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SCIENCE · INVESTIGATE AND EXPLAIN

Separate sinusoidal peak, rms and transformed power

A rotating coil in a uniform field has sinusoidally changing flux. The supplied ideal generator has50 turns, area .02 m² and field .1 T, so peak emf is NABω=.1ω with ω=2πf. One complete cycle has zero signed mean voltage, but nonzero mean squared voltage. A resistor therefore can absorb positive mean power even though the voltage alternates sign.

Learning outcome: Compare frequencies10,20,40 Hz at one turns ratio, then change turns ratio at one frequency. Retain peak, rms and signed mean separately. Explain why doubling secondary voltage across the same resistor quadruples power and changes the required primary current.

Starting recap

Use a right-handed coordinate system, oriented surface normal, signed charge/current and declared energy boundary. Separate an entry force from a trajectory, emf from current, and rms from signed average.

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Try

At f=20 Hz, generator peak emf=4π V and primary rms=2√2π≈8.885766 V. With turns ratio1.5, secondary rms=3√2π≈13.328649 V. Secondary mean power is18π²/100≈1.776529 W, and ideal primary mean power matches.

Open the experiment

Predict

Zero signed mean sinusoidal voltage across a resistor implies:

Experiment

Compare frequencies10,20,40 Hz at one turns ratio, then change turns ratio at one frequency. Retain peak, rms and signed mean separately. Explain why doubling secondary voltage across the same resistor quadruples power and changes the required primary current.

Compare sinusoidal peak, rms and ideal transformed power

Worked teaching model · synthetic values · Ideal rotating 50-turn coil area .02 m² in B=.1 T provides sinusoidal source. Ideal transformer drives a purely resistive 100 Ω secondary load. No winding/core losses, saturation or magnetizing current. Fixed rotation maintained externally; no mains equipment claim.

Reset keeps your prediction, captures and practice history.

One sinusoidal generator cycle

instantaneous emf (V)time (s)00.0512.5664-12.5664

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: maintained rotation (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for One sinusoidal generator cycle

maintained rotation

time (s)instantaneous emf (V)
00
0.001251.96581
0.00253.88322
0.003755.70501
0.0057.38633
0.006258.88577
0.007510.1664
0.0087511.1967
0.0111.9513
0.0112512.4117
0.012512.5664
0.0137512.4117
0.01511.9513
0.0162511.1967
0.017510.1664
0.018758.88577
0.027.38633
0.021255.70501
0.02253.88322
0.023751.96581
0.0251.53894e-15
0.02625-1.96581
0.0275-3.88322
0.02875-5.70501
0.03-7.38633
0.03125-8.88577
0.0325-10.1664
0.03375-11.1967
0.035-11.9513
0.03625-12.4117
0.0375-12.5664
0.03875-12.4117
0.04-11.9513
0.04125-11.1967
0.0425-10.1664
0.04375-8.88577
0.045-7.38633
0.04625-5.70501
0.0475-3.88322
0.04875-1.96581
0.05-3.07787e-15

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Angular frequency125.664rad/s
Generator peak emf12.5664V
Primary rms voltage8.88577V
Secondary rms voltage13.3286V
Secondary rms current0.133286A
Primary rms current0.19993A
Mean secondary resistor power1.77653W
Mean primary electrical power1.77653W
Cycle-mean signed generator emf0V

Zero cycle-mean voltage does not mean zero rms voltage or zero mean resistor power.

Peak emf=NABomega. For a sine, Vrms=Vpeak/sqrt2; a purely resistive load has mean power Vrms²/R. An ideal transformer gives Vs/Vp=Ns/Np and Ip/Is=Ns/Np, conserving mean power while changing voltage/current. The mechanism requires changing linked flux; steady DC does not produce the stated continuous secondary sinusoid. Ideal lossless assumptions do not certify real equipment behavior.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Zero cycle-mean voltage does not mean zero rms voltage or zero mean resistor power.

Peak emf=NABomega. For a sine, Vrms=Vpeak/sqrt2; a purely resistive load has mean power Vrms²/R. An ideal transformer gives Vs/Vp=Ns/Np and Ip/Is=Ns/Np, conserving mean power while changing voltage/current. The mechanism requires changing linked flux; steady DC does not produce the stated continuous secondary sinusoid. Ideal lossless assumptions do not certify real equipment behavior.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define vectors, source and system boundary

A rotating coil in a uniform field has sinusoidally changing flux. The supplied ideal generator has50 turns, area .02 m² and field .1 T, so peak emf is NABω=.1ω with ω=2πf. One complete cycle has zero signed mean voltage, but nonzero mean squared voltage. A resistor therefore can absorb positive mean power even though the voltage alternates sign.

Derive the quantitative comparison

For a sine wave, rms voltage is peak divided by √2. Across a purely resistive load, mean power is Vrms²/R, not the square of the signed mean voltage. The ideal transformer gives secondary-to-primary voltage ratio n=Ns/Np and primary-to-secondary current ratio n. With a100 Ω secondary load, calculate secondary current first, then use power conservation to infer primary current.

Separate physical evidence from a model extension

The mechanism requires changing linked flux. Steady DC does not produce the stated continuing sinusoidal secondary voltage. The model omits core losses, winding resistance, saturation, magnetizing current and mechanical loss; the external rotation source maintains the frequency while supplying the electrical output. These calculations neither describe a real power outlet nor authorize mains or transformer handling. They are an algebraic rms and energy-transfer account.

Worked original case

At f=20 Hz, generator peak emf=4π V and primary rms=2√2π≈8.885766 V. With turns ratio1.5, secondary rms=3√2π≈13.328649 V. Secondary mean power is18π²/100≈1.776529 W, and ideal primary mean power matches.

Investigation

Compare frequencies10,20,40 Hz at one turns ratio, then change turns ratio at one frequency. Retain peak, rms and signed mean separately. Explain why doubling secondary voltage across the same resistor quadruples power and changes the required primary current.

Changed-case practice

Try the changed case before feedback

A sine voltage has peak10 V across a20 Ω resistor. Find mean power in W.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and evidence boundary

Original prose and numerical cases checked 30 October 2026 against the relevant primary explanation. All records and diagrams are virtual. No real coil, powered wire, high-field exposure, motor, transformer, mains connection or inductor switching is requested. Retain first predictions, contrasting recipes and independent criterion review; public agreement is not practical certification or demonstrated mastery.

Apply

Zero signed mean sinusoidal voltage across a resistor implies:

Public practice checks a specific claim. It does not establish independent mastery.

Which independent observation could reject this model rather than merely refit its parameters?

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