Enter the applied current (amps) and the total resistance of the circuit (ohms) into the calculator to determine the Applied Voltage.
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Applied Voltage Formula
The Voltage tab uses Ohm's law to find the voltage applied across a resistive load from the current through it and its total resistance.
V = I * R
The Ohm's law + power tab solves for any two unknowns when you supply the other two values from V, I, R, and P. It uses these relations:
V = I * R P = V * I P = I^2 * R P = V^2 / R
- V = applied voltage, in volts (V)
- I = current through the load, in amperes (A)
- R = total resistance of the load, in ohms (Ω)
- P = power dissipated by the load, in watts (W)
In the Voltage tab, you enter current and resistance and get the applied voltage plus the power the load will dissipate. In the Ohm's law + power tab, you enter exactly two of the four quantities and the calculator solves for the other two using the matching pair of formulas above.
Reference Tables
Use these tables to sanity check your inputs and the result the calculator returns.
| Source | Typical applied voltage |
|---|---|
| AA alkaline cell | 1.5 V |
| USB port | 5 V |
| Car battery | 12 V |
| US wall outlet | 120 V |
| EU wall outlet | 230 V |
| US clothes dryer | 240 V |
| Unit | Equals |
|---|---|
| 1 kV | 1,000 V |
| 1 V | 1,000 mV |
| 1 mV | 0.001 V |
| 1 kΩ | 1,000 Ω |
| 1 MΩ | 1,000,000 Ω |
Examples and FAQ
Example 1. A 220 Ω resistor carries 0.05 A. The applied voltage is V = 0.05 × 220 = 11 V. Power dissipated is P = 11 × 0.05 = 0.55 W, so a 1/4 W resistor would burn out and you need at least a 1 W part.
Example 2. A heater is rated 1,500 W at 120 V. Its current draw is I = P / V = 1,500 / 120 = 12.5 A, and its resistance is R = V² / P = 14,400 / 1,500 = 9.6 Ω.
Is applied voltage the same as voltage drop? For a single resistor connected directly to a source, yes. In a series circuit, the applied voltage equals the sum of the voltage drops across each element.
Why does the solver tab require exactly two values? Two known quantities are enough to fix the other two through Ohm's law and the power equation. Entering one is underdetermined; entering three can conflict.
Does this work for AC circuits? Only for purely resistive loads, where you use RMS values for V and I. Circuits with significant inductance or capacitance need impedance, not just resistance.
What if my measured voltage is lower than calculated? The source likely has internal resistance or the wiring adds resistance. Add those to the total R, or measure current under load and recompute.
