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Power Dissipation in a Resistive Load

Key idea: Use Irms and Vrms to calculate mean power in resistors for sinusoidal AC, and relate this to instantaneous power p = vi in questions (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse mean power in resistive a.c. loads and half-wave rectification.
  • Represent sinusoidal a.c. and use peak and r.m.s. values.

1. Definitions (Must Know)

A. Instantaneous power, p

Instantaneous power transferred to a component is:

p = vi

where v and i are the instantaneous potential difference and current.

B. Mean power in a resistor, ⟨P⟩

For a resistor, mean power is the time average of instantaneous power:

⟨P⟩ = ⟨vi⟩

For a purely resistive load (voltage and current in phase), you can use r.m.s. values:

⟨P⟩ = Iᵣₘₛ²R = Vᵣₘₛ²/R = IᵣₘₛVᵣₘₛ

2. Key Ideas (What Earns Marks)

  • For a resistor on sinusoidal a.c., v and i are in phase: v = V₀ sin(ω t), i = I₀ sin(ω t)
  • Instantaneous power is always non-negative: p = vi = V₀I₀ sin² (ω t)
  • Because sin² (ω t) has two peaks per a.c. cycle, the power waveform repeats at 2f (period T/2).
  • Mean power over a full cycle: ⟨P⟩ = (1/2)V₀I₀ = (1/2)I₀²R = (1/2)V₀²/R
  • Using r.m.s. values: Iᵣₘₛ = I₀/(square root of 2), Vᵣₘₛ = V₀/(square root of 2) Therefore ⟨P⟩ = Iᵣₘₛ²R.
What you must state

These power formulas using Iᵣₘₛ assume a purely resistive load (voltage and current in phase).

3. Detailed Explanations

A. Why power varies as sin²

For a resistor connected to a sinusoidal source:

v = V₀ sin(ω t) i = I₀ sin(ω t)

So:

p = vi = V₀I₀ sin² (ω t)

Since sin² (ω t) ≥ 0, the resistor is always dissipating energy as heat.

Sinusoidal a.c., r.m.s. values and resistive powerTwo aligned graphs over one alternating-current cycle. The upper graph shows normalized voltage and current coinciding because they are in phase. The lower graph shows normalized power following sine squared, remaining non-negative, peaking twice per cycle, and oscillating around a mean of one-half.Purely resistive load: v and i are in phaseωtscaledv, iv/V₀ (solid)i/I₀ (dashed)same zero crossings and peaks0π/2π3π/22πp/(V₀I₀) = sin²(ωt)ωtscaled pmean = 1/2p ≥ 0 throughoutT/2 between power peakspower repetition frequency = 2f0π/2π3π/22π
Scroll diagram horizontally to read all labels.
For a resistor, voltage and current are in phase. Their product is proportional to sin²(ωt), so power is never negative, repeats twice per a.c. cycle, and has mean value one-half of its peak value.

B. Mean power over a cycle

The mean value of sin² over one full cycle is 1/2, so:

⟨P⟩ = V₀I₀⟨sin² (ω t)⟩ = (1/2)V₀I₀

Using V₀ = I₀R for a resistor:

⟨P⟩ = (1/2)I₀²R = (1/2)V₀²/R

C. Why r.m.s. values work

For a sinusoid:

Iᵣₘₛ = I₀/(square root of 2), Vᵣₘₛ = V₀/(square root of 2)

So:

Iᵣₘₛ²R = (I₀²/2)R = (1/2)I₀²R = ⟨P⟩

Review the equal-heating definition in Root-Mean-Square Values, then use the AC Waveform & Power Explorer to vary peak voltage, frequency and resistance.

4. Common Mistakes

  • Using peak values (V₀, I₀) directly in ⟨P⟩ = IV without converting to r.m.s.
  • Forgetting the “purely resistive” condition for IᵣₘₛVᵣₘₛ.
  • Mixing up instantaneous power p(t) with mean power ⟨P⟩.
  • Saying the power waveform has frequency f; for a sinusoidal resistor current it repeats at 2f.

5. Exam Tips

  • If the question gives mains values (e.g. 230 V), they are usually r.m.s. values.
  • Choose a method and stick to it:
    • r.m.s.: ⟨P⟩ = Iᵣₘₛ²R or ⟨P⟩ = Vᵣₘₛ²/R
    • peak: ⟨P⟩ = (1/2)V₀I₀
  • If asked for instantaneous power, write p(t) = v(t)i(t) first.
  • On a sketch, show two non-negative power peaks during each voltage cycle.

6. Worked Examples

Modelled example 1

Mean power from r.m.s. voltage

Core

Problem

A 60 Ω resistor is connected to a Vᵣₘₛ = 12 V supply. Find the mean power dissipated.
Study the worked solution

  1. Choose an r.m.s. power relation

    Method

    Use ⟨P⟩ = Vᵣₘₛ²/R.

