One waveform, two different questions: what is its net effect, and how much resistive heating does it produce? Predict, explore, and then calculate.
We use current i(t) in amperes. The same definitions apply to voltage. Positive and negative current refer to a chosen reference direction. All averages below use a complete repeating interval T.
The constant level with the same signed area over T. For current, it gives the net charge transferred per unit time. Positive and negative areas can cancel.
It is the DC component: Iav = Idc (also written IDC). The signed mean may be positive, zero, or negative.
The magnitude of a DC current producing the same average Joule heating in the same constant resistance. Squaring makes both current directions contribute positively.
It is nonnegative. Here RMS includes both DC and AC components.
Change one parameter at a time. Watch the signed area above and the squared waveform below.
Move the common cursor with the slider or click either plot.
Squaring the current also squares its unit; taking the square root returns a current unit. Dashed lines show averages, not instantaneous currents.
For a didactic constant resistor R = 2 Ω, the average dissipated power is:
The expressions and numerical substitution below follow the current settings. Integrate over the whole interval, including intervals where the current is zero.
Scroll sideways inside a formula or table if it is wider than the panel.
Use time t over one complete interval: 0 ≤ t < T. The expression below repeats every T. Use the same time unit for t and T; sine arguments are in radians.
Here T is the period of the rectified current: one positive lobe. The underlying unrectified sine has period 2T. Averaging over two lobes gives the same result.
The dimensionless shape g(t) rises from −1 to +1 in T/2, then falls back to −1. Its argument is the same time t as in the plots:
Integrate with respect to time from 0 to T, then divide by the full duration T. Split the integral wherever the waveform changes expression; keep the same divisor T even when the current is zero for part of the interval.
All results use the complete repeating interval and include DC. The amplitude is nonnegative. A “rectified sine” means the ideal waveform shown, not the current of every rectifier load.
For levels Ik lasting Δtk, use time-weighted contributions. The durations must add up to T. Do not average the RMS of separate intervals.
For any periodic waveform with finite RMS, the total mean square equals the square of its mean plus the mean square of its zero-mean AC part, defined as i(t) − ⟨i⟩. Every term uses the same averaging interval. Adding squared RMS values of arbitrary components requires a zero cross term (orthogonality), as for different harmonic orders averaged over a full fundamental period. In rectifiers, the extra subscript o denotes output: Io,dc is the mean of io(t), not necessarily the mean source current.
Measurement note: an AC-coupled RMS reading excludes DC. Also, the mean of the absolute value is not the signed mean and is generally not RMS. Check what a meter actually reports.
Specify both response and coupling: an AC-only true-RMS measurement rejects DC; total RMS requires AC+DC. For an ideal DC-coupled, sine-calibrated average-responding measurement, the indication is:
For a positive sine peak , a full-wave rectified sine gives , equal to its total RMS. A half-wave rectified sine gives , instead of its total RMS : about 29.3 % low. These examples preserve DC at the meter input.
An AC-coupled meter sees a different waveform after DC rejection, so do not transfer these indications to it. “True RMS” alone does not specify AC+DC; bandwidth and crest-factor limits also matter.
Technical reference: understand true-RMS measurements (Keysight).
Choose a short investigation. Make your prediction before loading the changed case, then compare the results in the explorer.
Start with a sine: peak 10 A, period 20 ms. Predict how the mean and RMS change when the negative half is reflected upward, without changing its magnitude.
Use 10 A unipolar pulses. Predict what happens to the mean, RMS, and average resistor power when the duty cycle changes from 25% to 50%.
Start with a DC level of 6 A and zero ripple. Predict what changes when a zero-mean triangular ripple of amplitude 4 A is added. The total current then ranges from 2 to 10 A.
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Choose one answer per question, then check it. Values are rounded. Feedback explains both the correct answer and the option you chose.
These are prescribed ideal periodic current waveforms, not a circuit simulator. Transients, switching losses, bandwidth limits, and temperature dependence are not modeled. Resistive-heating examples assume a constant 2 Ω resistor. Mean and RMS metrics use closed-form expressions; plot points only draw those expressions. Isolated ideal switching instants do not affect the integrals.
Optional manufacturer references support the heating interpretation and the distinction between AC-only and total RMS.
The TXT includes the source HTML identifier and export time. It keeps the existing numerical units and your notes exactly as written; it is a readable record, not an importable backup.