A.C. waveform and power explorer
Drag the crest of a sinusoidal supply to set its peak and period, then read r.m.s. values, resistive power and ideal half-wave output from synchronised graphs.
Learning goals
- Represent sinusoidal A.C. and use peak and r.m.s. values.
- Analyse mean power in resistive A.C. loads and half-wave rectification.
A 12.0 volt peak, 40 hertz supply across a 6.00 ohm resistor. Period 25.0 milliseconds, r.m.s. p.d. across the resistor 8.49 volts, mean power 12.0 watts.
- Period, T
- 25.0 ms
- ω = 2πf
- 251 rad s⁻¹
- Vrms across R
- 8.49 V
- Irms
- 1.41 A
- Peak power, P₀
- 24.0 W
- Mean power, ⟨P⟩
- 12.0 W
Try this
0 of 4 doneSet the supply to 120 V r.m.s., as in North American homes, with no diode. (not done yet)
Vrms = V₀/√2, so a 120 V supply peaks at 170 V, and Singapore's 230 V mains peaks at 325 V.
Drag the crest to make the period 12.5 ms. (not done yet)
f = 1/T = 80 Hz, and ω = 2πf = 503 rad s⁻¹: halving T doubles f and ω.
Make the mean power in the resistor 50 W, with no diode. (not done yet)
⟨P⟩ = V₀²/(2R) = ½P₀: a steady p.d. equal to Vrms would heat the resistor at the same rate.
Add the diode and compare the trace and the mean power. (not done yet)
The diode conducts only on the positive half-cycles: one pulse per cycle, Vrms = V₀/2 and the mean power halves.
You’ve tried every task.