Alternating Currents
21. Alternating Currents
5 min read 7 parts Lesson 1 of 1
Lesson summary
Alternating Currents — Cambridge AS & A Level Physics 9702 — CAMBRIDGE AS & A LEVEL PHYSICS 9702 — COMPLETE NOTES notes: Describe and calculate with alternating currents and voltages, use r.m.s.
- Alternating current (a.c.)
- Period and frequency
- Peak value
- Mains supply
- Angular frequency
- Using the equations
Part 1 of 7Alternating current and voltage
- Alternating current (a.c.) a current that varies periodically between positive and negative, reversing direction every half cycle.
- Waveformthe variation with time is sinusoidal, so the electrons in the wire move back and forth in simple harmonic motion.
- Period and frequencyT = 1/f.
- Peak valueI0 or V0, the maximum current or voltage, read as the amplitude of the graph.
- Mains supplypower stations produce a.c., and this is what comes from wall sockets.
- Time axisit is often in milliseconds, so check the unit.
Part 2 of 7Equations for a.c.
- Current — I = I0 sin ωt.
- Voltage — V = V0 sin ωt.
- Angular frequency — ω = 2π/T = 2πf, in rad s−1.
- Using the equations — read I0 and T from the graph, then substitute the time.
- A negative answer means the current is in the opposite direction, and its size is what matters.
- The calculator must be in radian mode.
Part 3 of 7Root-mean-square values
- R.m.s. current = the steady d.c. current that delivers the same power in a resistor as the alternating current.
- R.m.s. voltage = the steady d.c. voltage that delivers the same power in a resistor as the alternating voltage.
- Equations — Ir.m.s. = I0/√2 and Vr.m.s. = V0/√2.
- Ir.m.s. is about 0.707I0, around 70% of the peak value.
- These equations are not in the data booklet.
- Reading an equation — comparing I = 410 sin(100πt) with I = I0 sin ωt gives I0 = 410 A and ω = 100π rad s−1, so f = ω/2π = 50 Hz.
Part 4 of 7Mean and peak power
- Power formulaeP = IV = I2R = V2/R, and with r.m.s. values they apply to a.c. in a resistive load such as a lamp.What the symbols mean: P = power (W) · I = current (A) · V = potential difference (V)
- Mean powerPmean = (Ir.m.s.)2R.
- Peak powerPpeak = I02R = (√2 Ir.m.s.)2R = 2Pmean.
- Resultfor a sinusoidal supply, the mean power in a resistive load is half the maximum power.
Part 5 of 7Rectification
- Rectification converting alternating current and voltage into direct current and voltage, needed for electronics that require d.c.
- Diode action — a diode conducts only in the forward direction.
| Type | Circuit | Output |
|---|---|---|
| Half-wave | One diode in series with the load resistor | Only the positive half cycles, with zero during the negative halves, so less power is available |
| Full-wave | Bridge rectifier of four diodes, output across the load resistor | Both half cycles appear as positive bumps, giving more power |
- Half-wave output — a sine curve for the positive half cycles and a flat line (Vout = 0) on the negative cycles.
Part 6 of 7The bridge rectifier
- Positive half cycletwo diagonally opposite diodes are forward biased and conduct, and the other pair are reverse biased.
- Negative half cyclethe supply polarity swaps, so the other pair conduct and the first pair do not.
- Resultthe current always passes down through the load resistor in the same direction.
- Full-waveeach diode pair acts like a half-wave rectifier, and with two pairs the supply is used on every half cycle.
- Tracing currentfollow conventional current from the positive terminal through forward-biased diodes only.
Part 7 of 7Smoothing
- Smoothing = reducing the variation in the rectified output so it is closer to steady d.c.
- Set-up — a smoothing capacitor C in parallel with the load resistor R.
- The capacitor charges while the rectified voltage rises, holding the voltage high.
- When the supply voltage falls, the capacitor discharges gradually through R.
- The output shows a small ripple rather than dropping to zero.
- Better smoothing — a larger C or larger R makes the discharge slower and the ripples smaller.
- Design rule — the time constant RC must be much greater than the time between adjacent peaks of the rectified signal.
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