Example 1
A 1500 W space heater on a 120 V supply draws 12.5 A. This is the reference value for this page.
I = 12.5 A
Current from Power and Voltage Calculator
Enter power P in watts and voltage U in volts. The calculator divides power by voltage using I = P/U and reports the current in amperes. For a 1500 W heater at 120 V the result is 12.5 A, and for 60 W at 12 V it is 5 A. The model does not use cos φ or Ohm’s law; zero voltage is not used in the division.
Result
Enter power in watts and voltage in volts. Current, conversions, and the chart will appear here.
Power from voltage and current is the P = U·I calculator, and voltage from power and current is the U = P/I calculator.
How it works
P1500 W
÷ U120 V
I12.5 A
Current in a DC circuit follows from the definition of electric power. Power is the product P = U·I, so the current sought is the quotient I = P/U. In SI, one ampere is one watt per volt: 1 A = 1 W/V. This is the textbook rearrangement of electric power (Halliday, Resnick, and Walker; Young and Freedman), not Ohm’s law and not I = Q/t.
The calculator uses the two numbers you enter: power P in watts (field prad-z-mocy-i-napiecia-a) and voltage U in volts (field prad-z-mocy-i-napiecia-b). The result is their quotient. For a 1500 W heater at 120 V you get 1500 / 120 = 12.5 A. For 60 W at 12 V you get 5 A. The same 60 W at 230 V is 0.2609 A, because the voltage in the denominator is higher.
What the result panel shows. Besides current in amperes, the calculator reports the same value in milliamperes, repeats the power you entered (with kilowatts when |P| ≥ 1000 W), and repeats the voltage you entered. When power is not zero, it reports equivalent resistance R = U²/P, a check of the identity P = U²/R for a resistive load. For the 1500 W heater that resistance is 9.6 Ω. The panel also reports the current that same power would draw at 120 V and at 230 V. The chart holds the voltage you entered and shows current for power equal to 0, 0.25, 0.5, 0.75, 1, 1.5, and 2 times |P|. When power is zero, the power axis runs from 0 W to 2 W.
Limits of the model. The calculator does not solve Ohm’s law as a path to current, does not find voltage U = P/I, and does not compute current from charge, I = Q/t. It does not know the power factor cos φ. On alternating current, P = U·I·cos φ is real power. The quotient P/U is then the RMS current only for a resistive load, when cos φ = 1. The result does not judge conductor size or breaker selection.
A comma and a period in the fields are the same decimal separator. The power field expects watts: enter 1.5 kW as 1500. Entering 1.5 instead of 1500 at 120 V gives 0.0125 A instead of 12.5 A. Voltage must not be zero, because division by zero is undefined. Zero watts at a nonzero voltage gives I = 0 A. Negative values are allowed, and the current has the sign of the quotient.
Examples on this page: 1500 W and 120 V → 12.5 A; 60 W and 12 V → 5 A; 6 W and 5 V → 1.2 A; 0.1 W and 5 V → 0.02 A; 10 W and 5 V → 2 A; 1800 W and 120 V → 15 A; 65 W and 20 V → 3.25 A; 2300 W and 230 V → 10 A; 3.7 W and 3.7 V → 1 A; 60 W and 230 V → 0.2609 A. The button “Fill this example” enters those same numbers and refreshes the result panel and the chart. The fields start empty.
Formula
I = P / U
[I] = W/V = A; 1 kW = 1000 W; 1 mA = 0.001 A
Model: DC electric power P = U·I (the SI definition of the watt, BIPM), rearranged to current. André-Marie Ampère is the namesake of the ampere, and James Prescott Joule related heat in a conductor to I²R. With Ohm’s law, I²R = U·I. This calculator solves only for I. Order of calculation: I = P / U, with U ≠ 0, then milliamperes, equivalent resistance R = U²/P when P ≠ 0, and the current of the same power at 120 V and at 230 V.
Theory, derivation, and domain
Definition of current from power and voltage
DC electric power is the product of voltage and current: P = U·I. When power and voltage are known, and voltage is not zero, current is the quotient I = P/U. The SI ampere satisfies 1 A = 1 W/V = 1 C/s. That unit identity does not mean the calculator computes charge. Current I = Q/t is a different formula and has its own page.
When voltage and current vary with time, the instantaneous product p(t) = u(t)·i(t) is instantaneous power. This calculator takes one number P and one number U, so it describes the steady case, or one pair of values at a power factor of one.
Historical note: Ampère, Volta, Ohm, and Joule
André-Marie Ampère described the force between current-carrying conductors in the 1820s. In 1826, he published Mémoire sur la théorie mathématique des phénomènes électrodynamiques uniquement déduite de l’expérience. The ampere is named for Ampère. Ampère did not write I = P/U. That formula is a later rearrangement of the definition of power.
