Instantaneous AC Voltage

v = Vmax · sin(θ)

Quick answer

This calculator solves v = Vmax × sin(θ) (θ = ωt) for Instantaneous Voltage. With Peak Voltage 26.4 V, Phase Angle 45 °, the result is 18.67 V. It solves in 3 directions, so you can enter the values you have and get the one you need.

Calculator

Instantaneous Voltage (V)
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Formula

v = Vmax × sin(θ) (θ = ωt)

Description

The instantaneous value of a sinusoidal voltage at any point in its cycle is the peak amplitude times the sine of the phase angle. This is the fundamental description of an AC waveform from which RMS, average, and power values are derived. Evaluating it at a given angle (or time, since θ = ωt) gives the actual voltage at that instant.

Variables

  • v — Instantaneous voltage (V)
  • Vmax — Peak amplitude (V)
  • θ — Phase angle ωt (degrees)

Practical Notes

At θ = 90° the voltage is at its peak; at 0° and 180° it crosses zero. Replace θ with ωt = 2πft to get the time-domain waveform.

Symbols and units

Symbol Meaning Default unit Accepted units
v Instantaneous Voltage V pV, nV, µV, mV, V, kV
Vmax Peak Voltage V pV, nV, µV, mV, V, kV
theta Phase Angle ° °

Step by step

  1. 1. What you know: Peak Voltage 26.4 V, Phase Angle 45 °
  2. 2. The relation: v = Vmax × sin(θ) (θ = ωt)
  3. 3. Substitute: Vmax = 26.4 V, theta = 45 °
  4. 4. Result: Instantaneous Voltage = 18.67 V

Worked examples

Minimum range

Given: Peak Voltage 0.726 V, Phase Angle 1.5 °

Result: Instantaneous Voltage = 19 mV

Low range

Given: Peak Voltage 7.26 V, Phase Angle 15 °

Result: Instantaneous Voltage = 1.879 V

Typical

Given: Peak Voltage 26.4 V, Phase Angle 45 °

Result: Instantaneous Voltage = 18.67 V

High range

Given: Peak Voltage 0.506 kV, Phase Angle 75 °

Result: Instantaneous Voltage = 0.4888 kV

Maximum range

Given: Peak Voltage 5.06 kV, Phase Angle 750 °

Result: Instantaneous Voltage = 2.53 kV

Solving for each variable

Solve for Instantaneous Voltage

Peak Voltage 0.726 V, Phase Angle 1.5 ° → 19 mV

Solve for Peak Voltage

Instantaneous Voltage 0.33 V, Phase Angle 1.5 ° → 12.61 V

Solve for Phase Angle

Instantaneous Voltage 0.33 V, Peak Voltage 0.726 V → 27.04 °

How Instantaneous Voltage changes with Peak Voltage

Peak Voltage (input) Instantaneous Voltage (result)
5.81 V 5.738 V
14.53 V 14.35 V
29.05 V 28.69 V
43.58 V 43.04 V
58.1 V 57.38 V
87.15 V 86.08 V
116.2 V 114.8 V
0.2324 kV 0.2295 kV
0.4067 kV 0.4017 kV
0.581 kV 0.5738 kV

The same result in other units

Unit Value
mV 18670
V 18.67
kV 0.01867

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Frequently asked questions

How do you calculate Instantaneous AC Voltage?

Use v = Vmax × sin(θ) (θ = ωt). With Peak Voltage 26.4 V, Phase Angle 45 °, Instantaneous Voltage works out to 18.67 V.

How do you calculate if the rms voltage is e= 8sinwt + 6 sin 2wtwaht is the instantaneous voltage?

Rearrange for Instantaneous Voltage: with Peak Voltage 7.26 V, Phase Angle 15 °, Instantaneous Voltage is 1.879 V.

How do you calculate instantaneous value of phase a voltage?

Rearrange for Instantaneous Voltage: with Peak Voltage 26.4 V, Phase Angle 45 °, Instantaneous Voltage is 18.67 V.

What is the formula for Instantaneous AC Voltage?

The formula is v = Vmax × sin(θ) (θ = ωt). The instantaneous value of a sinusoidal voltage at any point in its cycle is the peak amplitude times the sine of the phase angle.

How do you find Peak Voltage when you know the other values?

Rearrange for Peak Voltage: with Instantaneous Voltage 0.33 V, Phase Angle 1.5 °, Peak Voltage is 12.61 V.

How do you find Phase Angle when you know the other values?

Rearrange for Phase Angle: with Instantaneous Voltage 0.33 V, Peak Voltage 0.726 V, Phase Angle is 27.04 °.

Related Concepts

All AC Circuits formulas →
Resonant Frequency Impedance (RLC Series) Quality Factor (Bandwidth) Quality Factor (RLC) Power Factor Apparent Power Reactive Power RLC Circuit Bandwidth

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