Represent an alternating current with a sinusoidal equation
Use x = x₀ sin ωt to connect an AC waveform's peak, angular frequency and value at a given time.
Section 21 · Alternating currents
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 21. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use period, frequency, sine graphs, current, voltage, power, diodes, capacitors and SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In x = x₀ sin ωt, what is x₀?
What you will learn
- Use x = x₀ sin ωt for sinusoidally alternating current or voltage
- Relate angular frequency to period and frequency
- Evaluate the instantaneous value at a specified time
The sine equation encodes peak and cycle timing
A sinusoid that starts at zero and initially rises can be represented by x = x₀ sin ωt. Here x is the instantaneous current or voltage, x₀ is its peak value, t is time, and ω is angular frequency in rad s⁻¹. The phase argument ωt is measured in radians.
One complete cycle is 2π radians, so ω = 2π/T = 2πf. At t = 0, x = 0; at a quarter period, the sine reaches +1 and x = +x₀; at half a period, x returns to zero. A waveform with a different starting phase needs an added phase term, but the syllabus form here is x = x₀ sin ωt.
Evaluate the sine argument before multiplying by the peak
To find an instantaneous value, calculate ωt in radians, evaluate sin(ωt), then multiply by x₀. Keep the sign: the sign identifies the current direction or voltage polarity relative to the chosen reference.
Check limiting points against the waveform. The value cannot exceed x₀ in magnitude; the current or voltage alternates above and below zero. If the time is one full period later, the value repeats because the sine argument has increased by 2π.
Pause and explain
An AC equation has ω = 80π rad s⁻¹. What is its frequency?
Worked example
A current is i = 6.0 sin(100πt) A. Find its peak value, frequency and current at t = 0.0050 s.
Show the worked solution
- Read x₀ and ω from the equation.
- Use f = ω/(2π).
- Evaluate the sine argument in radians.
Answer I₀ = 6.0 A, f = 100π/(2π) = 50 Hz, and at 0.0050 s the argument is π/2, so i = 6.0 A.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A voltage is v = 12 sin(200πt) V. Find its frequency, period and value at t = 0.0025 s.
Give me a hint
Compare ω with 2πf and evaluate 200πt in radians.
Compare my reasoning
- f = ω/(2π) = 200π/(2π) = 100 Hz, so T = 0.010 s.
- At 0.0025 s, 200πt = π/2.
- v = 12 sin(π/2) = +12 V.
f = 100 Hz, T = 0.010 s and v = +12 V at 0.0025 s.
Look for these in your work
- Frequency found
- Period found
- Instantaneous value evaluated with sign
An alternating current has peak value 4.0 A and frequency 25 Hz. Write an equation in the form i = I₀ sin ωt.
Give me a hint
Calculate ω = 2πf and retain the stated sine form.
Compare my reasoning
- I₀ = 4.0 A.
- ω = 2π × 25 = 50π rad s⁻¹.
- Substitute into i = I₀ sin ωt.
i = 4.0 sin(50πt) A.
Look for these in your work
- Peak value assigned
- Angular frequency calculated
- Equation written with units/context
For v = 100 sin(100πt) V, compare the voltage at t = 0.0025 s and t = 0.0125 s.
Give me a hint
Evaluate each argument in radians and use the sine's sign.
Compare my reasoning
- At 0.0025 s, the argument is π/4, so v ≈ 100 × 0.707 = +70.7 V.
- At 0.0125 s, the argument is 5π/4, so v ≈ 100 × (−0.707) = −70.7 V.
- The equal magnitudes and opposite signs occur half a period apart.
The voltages are approximately +70.7 V and −70.7 V.
Look for these in your work
- Arguments evaluated
- Signs retained
- Waveform timing interpreted
Common mistakes
- Treating ω as frequency in hertz
- Evaluating the sine angle in degrees when the equation uses radians
- Dropping the sign of an instantaneous value
- Using peak-to-peak value as x₀
What can you explain now?
In x = x₀ sin ωt, what is x₀?
Compare with the explanation
The peak value
The sine ranges from −1 to +1, so x₀ sets the maximum magnitude.
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Make it stick
How do x₀, ω and t determine an instantaneous alternating current or voltage?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.