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Cambridge International AS & A Level · 21 · structured

Sinusoidal current and average heating

Sinusoidal current and average heating · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
6
Topics
2
Answer
Complete
structured6 marks

A purely resistive 5.00 Ω load carries current I=2.00 sin(100πt) A, where t is in seconds. Use radians.

  1. (a)

    Determine the instantaneous current at t=5.00 ms and identify the angular frequency and frequency.

    3 marks
  2. (b)

    Determine maximum and mean power. Explain why a zero mean current does not imply zero mean heating.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overview(a) I=2.00 A; ω=100π rad s⁻¹ and f=50.0 Hz. (b) P_max=20.0 W and mean P=10.0 W; heating depends on I², whose sine-squared mean is positive.
01

(a)

3 marks

Determine the instantaneous current at t=5.00 ms and identify the angular frequency and frequency.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Convert t=5.00×10⁻³ s; phase=100π×0.00500=π/2.

  2. 2

    I=2.00 sin(π/2)=2.00 A.

  3. 3

    ω=100π rad s⁻¹ and f=ω/(2π)=50.0 Hz.

02

(b)

3 marks

Determine maximum and mean power. Explain why a zero mean current does not imply zero mean heating.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Maximum power is I₀²R=2.00²×5.00=20.0 W.

  2. 2

    For a sinusoidal current in a resistor, mean power is half the maximum: 10.0 W.

  3. 3

    I reverses sign but I²R stays nonnegative; the mean of sin² over a complete cycle is 1/2.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

  1. (a) Convert t=5.00×10⁻³ s; phase=100π×0.00500=π/2.
  2. (a) I=2.00 sin(π/2)=2.00 A.
  3. (a) ω=100π rad s⁻¹ and f=ω/(2π)=50.0 Hz.
  4. (b) Maximum power is I₀²R=2.00²×5.00=20.0 W.
  5. (b) For a sinusoidal current in a resistor, mean power is half the maximum: 10.0 W.
  6. (b) I reverses sign but I²R stays nonnegative; the mean of sin² over a complete cycle is 1/2.

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Your self-assessment

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