☰ Year plan · Physics

The sine wave, the DC generator and the transformer

Week 10 · Chapter 3 (C): the sine wave of a generator and its frequency (50 Hz at home), the effective value (0.707 of the maximum), rectification and the DC generator (split-ring commutator), the transformer (turns, voltage and current ratios, step-up and step-down) · 3 lessons
1

The generator’s sine wave: frequency and effective value

Physics Sec 3 Term 1, Chapter 3: pp. 73–79
GoalUse frequency and period, and the effective value Ieff = 0.707 Imax, Veff = 0.707 Vmax.
New words
frequency (f) and period (T) · f is the number of cycles per second, in hertz (Hz); T = 1 ÷ f is the time of one cycleeffective value · the value of a steady direct current that would heat the same resistor at the same rate as the alternating currentpeak (maximum) value · the largest value of the emf or current in a cycle: Imax or Vmax
Explanation

In an AC generator one turn of the coil is one cycle of the emf, so the frequency f in hertz equals the number of turns per second, and the period is T = 1 ÷ f. The emf, and the current in a resistor, reverse direction every half turn and follow a sine curve. The home supply has f = 50 Hz, so T = 0.02 s. Over a whole cycle the average current is zero, yet a resistor still gets hot, because the heating depends on I², which is never negative. The effective value is the steady current that gives the same heating: Ieff = Imax ÷ √2 = 0.707 Imax and Veff = 0.707 Vmax. Meters and labels use effective values: a 220 V supply is 220 V effective, with a peak of 220 ÷ 0.707 = 311 V. For a resistor R the average power is Ieff² R.

t (ms)V (V)10203040−400−300−200−100100200300400050 Hz supplypeak 311 VVeff = 220 V
Worked examples
A generator turns at 1500 turns per minute. That is 1500 ÷ 60 = 25 turns per second, so f = 25 Hz and T = 1 ÷ 25 = 0.04 s.
Imax = 4.0 A gives Ieff = 0.707 × 4.0 = 2.83 A. Vmax = 170 V gives Veff = 0.707 × 170 = 120 V. A 220 V supply has Vmax = 220 ÷ 0.707 = 311 V.
A resistor of 8.0 Ω carries an alternating current from a source of peak voltage 24.0 V. Imax = 24.0 ÷ 8.0 = 3 A, Ieff = 0.707 × 3.0 = 2.12 A, and the average power is Ieff² R = 2.121² × 8.0 = 36 W (a steady 16.97 V would give the same).
Class activity

Paper practice. Sketch two cycles of a 50 Hz sine wave and mark the times 0, 5, 10, 15 and 20 ms on the axis. Mark where the emf is zero and where it is at its positive and negative peaks. Then mark the effective value on the vertical axis for a peak of 100 V.

Practice
1
A generator turns at 900 turns per minute. Find f and T.
2
An alternating current has a maximum of 7.5 A. Find its effective value.
3
A supply is marked 120 V effective. Find its peak voltage.
4
An emf is e = Emax sin 2πft with Emax = 10 V and f = 50 Hz (π = 3.14). Find e at t = 0.0025 s.