electrical power associated with those values; it does not, by itself, establish what power rating a particular component should have.

For an ideal resistor, the electrical power is dissipated mainly as heat. In other devices, energy may be converted into light, motion, sound, or other forms. A simple resistance calculation may not capture all behavior in a device whose properties change with temperature, frequency, or operating conditions.

Reading Inputs, Units, and Results

Enter current in amperes and resistance in ohms. The result is in watts. If your measurements use different units, convert them before calculating; the calculator does not infer or convert units from the numbers alone.

• 1 milliampere (mA) = 0.001 A. For example, 250 mA = 0.25 A.
• 1 kilo-ohm (kΩ) = 1,000 Ω. For example, 4.7 kΩ = 4,700 Ω.
• 1 kilowatt (kW) = 1,000 W; 1 milliwatt (mW) = 0.001 W.

With 250 mA through 4.7 kΩ, first convert to 0.25 A and 4,700 Ω. Then P = 0.25² × 4,700 = 293.75 W. This illustrates why unit conversion is essential: entering 250 and 4.7 as though they were amperes and ohms would describe a different calculation entirely.

In U.S. customary contexts, electrical quantities still use the same units: amperes, ohms, volts, and watts. There is no separate customary-unit conversion needed for this formula. If a result is expressed in horsepower, use the appropriate power-unit conversion only after calculating watts.

Common Mistakes and Limits

• Forgetting to square current: P = I²R means I × I × R, not I × R.
• Mixing prefixes: Convert milliamperes and kilo-ohms to base units, or carefully account for their factors.
• Using the wrong circuit model: The formula applies directly to resistance. For alternating-current circuits with changing voltage and current, phase relationships and power factor may matter; simple P = I²R may describe resistive heating but not total real power in every case.
• Treating resistance as constant: Some materials and components change resistance as temperature or operating conditions change. A calculation using one resistance value is only as representative as that value.
• Confusing calculated power with a rating: The formula calculates power from inputs. It does not verify whether a component, supply, or circuit can safely operate at that power.

A resistance of zero gives zero in this formula for any finite current, while a zero current also gives zero power for any finite resistance. Real circuits can involve sources, contacts, and components whose behavior is not represented by a single resistance, so use measured or specified values appropriate to the part of the circuit being analyzed.

Frequently Asked Questions

What is the power formula with current and resistance?
P = I²R. Square current in amperes and multiply by resistance in ohms to obtain power in watts.

Why is current squared in P = I²R?
Because power is P = VI and Ohm’s law gives V = IR. Substituting IR for voltage produces P = I × IR = I²R.

Can I calculate power from voltage and resistance instead?
Yes. Use P = V²/R when voltage and resistance are known. Keep voltage in volts and resistance in ohms to get watts.

What happens to power if current doubles?
If resistance stays constant, power becomes four times as large because current is squared: (2I)²R = 4I²R.

Does P = I²R apply to AC circuits?
It can calculate resistive power from an appropriate RMS current and resistance, but it does not account for every AC circuit’s phase shift or reactive behavior. In those cases, distinguish real power from apparent power.

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in i | Current | A | 2 | 0 | 1000000
in r | Resistance | ohm | 10 | 0 | 1000000
out p | Power | W | i ^ 2 * r | 2
out v | Voltage | V | i * r | 2