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Alternating voltages and currents (AC)
Page 88
DC- Direct current the flow in one direction only
AC-Alternating current The quantity is constantly changing in
direction
Introduction
Advantages of AC All large scale energy supplies around the
world are AC signals. This is because:
◦ Transformers are AC only.◦ AC motors are simpler and cheaper.◦ AC systems are easier to keep stable.◦ Rotating machines naturally generate AC.
L N
E
Sine Wave
Vp
T
)(SinVV p
)2( ftSinVV p
)( tSinVV p
t
V
The time taken for an alternating quantity to complete one cycle is called the time period , T.
The number of cycles completed in one second is called the frequency.
The frequency of the signal is given by f = 1/Period = 1/T (Hz)
Time period (T)
Determine the periodic time for frequencies (a) 50 Hz and (b) 20 kHz
The instantaneous value of V is given by V=Vp sin ().
The angle increases with time. So that = 2ft (radians). The term 2f is sometimes given the symbol
.
Peak value-the largest value reach in a half cycle
Peak-to-peak value- difference between maximum and minimum values
Average value- measured over a half cycle,
For a sine waveform,
p
avg
VV
2
pavg VV 636.0
RMS Value The root mean square of the signal (RMS)
is the DC voltage that has the same heating effect as the AC wave form.
It is the rms value that is used almost exclusively in circuit calculations.
All waveforms have an RMS value given by,
For a pure sine wave,
prms VV 707.02p
rms
VV
T
rms dtVT
V0
21
The form factor for a signal is,
For a pure sine wave this is
Form factor
avg
rms
V
V
valueaverage
valuermsfactorForm
pavg VV 636.0 prms VV 707.0
11.1636.0
707.0factorForm
p
p
V
V
1.A sinusoidal voltage has a maximum value of 120 V. Calculate its rms and average values and the form factor.
2. find the peak value for a 230 V mains supply.
This is single line that is imagined to spin at a speed equal to the frequency.
When we draw a phasor we imagine it is frozen at a point in time.
Phasors always spin anti-clockwise.
http://www.rkm.com.au/animations/animation-sine-wave.html
http://www.youtube.com/watch?v=IyhQnt5cgZs
Phasor Diagram
Phasor Diagram
When signals are in phase they may be dealt with in terms very similar to DC calculations but if a phase shift is involved which is very often the case we invoke the idea of the phasor.
Resistor
Inductor
The voltage across coils is not caused by resistance it is caused by the back emf generated due to Faraday's law.
XL Inductive Reactance
reactanceinductive)(2 LL
L XfLi
V
Derivation of XL Inductive Reactance The voltage across a coil
If Then
This is an equation for the instantaneous value.
It follows that:
reactanceinductive)(22
,
, Lp
p
rmsL
rmsL XfLI
fLI
i
V
dt
diLV L
L
)2( ftSinIi pL
)2(2)]2([
ftCosfLIdt
ftSinIdLv p
pL
rmsL
rmsL
pL
pL
i
V
i
V
,
,
,
,
In a purely inductive AC circuit, the current IL lags the applied voltage VL by 900
In a purely inductive circuit the opposition to the flow of AC current is called the inductive reactance XL.
Where f is the frequency and Lis the inductance.
reactanceinductive)(2 fLX L
Calculate the reactance of a coil of inductance 0.32 H when it is connected to 50 Hz supply.
A coil has a reactance of 124 ohm in a circuit with a supp ly of frequency 5 kHz. Determine the inductance of the coil.
Capacitor
The voltage across a capacitor is not due due to resistance it built up due to charge accumulating upon the capacitor plates.
Xc capacitive reactance
reactanceCapacitive)(2
1 C
L
L XfCi
V
Derivation of Xc capacitive reactance
This is an equation for instantaneous value. It follows that :
CVq
dt
dvCi
dt
dq cc
)2( ftSinVv pc
)2(2)]2([
ftCosfCVdt
ftSinVdCi p
pC
rmsL
rmsL
pL
pL
i
V
i
V
,
,
,
,
reactanceCapacitive)(2
1
2,
, Cp
p
rmsL
rmsL XfCfLV
V
i
V
In a purely inductive AC circuit, the current IC leads the applied voltage VC by 900
In a purely inductive circuit the opposition to the flow of AC current is called the capacitive reactance Xc.
Where f is the frequency and C is the capacitance.
reactanceCapacitive)(2
1
fCXC
Leading and Lagging Phasors
Determine the capacitive reactance of a capacitor of 10 uF when connected to a circuit of frequency (a) 50 Hz (b) 20 kHz