Instantaneous & Average Power

How power flows in AC circuits — oscillating instant by instant, yet delivering useful energy at a steady average rate.

Circuit AnalysisElectrical Engineering Year 2Free preview
⏱️ About 16 min

A light bulb connected to AC glows steadily — yet the power delivered to it reverses direction 120 times per second. How does that work?

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The big idea: In AC circuits, instantaneous power oscillates at twice the source frequency, but only its constant component — the average power — represents energy converted to heat or work.
🎯 By the end, you'll be able to
  • Express instantaneous power p(t) for sinusoidal voltage and current using the product-to-cosine identity
  • Identify the constant and double-frequency terms in the expanded power expression
  • Calculate average (real) power P from peak voltage, peak current, and phase angle difference
  • Explain why the double-frequency term contributes zero net energy over one period
📎 Helpful to know first

Instantaneous power: the product rule

In DC circuits, power is simply P = VI — a constant. In AC circuits, both voltage and current vary with time, so the power delivered at any instant is the product of the instantaneous voltage and instantaneous current:

The instantaneous power p(t) tells you exactly how much energy per second is flowing into the load at moment t. For a resistor, this power is always non-negative (the resistor never returns energy). For reactive elements like capacitors and inductors, p(t) swings positive and negative — energy flows in during part of the cycle and back out during another part.

\[ p(t) = v(t)\,i(t) \]
Instantaneous power p (watts, W) is the product of instantaneous voltage and instantaneous current at every moment in time.

Expanding p(t) for sinusoidal signals

When voltage and current are sinusoids — v(t) = V_m cos(ωt + θ_v) and i(t) = I_m cos(ωt + θ_i) — we can expand the product using the trigonometric identity cos(A)cos(B) = ½[cos(A−B) + cos(A+B)]. This splits p(t) into two distinct terms with very different physical meanings.

\[ p(t) = \frac{V_m I_m}{2}\Big[\cos(\theta_v - \theta_i) + \cos(2\omega t + \theta_v + \theta_i)\Big] \]
The first term is a constant — the average power. The second term oscillates at twice the source frequency (2ω) and averages to zero over a full period.

The two terms and what they mean

The constant term, (V_m I_m/2) cos(θ_v − θ_i), represents the net energy per cycle that is converted to heat, light, mechanical work, or otherwise dissipated. This is the real or average power.

The oscillating term, (V_m I_m/2) cos(2ωt + θ_v + θ_i), swings positive and negative at twice the source frequency. Over one complete period T = 2π/ω, the cosine of 2ωt goes through two full cycles, so its integral — and hence its average — is exactly zero. It represents energy sloshing back and forth between source and load without net delivery.

Average (real) power

Because the double-frequency term integrates to zero over one period, the average power is simply the constant term from the expansion:

\[ P = \frac{V_m I_m}{2}\cos(\theta_v - \theta_i) \]
Average power P (watts, W) depends on the peak amplitudes and the phase difference between voltage and current. When voltage and current are in phase (θv = θi), cos(0) = 1 and power is maximized.

The phase angle's role

The factor cos(θ_v − θ_i) is called the power factor. It ranges from 0 to 1 and tells you what fraction of the maximum possible power is actually delivered. A purely resistive load has θ_v = θ_i (zero phase difference), giving a power factor of 1. A purely reactive load (ideal capacitor or inductor) has a 90° phase difference, giving a power factor of 0 — no net power is delivered, even though current flows.

📝 Worked example: Given v(t) = 100 cos(ωt + 20°) V and i(t) = 4 cos(ωt − 10°) A, find the complete instantaneous power p(t) and the average power P.
  1. Identify the parameters: Vm = 100 V, Im = 4 A, θv = 20°, θi = −10°.
  2. Compute the amplitude factor: Vm Im / 2 = (100 × 4) / 2 = 200.
  3. Compute the phase difference: θv − θi = 20° − (−10°) = 30°.
  4. Constant term: 200 cos(30°) = 200 × 0.8660 = 173.2 W.
  5. Double-frequency term: 200 cos(2ωt + θv + θi) = 200 cos(2ωt + 20° + (−10°)) = 200 cos(2ωt + 10°) W.
  6. Assemble p(t) = 173.2 + 200 cos(2ωt + 10°) W.
  7. Average power equals the constant term: P = 173.2 W.
✓ p(t) = 173.2 + 200 cos(2ωt + 10°) W, and the average power P = 173.2 W.
✏️ Practice: For v(t) = 50 cos(ωt + 30°) V and i(t) = 2 cos(ωt − 30°) A, find the average power P in watts.
W
Solution
  1. Amplitude factor: Vm Im / 2 = (50 × 2) / 2 = 50.
  2. Phase difference: θv − θi = 30° − (−30°) = 60°.
  3. P = 50 cos(60°) = 50 × 0.5 = 25 W.

Check your understanding

1. The instantaneous power p(t) in an AC circuit contains a term oscillating at 2ω. Why doesn't this term contribute to the average power?
The double-frequency term is a cosine (or sine) over complete periods, and the integral of any sinusoid over a full period is zero. So its average contribution vanishes, leaving only the constant term.
2. If the voltage and current are 90° out of phase, what is the average power delivered?
cos(90°) = 0, so P = (Vm Im / 2) × 0 = 0. This is the case for a purely reactive load — energy sloshes back and forth but none is net delivered.
✅ Key takeaways
  • Instantaneous power p(t) = v(t)·i(t) expands into a constant term plus a double-frequency (2ω) oscillating term.
  • The constant term (Vm Im / 2) cos(θv − θi) is the average power P — the only part that delivers net energy.
  • The oscillating term averages to zero over one period because the integral of cosine over full cycles is zero.
  • The power factor cos(θv − θi) ranges from 0 (purely reactive) to 1 (purely resistive), governing how much power is actually delivered.
➡️ The average power formula P = (Vm Im / 2) cos(θv − θi) uses peak amplitudes — but engineers prefer working with RMS values, which simplify the expression and match how real-world voltages are rated.
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