Voltage to Power Converter
Our free signal & frequency converter handles voltage power conversions. See tables, ratios, and examples for quick reference.
Voltage to Power Converter
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Formula: Power = Vrms^2 / Impedance
Worked example โ 120V RMS into 15 ohms = 960 W | 8 A RMS
Formula
Power = Vrms^2 / Impedance
Power in watts equals the RMS voltage squared divided by impedance in ohms. For peak voltage, divide by sqrt(2) first. For peak-to-peak, divide by 2*sqrt(2). dBm = 10*log10(P_mW). Current = Vrms / Z.
Worked Examples
Example 1: US Household Outlet Power
Problem:Calculate the maximum power from a 120V RMS outlet into a 15 ohm heater element.
Solution:Power = V^2 / R P = 120^2 / 15 = 14,400 / 15 = 960 W Current = V / R = 120 / 15 = 8 A Peak voltage = 120 * 1.414 = 169.7 V
Result:120V RMS into 15 ohms = 960 W | 8 A RMS
Example 2: RF Signal Power Measurement
Problem:An oscilloscope shows a 2V peak-to-peak signal on a 50 ohm system. What is the power in dBm?
Solution:Vrms = Vpp / (2 * sqrt(2)) = 2 / 2.828 = 0.7071 V Power = 0.7071^2 / 50 = 0.01 W = 10 mW dBm = 10 * log10(10) = 10 dBm
Result:2 Vpp into 50 ohms = 10 mW = 10 dBm
Frequently Asked Questions
How do you convert voltage to power?
Power equals voltage squared divided by impedance: P = V^2 / Z, where V is the RMS voltage and Z is the impedance in ohms. This comes from combining Ohm's law (V = IR) with the power formula (P = IV). For example, 120V RMS across 50 ohms produces 120^2/50 = 288 watts. Always use RMS voltage for power calculations with AC signals, as peak voltage will give incorrect results.
What is the difference between RMS, peak, and peak-to-peak voltage?
For a sinusoidal AC signal, these three measurements are related by fixed ratios. RMS (root mean square) voltage equals peak voltage divided by the square root of 2 (approximately 1.414). Peak-to-peak voltage is twice the peak voltage. A US wall outlet at 120V RMS has a peak voltage of 169.7V and peak-to-peak of 339.4V. RMS is used for power calculations because it represents the DC equivalent voltage that would deliver the same power.
What are dBV and dBuV?
dBV and dBuV are logarithmic voltage units. dBV uses 1 volt RMS as the reference: dBV = 20*log10(Vrms/1V). So 0 dBV = 1V, -20 dBV = 0.1V, and 20 dBV = 10V. dBuV (decibels relative to 1 microvolt) uses 1 microvolt as reference: dBuV = 20*log10(Vrms/1uV). The relationship is dBuV = dBV + 120. dBuV is commonly used in EMC testing and antenna measurements where signal levels are very small.
Why does impedance matter in voltage-to-power conversion?
Impedance determines how much current flows for a given voltage, and therefore how much power is dissipated. The same voltage across different impedances produces different power levels. For instance, 1V RMS across 50 ohms produces 20 mW (13 dBm), but across 600 ohms it produces only 1.67 mW (2.2 dBm). In RF systems, impedance matching is critical because maximum power transfer occurs when source and load impedances are equal.
How do you convert watts into dBm?
Express the power in milliwatts, then take ten times its base-ten logarithm, so dBm equals 10 x log10(P in mW). One watt is 1000 mW and therefore exactly 30 dBm, while one milliwatt is 0 dBm. A 288 mW result works out at 10 x log10(288) = 24.6 dBm. The multiplier is 10 rather than 20 because dBm expresses a power ratio, not a voltage ratio.
How do you get back from dBm to a voltage across a known impedance?
First convert to watts with P equal to ten raised to the power of dBm minus 30, all divided by 10. Then take the square root of P times the impedance, since P equals V squared over Z. At 0 dBm into 50 ohms that is 0.001 W, and the square root of 0.001 x 50 gives 0.2236 V RMS. The answer moves with impedance, so a dBm figure alone never fixes a voltage.
Why does doubling the voltage quadruple the power?
Because power varies with the square of voltage at a fixed impedance, so doubling V multiplies V squared by four. In decibel terms a factor of two in power is 3 dB, while a factor of two in voltage is 6 dB, since amplitude ratios are scaled by 20 and power ratios by 10. This square law is also why a small voltage error becomes a much larger power error.
What is the difference between real power and apparent power?
The V squared over Z relationship returns real power in watts only when the load is purely resistive. With a reactive load, current and voltage are out of phase, so voltage times current gives apparent power in volt-amperes instead. Real power is apparent power times the power factor, meaning an inductive or capacitive load dissipates less heat than the VA figure suggests. Use the resistive part of the impedance when you need true watts.
Why do RF systems use 50 ohms while older audio lines use 600 ohms?
Fifty ohms became the RF standard as a practical compromise between the coaxial geometries that minimise loss and those that maximise power handling, and keeping sources, cables and instruments matched holds reflections small. The 600 ohm figure comes from telephone line practice, where matched impedance mattered over long transmission runs, and it survives today as the reference impedance behind the dBu voltage.
How does crest factor affect a peak-to-power conversion?
Crest factor is peak voltage divided by RMS voltage, and the familiar shortcut of dividing peak by the square root of two holds only for a sine wave. A square wave has a crest factor of 1, so peak and RMS are equal, while speech, music and modulated RF have much higher crest factors. Converting a measured peak on such a signal with the sine assumption overstates the RMS value and therefore the power.
References
Reviewed for accuracy by Manoj Kumar, Mathematics Educator ยท Editorial policy
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