Sinusoidal a.c., peak and r.m.s. values
Key idea: H2 Physics lessons on drift velocity, electrical energy, sinusoidal supplies and half-wave rectification.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: How do peak and r.m.s. values describe a sinusoidal supply?
A sinusoidal current changes direction and follows I = I₀sinωt, with period T = 2π/ω. The r.m.s. value is the direct current producing the same mean heating power: Irms = I₀/√2 and Vrms = V₀/√2 for a sine wave.
Read the waveform before using a formula
A sinusoidal current can be written I = I₀ sin ωt, where I₀ is the peak value and ω = 2πf = 2π/T. Peak-to-peak current is 2I₀. The sign shows direction; it does not mean that the current's magnitude is physically negative.
The mean current over a whole cycle is zero because positive and negative halves cancel. That does not mean zero heating, because resistor power I²R is positive in both halves.
Check your understanding: A sine wave has period 20 ms and peak voltage 12 V. Find frequency and angular frequency.
f = 1/0.020 = 50 Hz and ω = 2πf = 314 rad s⁻¹.
Define r.m.s. by equal heating
The r.m.s. current is the steady direct current that would produce the same mean power in a resistor. For a sine wave, I_rms = I₀/√2 and V_rms = V₀/√2. The √2 relation is not valid for an arbitrary waveform.
The name describes the calculation: square the instantaneous values, average them over a cycle, then take the square root. This prevents the two directions from cancelling.
Check your understanding: Why is r.m.s. not the average magnitude of a sine wave?
It is defined through mean squared value and equal heating power; the average magnitude is a different number.
Key ideas to keep
- Peak-to-peak value is twice the peak value.
- R.m.s. is not the simple mean of a sine wave, which is zero over a full cycle.
- The √2 relations apply specifically to sinusoidal waveforms.
See the reasoning
Worked example
Build a sinusoidal-current equation from r.m.s. data
Question: A 60 Hz sinusoidal current has Iᵣₘₛ = 3.0 A and is zero, increasing positively, at t = 0. Write i(t).
Step 1: Recover the peak
Why: For a sine wave, r.m.s. is peak divided by √2.
Working: I₀ = √2(3.0) = 4.24 A.
Step 2: Convert frequency to angular frequency
Why: The sine argument uses radians.
Working: ω = 2πf = 2π(60) = 120π rad s⁻¹.
Step 3: Use the stated initial phase
Why: Zero and increasing at t = 0 matches a positive sine with no phase constant.
Working: i = 4.24 sin(120πt) A.
Answer: I₀ = √2 Iᵣₘₛ = 4.24 A and ω = 2πf = 120π rad s⁻¹, so i = 4.24 sin(120πt) A.
Check: At t = 0 the expression gives zero, and its initial gradient is positive.
Use a hint if needed
Practise with support
Try this
A sinusoidal voltage has period 8.0 ms and peak 20 V. Find f, ω and Vᵣₘₛ.
Hint: Convert milliseconds before taking the reciprocal.
Check your answer
f = 1/T = 125 Hz, ω = 2πf = 785 rad s⁻¹ and Vᵣₘₛ = 14.1 V.
Now work without the hint
Practise independently
Your turn
Define period, frequency, peak and r.m.s. value, and explain the physical meaning of r.m.s.
Check your answer
T is time per cycle, f = 1/T is cycles per second, and peak is maximum magnitude. The r.m.s. current or voltage is the steady d.c. value producing the same mean power in a resistor; for a sinusoid it is peak/√2.
Avoid these traps
Common mistakes
Common mistake
The r.m.s. value is the arithmetic mean of a sinusoid.
What is wrong with this reasoning?
Show better thinking
A full sinusoid has zero arithmetic mean; r.m.s. is the d.c.-equivalent value for mean resistive power.
Common mistake
Peak/√2 applies to every alternating waveform.
What is wrong with this reasoning?
Show better thinking
The stated peak/√2 relation is for a sinusoidal current or voltage.
Write for the examiner
Exam guidance
Mark peak, period and zero crossings on the waveform before extracting values.
Exam-style practice [6 marks]
A 50 Hz supply has Vᵣₘₛ = 230 V. Find V₀, T and ω, then write v(t) for zero phase.
Plan before you answer
- Convert r.m.s. to peak.
- Find T and ω from f.
- Use the zero-phase condition.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
V₀ = √2(230) = 325 V, T = 0.020 s, ω = 100π rad s⁻¹ and v = 325 sin(100πt) V.
Come back in three days
Check what stayed with you
Recall question
For i = 5.0 sin(400πt) A, find f and Iᵣₘₛ.
Check the answer
ω = 400π rad s⁻¹, so f = 200 Hz. Iᵣₘₛ = 5.0/√2 = 3.54 A.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 15 states no explicit exclusions. q in I = nAvq is treated as carrier-charge magnitude when calculating current magnitude; electron drift is opposite conventional current. E.m.f. is energy supplied per unit charge, while p.d. is energy transferred from electrical form per unit charge. The peak/√2 and half-maximum-power results are restricted to sinusoidal waveforms and a resistive load; a single ideal diode gives unsmoothed half-wave rectification.
- GCE A-Level H2 PhysicsTopic 15(f) / Topic 15(g) / Topic 15(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027