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Magnetic fields and alternating currents (A Level)

21. Alternating Currents

5 min read 7 parts Lesson 24 of 24 Unit quiz

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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

Section 1 of 7Alternating current and voltage

  • Alternating current (a.c.) = a current that varies periodically between positive and negative, reversing direction every half cycle.
  • Waveform
    • the variation with time is sinusoidal, so the electrons in the wire move back and forth in simple harmonic motion.
  • Period and frequency
    • T = 1/f.
      What the symbols mean: T = period, the time for one complete wave (s) · f = frequency (Hz)
  • Peak value
    • I0 or V0, the maximum current or voltage, read as the amplitude of the graph.
  • Mains supply
    • power stations produce a.c., and this is what comes from wall sockets.
  • Time axis
    • it is often in milliseconds, so check the unit.

Section 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.
      What the symbols mean: ω = angular frequency (rad s−1) · f = frequency (Hz) · T = period (s)
  • 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.

Section 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.

Section 4 of 7Mean and peak power

  • Power formulae
    • P = 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 power
    • Pmean = (Ir.m.s.)2R.
  • Peak power
    • Ppeak = I02R = (√2 Ir.m.s.)2R = 2Pmean.
  • Result
    • for a sinusoidal supply, the mean power in a resistive load is half the maximum power.

Section 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.
TypeCircuitOutput
Half-waveOne diode in series with the load resistorOnly the positive half cycles, with zero during the negative halves, so less power is available
Full-waveBridge rectifier of four diodes, output across the load resistorBoth 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.

Section 6 of 7The bridge rectifier

  • Positive half cycle
    • two diagonally opposite diodes are forward biased and conduct, and the other pair are reverse biased.
  • Negative half cycle
    • the supply polarity swaps, so the other pair conduct and the first pair do not.
  • Result
    • the current always passes down through the load resistor in the same direction.
  • Full-wave
    • each diode pair acts like a half-wave rectifier, and with two pairs the supply is used on every half cycle.
  • Tracing current
    • follow conventional current from the positive terminal through forward-biased diodes only.

Section 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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