Alessandro Volta presented the electric pile in 1800. Georg Simon Ohm, in Die galvanische Kette, mathematisch bearbeitet (1827), gave the relation of voltage, current, and resistance. In 1841, in the Philosophical Magazine, James Prescott Joule showed that heat released in a metallic conductor is proportional to the square of the current, to the resistance, and to time. Ohm’s law makes P = I²R = U·I. James Watt did not write the electrical formula. The name of the watt moved from mechanics into electrical engineering. The BIPM SI Brochure states the definitions of the ampere, the volt, and the watt.
Assumptions: the product U·I and a nonzero voltage
The formula I = P/U is valid when the power you enter is the product of that voltage and that current. For direct current this is the arrow convention in which the sign of power says whether the element absorbs or delivers energy. Voltage must not be zero.
The calculator does not assume the element is linear. Resistance may depend on temperature, and the result is still the quotient of the two numbers you enter. The calculator does not check whether that pair appears on a nameplate.
Derivation from P = U·I and equivalent resistance
Step 1. Definition: P = U·I. Step 2. When U ≠ 0, divide both sides by U and I = P/U remains. Step 3. Units: a watt divided by a volt is an ampere. Step 4. For 1500 W and 120 V the current is 1500 / 120 = 12.5 A. For 60 W and 12 V the current is 5 A. For the same 60 W at 230 V the current is 0.2609 A.
Step 5. If the element obeys Ohm’s law, P = U²/R. When P ≠ 0, resistance can be recovered as R = U²/P. For the 1500 W heater, 120² / 1500 = 9.6 Ω. The panel shows that quotient as a check. A diode, a motor, and a switching supply are not resistors, so the ohm on the panel does not replace a resistance measurement or the Ohm’s-law page.
Alternating current, RMS value, and power factor
For sinusoidal voltage and current, real power is P = U_rms · I_rms · cos φ, where φ is the phase shift and U_rms and I_rms are RMS values. From that equation, P / U_rms = I_rms · cos φ.
The quotient from this calculator is the RMS current only when cos φ = 1, that is for a resistive load such as a heater or a classic incandescent lamp. A motor often has a power factor below one, and the same quotient is not the current drawn from the supply. The calculator has no phase-angle field and no reactive-power field.
Units: ampere, watt, volt, and the kilowatt trap
1 kW = 1000 W and 1 mA = 0.001 A. The power field expects watts. Entering 1.5 instead of 1500 at 120 V gives 0.0125 A instead of 12.5 A, a result a thousand times too small. A comma and a period are the same separator.
The panel also reports the current of the same power at 120 V and at 230 V. For 1500 W those currents are 12.5 A and 6.522 A. They are the quotients P/120 and P/230, not a breaker selection and not a judgment that an installation will “hold.”
Sources
The definition, the units, and the derivation come from general-physics textbooks and from metrology documents: D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics (current, resistance, and power); H. D. Young and R. A. Freedman, University Physics; A.-M. Ampère, Mémoire sur la théorie mathématique des phénomènes électrodynamiques uniquement déduite de l’expérience, 1826; J. P. Joule, Philosophical Magazine, 1841; G. S. Ohm, Die galvanische Kette, mathematisch bearbeitet, 1827; BIPM, The International System of Units (SI), 9th edition; NIST Special Publication 330. The page is educational.
How to use
- Enter power P in watts, for example 1500. Enter 1.5 kW as 1500.
- Enter voltage U in volts, for example 120 or 12. The calculator rejects zero volts.
- Choose Calculate, or wait for the result after both numbers are entered. 1500 W and 120 V give 12.5 A. 60 W and 12 V give 5 A.
- Read the current, the milliamperes, the repeated inputs, the equivalent resistance, and the current of the same power at 120 V and at 230 V. The chart shows I at fixed U.
- When you need watts, open the P = U·I page. When you need volts, open the U = P/I page. When you know charge and time, open the I = Q/t page.
Symbol glossary
The symbols below belong to I = P/U. For 1500 W and 120 V the result is 12.5 A, and for 60 W and 12 V it is 5 A. Voltage must not be zero. Current I = Q/t is a different quantity and has its own calculator.
- I
- The current sought, in amperes. 1500 / 120 = 12.5 A. This is not I = Q/t.
- P
- Power in watts. 1500 W at 120 V gives 12.5 A. Convert kilowatts to watts before you enter them.
- U
- Voltage in volts. U = 0 cannot stand in the denominator. At 60 W, 12 V gives 5 A, and 230 V gives 0.2609 A.
Worked examples
Example 2
A 60 W lamp on a 12 V system draws 5 A.
I = 5 A
Example 3
A USB charger delivers 6 W at 5 V. The current is 1.2 A.
I = 1.2 A
Example 4
An indicator LED draws 0.1 W at 5 V. The current is 0.02 A, or 20 mA.
I = 0.02 A
Example 5
A 10 W LED strip supplied at 5 V draws 2 A.
I = 2 A
Example 6
An 1800 W load at 120 V draws 15 A. The calculator does not decide whether a breaker is suitable.
I = 15 A
Example 7
A laptop supply rated 65 W at 20 V draws 3.25 A.
I = 3.25 A
Example 8
A European kettle rated 2300 W at 230 V draws 10 A.
I = 10 A
Example 9
A lithium-ion cell at 3.7 W and 3.7 V draws 1 A.
