1 AC Electricity. Time variation of a DC voltage or current 2 I V Current Voltage time t.
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Transcript of 1 AC Electricity. Time variation of a DC voltage or current 2 I V Current Voltage time t.
1
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
4
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.
6
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
23
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