Voltage Calculator
Calculate voltage from current and resistance, power and current, power and resistance, energy and charge, or the output of a voltage divider.
Voltage is the energy each coulomb of charge gains or loses between two points. This calculator works it out five ways: from current and resistance (Ohm's law), from power and current, from power and resistance, from energy and charge, and as the output of a two-resistor voltage divider. Pick the method that matches the numbers you have; the other boxes are ignored.
How to use this tool
- Choose the method.
- Enter the values that method uses — the note under each box says which method needs it.
- For a voltage divider, enter the input voltage and both resistors; R2 is the one the output is taken across.
- Read the voltage and the related values shown with it.
Formula
V = I × R · V = P ÷ I · V = √(P × R) · V = W ÷ Q · V_out = V_in × R2 ÷ (R1 + R2)
I is current (A), R resistance (Ω), P power (W), W energy (J) and Q charge (C). One volt equals one joule per coulomb. The divider formula assumes nothing draws current from the output; a load in parallel with R2 lowers V_out. Power/resistance methods use positive magnitudes for ohmic loads. Energy/charge can be signed and refers to potential-energy change per charge. AC values must use compatible RMS quantities and the applicable power factor.
Example
A divider with R1 = 10 kΩ and R2 = 4.7 kΩ on a 12 V supply gives V_out = 12 × 4,700 ÷ 14,700 ≈ 3.84 V. About 0.82 mA flows through both resistors, and the remaining 8.16 V is dropped across R1.
Frequently asked questions
What is voltage, in plain terms?
It is the push behind the current: energy per unit of charge. A 9 V battery gives every coulomb that passes through it 9 joules of energy.
Can I power a circuit from a voltage divider?
Not anything that draws real current. The load sits in parallel with R2 and drags the output down. Dividers suit reference voltages and sensor inputs; use a regulator to supply power.
Does V = P ÷ I work for AC?
For resistive loads, yes, with real power and RMS current. For motors and other loads with a power factor below 1, the voltage is P ÷ (I × PF).
Last updated: 2026-10-30