Power in Ac Circuits

Physics Electricity and Magnetism • Ac Circuits and Electromagnetic Waves

Written by STEM Calculators Team Published January 18, 2026 Updated February 24, 2026

Input mode

RMS + phase

Voltage V_rms (V) Current I_rms (A) Phase angle φ (deg)

Use a negative φ for leading current (capacitive load). Example: φ = −30°. Inputs accept pi, sqrt(2), 1e-3. Use * for multiplication.

Plot Wave frequency f (Hz) Cycles shown Graph text size

Pan/zoom: drag to pan • mouse wheel/trackpad/pinch to zoom in/out • “Reset view” restores default. The p(t) plot uses RMS-based sinusoidal power: it averages to the real power P_avg (the billable power).

Ready

Enter values and click “Calculate”.

Interactive plot

What you’re seeing

computed curve
average power line (p(t) mode)
marker point

Rate this calculator

0.0 /5 (0 ratings)

Be the first to rate.

Your rating

Name (optional) Review (optional)

You can update your rating any time.

Theory

Real, reactive, and apparent power

In sinusoidal steady state, voltage and current are represented by RMS values and a phase difference φ. Power in AC is commonly described using three related quantities: real (average) power, reactive power, and apparent power.

Definitions (RMS)

\[ \begin{aligned} S &= V_{\mathrm{rms}}\cdot I_{\mathrm{rms}} \\ P_{\mathrm{avg}} &= V_{\mathrm{rms}}\cdot I_{\mathrm{rms}}\cdot \cos\varphi \\ Q &= V_{\mathrm{rms}}\cdot I_{\mathrm{rms}}\cdot \sin\varphi \end{aligned} \]

The power factor is \( \mathrm{pf}=|\cos\varphi| \). Inductive loads typically have φ > 0 (lagging current, Q > 0), while capacitive loads have φ < 0 (leading current, Q < 0).

Complex power

\[ \begin{aligned} \tilde S &= P_{\mathrm{avg}} + jQ \end{aligned} \]

Instantaneous power waveform and why the average matters

If \(v(t)=V_m\cos(\omega t)\) and \(i(t)=I_m\cos(\omega t-\varphi)\), then the instantaneous power is \(p(t)=v(t)\,i(t)\). Using trig identities, it splits into a constant part (the average) plus a ripple at \(2\omega\).

Derivation

\[ \begin{aligned} p(t) &= V_m\cos(\omega t)\cdot I_m\cos(\omega t-\varphi) \\ &= \frac{V_m I_m}{2}\Big(\cos\varphi + \cos(2\omega t-\varphi)\Big) \end{aligned} \]

Since \(V_m=\sqrt{2}\,V_{\mathrm{rms}}\) and \(I_m=\sqrt{2}\,I_{\mathrm{rms}}\), the constant term becomes:

\[ \begin{aligned} \langle p(t)\rangle &= \frac{V_m I_m}{2}\cos\varphi \\ &= V_{\mathrm{rms}}\cdot I_{\mathrm{rms}}\cdot \cos\varphi \\ &= P_{\mathrm{avg}} \end{aligned} \]

Practical note: utility billing and real energy usage are tied to \(P_{\mathrm{avg}}\), not to \(S\) or \(Q\) alone.

From impedance (φ = arg Z)

If the load impedance is \(Z=R+jX\), the phase angle is the argument of Z: \( \varphi = \arg(Z) \). For a supply voltage \(V_{\mathrm{rms}}\), the RMS current magnitude is \(I_{\mathrm{rms}}=V_{\mathrm{rms}}/|Z|\).

Magnitude and phase

\[ \begin{aligned} |Z| &= \sqrt{R^2 + X^2} \\ \varphi &= \tan^{-1}\!\left(\frac{X}{R}\right) \end{aligned} \]

Then compute S, P, Q

\[ \begin{aligned} I_{\mathrm{rms}} &= \frac{V_{\mathrm{rms}}}{|Z|} \\ S &= V_{\mathrm{rms}}\cdot I_{\mathrm{rms}} \\ P_{\mathrm{avg}} &= S\cos\varphi \\ Q &= S\sin\varphi \end{aligned} \]

University extension: three-phase power

For a balanced three-phase system (line-to-line voltage \(V_L\), line current \(I_L\)), total powers are:

\[ \begin{aligned} P_{3\phi} &= \sqrt{3}\,V_L\cdot I_L\cdot \cos\varphi \\ Q_{3\phi} &= \sqrt{3}\,V_L\cdot I_L\cdot \sin\varphi \\ S_{3\phi} &= \sqrt{3}\,V_L\cdot I_L \end{aligned} \]

Rate this calculator

Related calculators