I = 1 A
Example 10
The same 60 W at 230 V draws 0.2609 A, not the 5 A it draws at 12 V.
I = 0.2609 A
Ways to use it
- Check that a 1500 W nameplate at 120 V corresponds to 12.5 A.
- Read the current of a 60 W lamp on a 12 V system.
- Compare the same power at 120 V and at 230 V by two quotients, without mental arithmetic.
Common questions
What does this current calculator compute?
In this model current is power divided by voltage: I = P/U, because DC electric power is P = U·I. The calculator divides the power you enter in watts by the voltage you enter in volts and reports the result in amperes and in milliamperes. When power is not zero, the panel also reports the equivalent resistance R = U²/P. It separately shows the current that same power would draw at 120 V and at 230 V.
Which units should I enter for power and voltage?
Enter power in watts [W] and voltage in volts [V]. The result is in amperes [A]. A kilowatt is not the unit of the power field: enter 2.3 kW as 2300 W. A comma and a period are the same decimal separator, so 120 and 120.0 are the same voltage. The milliampere figure is a conversion of the result, not a separate field.
Why does zero voltage produce no result?
Division by zero is undefined. At U = 0 the formula I = P/U has no value, whatever the power is. The calculator shows a message and does not draw the chart. Zero watts at a nonzero voltage is valid and gives 0 A. Equivalent resistance at zero power does not exist, because R = U²/P would also divide by zero.
How is this different from the power and voltage pages?
Here the unknown is current, and the inputs are power and voltage. On the power page the unknown is watts, and the inputs are voltage and current. On the voltage page the unknown is volts, and the inputs are power and current. It is the same trio P, U, and I and the same equation, with a different unknown.
How is this different from Ohm’s law and from I = Q/t?
Ohm’s law ties voltage, current, and resistance: U = I·R. This calculator has no resistance field. Equivalent resistance R = U²/P appears only as a check when the power you enter is not zero and the element can be treated as a resistor. Current I = Q/t uses charge and time and has its own page. This page has no charge field and no time field.
Does the formula apply to alternating current?
For direct current, I = P/U is the current in the element when the power you enter is the product U·I. For a sine wave and a resistive load the same quotient is the RMS current, because the power factor is 1. When cos φ is below 1, real power divided by RMS voltage equals I·cos φ, not the RMS current itself. The calculator has no phase-angle field.
What do a negative power or a negative voltage mean?
The calculator keeps the sign of the quotient. Two negative signs give a positive current, and exactly one negative sign gives a negative current. The sign follows the arrow convention you chose. Nameplate values are usually entered as absolute numbers.
Will the calculator size a wire or a breaker?
No. The result is a current from two numbers. It does not account for conductor area, length, temperature, insulation, or the trip curve of a breaker. The 120 V and 230 V rows are only that same power divided by those voltages. Wire resistance is a separate page, from resistivity, length, and cross-sectional area.
What does the chart under the result show?
The chart holds the voltage you entered and places power at 0, 1/4, 1/2, 3/4, 1, 3/2, and 2 times the absolute value of the power you entered. The vertical axis is current I = P/U at that voltage. When the entered power is zero, the power axis runs from 0 W to 2 W so the slope 1/U is visible.
Where does the formula come from?
The equation P = U·I is the definition of electric power: 1 W = 1 V·A, so 1 A = 1 W/V. André-Marie Ampère described the force between current-carrying wires, and the ampere is named for him. James Prescott Joule related heat in a conductor to I²R, and Ohm’s law makes I²R equal to U·I. The browser does the arithmetic, and Hubert Ligocki reviewed the page.
How do I calculate current with I = P/U?
Divide power in watts by voltage in volts. The result is in amperes. For 1500 W and 120 V: I = 1500/120 = 12.5 A.
What is I for a 1500 W heater at 120 V?
12.5 A. For a resistive load that is a typical power–voltage pair on a 120 V supply.
What is I at 60 W and 12 V?
5 A. The same power at a higher voltage draws less current: at 24 V it would be 2.5 A.
How do I enter 1.5 kW in the power field?
Convert to watts: 1.5 kW = 1500 W. The field expects watts, not kilowatts.
Does I = P/U include the power factor cos φ?
No. On AC, real power is P = U·I·cos φ, so P/U equals I only when cos φ = 1. A motor with cos φ < 1 needs a separate calculation.
Sources
The definition, the unit, the history, and the derivation come from physics textbooks and from metrology documents.
- Wikipedia — Electric current
- Wikipedia — Electric power
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- NIST — SI units: electric current
- Halliday, D.; Resnick, R.; Walker, J., Fundamentals of Physics — current, resistance, and power.
- Young, H. D.; Freedman, R. A., University Physics — electric power.
- Ampère, A.-M., Mémoire sur la théorie mathématique des phénomènes électrodynamiques uniquement déduite de l’expérience, 1826.
- Joule, J. P., Philosophical Magazine, 1841.
- Ohm, G. S., Die galvanische Kette, mathematisch bearbeitet, 1827.
Page updated November 3, 2026.