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AC Electricity

Time variation of a DC voltage or current

2

I

V

Current

Voltage

time t

time t

Sinusoidal Waveform

3

voltageor

current

Sinusoidal Wave

-

+

0

T

t

The relationship between frequency, f, in hertz and period, T, in seconds, is given by the expression:

f = 1/T

Mathematically, we can represent the sinusoid as a function of time using the equation

πft θVπft VtV 2,sinˆ2sinˆ

is the peak value of the waveform and f is its frequency in Hz.

This is simply the sine function of an angle measured in radians and where the angle varies with time in accordance with the angular frequency = 2f, ie

ftπft θ 2,2

is measured in radians per second

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coil thelinking in webers) (flux in change ousinstantane theis

and coil theof turnsofnumber theis N Where

volts voltageo/p that thegives Law sFaraday'

dtd

dt

dN e

5

ab

c

At point a the magnetic flux linking the coil is maximum, the rate of change of magnetic flux is zero thus the induced voltage is zero.

At point c the magnetic flux linking the coil is zero, the rate of change of magnetic flux being cut is maximum thus the induced voltage is maximum

At point b the magnetic flux linking the coil is NØ = N A B cos θ, θ = ωt = 2πft = 45o, thus the induced voltage v = -NdØ/dt = (ωNAB) sin θ, θ = ωt = 2πft = 45o.

where N = the number of turns on the coil, A is its cross-sectional area, and B is the magnetic flux density between the poles of the magnet.

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00

7

450

8

900

9

1350

10

1800

11

2250

12

2700

13

3150

14

3600 Or 00

15

00 from angle phase the is

wave the of value maximum the is ˆ

wave the of value ousinstantane the is where

sin

V

v

Vv

00 3600

V̂

V̂

V0

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Definitions

1 Waveform

2 Instantaneous value

3 Peak value

4 Peak to peak (p/p) value

5 Periodic waveform

6 Period (T)

7 Cycle

8 Frequency (f)

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Important parameters for a sinusoidal voltage.

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Defining the cycle and period of a sinusoidal waveform.

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Demonstrating the effect of a changing frequency on the period of a sinusoidal waveform.

The unit of measurement for frequency is the hertz (Hz) where

1 hertz (Hz) = 1 cycle per second (c/s)

f T

T f

1 Period

1 frequency

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Determine

1 Period

2 frequency

3 Amplitude

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FIGURE 13.18 Basic sinusoidal function.

tVv

Vv

sinˆ

sinˆ

T PeriodHertz

1

Tf

V̂

f 2

22

V̂

V̂

V0

tVv sinˆ

)30sin(ˆ oωtVv

o30

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090

tVv sinˆ

tVv cosˆ

ωtVωtVv o cosˆor 90sinˆ waveBlue

Power dissipation in a DC circuit

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I

(I2R)Power

Current

t

t(a) Current variation through a resistance R in a DC circuit

(b) Power dissipation in a resistance R in a DC circuit

 Power dissipation in an AC circuit

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i2 (t)Irms 2Average of squared current = 1/2 Ip

2

Ip 2

Time, t

Current

i (t)

p

p

rms II

IR 707.02

,.IRI2

1P 2

rms

2

pav

RtI P 222 sinRtitP

Power dissipation is proportional to voltage (or current) squared)

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0

-2

+2

+4

-4

Imagine the sinewave shown below A = 2.sin α is squared.

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The waveform shown in red below is obtained

Note that it is all positive

And twice the frequency of the blue waveform

0

-2

+2

+4

-4

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0

-2

+2

+4

-4

Peak Value = 2

Peak Squared Value = 4

Mean Square Value = 2

Root Mean Square Value 2

2

22

valuepeak

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The root mean square value of an ac voltage or current is the equivalent d.c.voltage or current that will produce the same power (heating effect).

.1.414 2 by value peak the dividing

by obtained is value rms the waveforms sinusoidalFor

ROOT MEAN SQUARE (r.m.s.) value

0.707 2

1 e by peak valulying theely multipor convers

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AC Voltage & Current In a Resistor

For a purely resistive element (i.e. one that does not contain any capacitive or inductive elements) the voltage across and the current through the element are in phase, with the ratio of voltage across to the current through giving the resistance.

i.e. the same as for a d.c. circuit

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ωtIitVv mRmR sin andsin

sin

sin

resistance flowcurrent to opposition

m

m

m

m

R

R

I

V

ωtI

tV

i

v

For the circuit shown , calculate:(a) The rms current supplied by the generator.(b) The rms voltage across the 2.2 kΩ resistor.(c) The average power dissipated in the 6.8kΩ resistor.

V

18 V rms1 kHz

R1= 6.8 k

R2= 2.2 k

22 iRV

1

2RiP

Solution

(a) The total circuit resistance, RT = 6.8 + 2.2 = 9kThe rms current, I = v/R

= 18/(9 × 103) = 2mA

(b) Let v2 = rms voltage across R2 = 2.2k

= 27.2mW

= 2 × 10-3 ×2.2 × 103 = 4.4V

(c)Let P = average power dissipation in R1 = 6.8k

= (2 × 10-3)2 × 6.8 × 103

= 27.2 × 10-3