    Reason

    R.m.s. voltage gives the same mean heating as the equivalent d.c. voltage.

    Working

    ⟨P⟩ = 12²/60

  2. Evaluate

    Method

    Obtain 2.4 W.

    Reason

    Both voltage and resistance are already in SI units.

    Working

    ⟨P⟩ = 2.4 W

Guided practice 2

Mean power from peak current

About 4 min

Problem

A resistor of R = 10 Ω has current i = 5.0 sin(100π t) A. Find the mean power.

Try this before viewing the solution

Unit: W

Hints

Hint 1: identify the current value
The coefficient outside the sine is I₀, not Iᵣₘₛ.
Hint 2: use the peak-value form
For sinusoidal current, ⟨P⟩ = 1/2 I₀²R.
View solution step by step

  1. Identify the peak current

    Method

    Read I₀ = 5.0 A.

    Reason

    The sine factor has a maximum magnitude of one.

    Working

    I₀ = 5.0 A.

  2. Calculate mean power

    Method

    Apply the peak-value form.

    Reason

    The time average of sin² over a cycle is 1/2.

    Working

    ⟨P⟩ = (1/2)(5.0)²(10) = 125 W

Common misconception 3

Instantaneous power at a given time

Find and correct the mistake

Learner claim

A resistor has v = 20 sin(100π t) V and i = 2.0 sin(100π t) A. A learner uses (1/2)V₀I₀ = 20 W as the power at t = 5.0 ms. Diagnose the error and find the instantaneous power.

Try this before viewing the solution

What does 20 W represent?

View solution step by step

  1. Form instantaneous power

    Method

    Multiply the instantaneous voltage and current.

    Reason

    The question asks for p(t), not the cycle average.

    Working

    p = vi = 40 sin² (100π t)

  2. Evaluate the phase

    Method

    Substitute t = 5.0 × 10⁻³ s.

    Reason

    100π t = π/2, so sin² (π/2) = 1.

    Working

    p = 40(1) = 40 W

Examiner practice 4

Mean power from peak voltage

3 marks

Examination question

A 100 Ω resistor is connected to a sinusoidal supply of peak voltage V₀ = 50 V. Find the mean power dissipated. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Select a peak-value relation

    1 mark

    Method

    Use ⟨P⟩ = (1/2)V₀²/R.

    Reason

    The supplied voltage is peak, not r.m.s.

    Working

    ⟨P⟩ = (1/2)50²/100

  2. Evaluate

    1 mark

    Method

    Obtain 12.5 W.

    Reason

    The factor 1/2 converts the squared sinusoidal peak to its mean.

    Working

    ⟨P⟩ = 12.5 W.

  3. State the condition

    1 mark

    Method

    Identify the load as purely resistive.

    Reason

    The peak voltage and current are in phase only under this stated model.

    Working

    Condition: purely resistive load.

Challenge 5

Mean power using mains r.m.s. voltage

Minimal support

Independent transfer

A 60 Ω resistive heater is connected to mains Vᵣₘₛ = 230 V. Estimate its mean power and r.m.s. current.

Try this before viewing the solution

Hints

Hint 1: mains rating
The stated mains voltage is already an r.m.s. value.
View solution step by step

  1. Find mean power

    Method

    Use Vᵣₘₛ²/R.

    Reason

    R.m.s. voltage directly represents heating effectiveness.

    Working

    ⟨P⟩ = 230²/60 = 8.82 × 10² W

  2. Find r.m.s. current

    Method

    Use consistent r.m.s. values in Ohm’s law.

    Reason

    The load is resistive.

    Working

    Iᵣₘₛ = 230/60 = 3.83 A

7. Mind Stretchers

Mind stretcher 1: Why instantaneous power is never negativeExtension

Explain why instantaneous power in a resistor on sinusoidal a.c. never becomes negative.

Show Answer

For a resistor, v and i are in phase: v = V₀ sin(ω t) and i = I₀ sin(ω t).

So p = vi = V₀I₀ sin² (ω t), and sin² (ω t) ≥ 0 always. Therefore p is never negative.

Mind stretcher 2: Consistency of peak vs r.m.s. power formsExtension

Show that ⟨P⟩ = (1/2)V₀I₀ is consistent with ⟨P⟩ = VᵣₘₛIᵣₘₛ for a sinusoid.

Show Answer

For a sinusoid, Vᵣₘₛ = V₀/square root of 2 and Iᵣₘₛ = I₀/square root of 2.

So:

VᵣₘₛIᵣₘₛ = V₀I₀/2 = (1/2)V₀I₀ = ⟨P⟩

Mind stretcher 3: Optional (Enrichment)Extension

A. Non-resistive loads (beyond this lesson)

If the circuit includes capacitors/inductors, v and i are not in phase and IᵣₘₛVᵣₘₛ is not the mean power. You then need power factor ideas, which are beyond this resistive-load lesson.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027