21. Alternating currents

Exam analysis

Most tested topics

21.1 Characteristics of alternating currents

An alternating voltage or current varies periodically, with period T, frequency f=1/T, angular frequency ω=2πf and peak magnitude I₀ or V₀.

Read peak from the centre line to an extreme, not peak-to-peak, and distinguish the waveform’s cycle time from its angular frequency.

A 50 Hz mains waveform has T=20 ms and ω=100π rad s⁻¹.

The mean of a symmetric AC waveform can be zero while its heating effect is not zero.

$x=x0 sin(ωt+φ), where ω=2πf=2π/T$

Parameter	Controls
x0	peak magnitude
ω	cycle rate / period
φ	value and direction at t=0

For φ=0	t=0	T/4	T/2	3T/4	T
x	0 rising	+x0	0 falling	-x0	0 rising

For v=12 sin(100πt) V, V0=12 V, f=50 Hz and T=0.020 s. At t=5.0 ms=T/4, v=+12 V; at 10 ms, v=0 and falling.

A sinusoid of peak 18 V that starts at +18 V is represented conveniently by v=18 cosωt, or equivalently v=18 sin(ωt+π/2).

Use angular frequency ω—not f—inside the trigonometric phase unless a factor 2π is included. x is instantaneous, x0 is peak, and neither is automatically an rms value.

With i=I₀sinωt and fixed resistance R, instantaneous power i²R varies from zero to I₀²R; its cycle average is ½I₀²R.

Use average over a complete cycle and distinguish peak power from mean heating power.

If peak current doubles, mean resistive power quadruples because it depends on I₀².

A zero mean current does not imply zero mean power, since power depends on current squared.

For a sinusoidal AC signal, rms current or voltage produces the same mean power in a resistor as a DC value: I_rms=I₀/√2 and V_rms=V₀/√2.

Use rms values in P=VI or P=I²R for resistive loads, and specify whether a quoted AC value is rms or peak.

A 10 A peak sinusoidal current has I_rms≈7.07 A.

RMS is not the arithmetic average of a symmetric waveform, which is zero.

21.2 Rectification and smoothing

Half-wave rectification passes one polarity half-cycle; full-wave rectification flips both half-cycles to the same load polarity.

Read the output graph: half-wave has gaps each alternate half-cycle, while full-wave pulses occur twice per input cycle and have higher ripple frequency.

A bridge rectifier uses four diodes to produce full-wave output without a centre-tapped transformer.

Rectification does not by itself make perfectly steady DC; smoothing capacitors reduce ripple after the diode stage.

A diode conducts mainly in forward bias and blocks reverse bias, so one diode passes only one half-cycle of an AC input.

Read diode orientation and output polarity, then identify the gaps in the load voltage waveform.

A positive half-wave output has pulses separated by zero-voltage intervals during the negative input half-cycle.

A diode does not convert AC directly into smooth DC; the output is pulsating and needs smoothing.

A bridge has four diodes in a diamond. Connect the AC source to one pair of opposite corners and the load to the other opposite pair. Both diodes meeting the positive load corner point toward it; both connected from the negative load corner point away from it.

Input half-cycle	Conducting path	Load result
first AC terminal positive	one diagonal pair of diodes conducts	current crosses load from its fixed + side to fixed - side
second AC terminal positive	the other diagonal pair conducts	current crosses load in the same direction

Both input halves become same-polarity output pulses, so the rectified output frequency is twice the input frequency. Each conducting route contains two forward-biased diodes.

To check a drawn bridge, trace conventional current from either AC corner through one diode, the load in the chosen direction, and a second diode back to the other AC corner. Repeat after swapping which AC corner is positive.

Only two opposite-path diodes conduct on each half-cycle, not all four. The bridge rectifies but does not itself remove ripple; a smoothing stage is separate.

Connect the smoothing capacitor in parallel with the load resistor, with correct polarity for a polarised capacitor. Its voltage is therefore the output voltage.

Part of rectified cycle	Capacitor action	Output shape
rising input exceeds capacitor voltage near a peak	diode conducts; capacitor charges rapidly toward peak	rapid rise/recharge
input falls below capacitor voltage	diode turns off; capacitor discharges through load	curved exponential fall
next peak arrives	diode conducts and recharges capacitor	ripple repeats above zero

$Between peaks, V≈V_peak e^(-t/(R_load C)); discharge timescale τ=R_load C.$

Change	Effect on discharge/ripple
increase C	larger τ, slower fall, smaller ripple, higher minimum output
increase R_load (lighter load)	larger τ, smaller load current, smaller ripple, higher minimum output
use full-wave rather than half-wave at same input f	peaks arrive twice as often, less discharge time, smaller ripple

Decreasing C or R_load produces faster decay, a lower minimum and larger peak-to-peak ripple. Very large C reduces ripple but causes large short charging-current pulses near peaks.

Smoothing does not create perfectly constant voltage and the capacitor must not be placed in series with the load. Increasing load resistance reduces ripple; increasing load current does the opposite.

21. Alternating currents

Exam analysis

Most tested topics

21.1 Characteristics of alternating currents

An alternating quantity is described by period, frequency, angular frequency and peak value

Use x=x0 sin(ωt+φ) for sinusoidal alternating current or voltage

For a sinusoidal current in a resistor, mean power is half the maximum instantaneous power

RMS values give the DC-equivalent heating effect: I_rms=I₀/√2 and V_rms=V₀/√2

21.2 Rectification and smoothing

Half-wave and full-wave rectification differ in which half-cycles reach the load

A single diode performs half-wave rectification by passing one polarity half-cycle

A four-diode bridge routes both AC half-cycles through the load in one direction

A capacitor across the load smooths rectified voltage by charging and discharging