AC CIRCUITS 90523. NCEA Level 3 Physics. AC CIRCUITS. RMS Values  Mains electricity  Power  Root mean square AC in capacitance  Intro  Reactance  Phase relationship for capacitance AC in inductance  Intro  Reactance  Phase relationship for reactance
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 Mains electricity
 Power
 Root mean square
 Intro
 Reactance
 Phase relationship for capacitance
 Intro
 Reactance
 Phase relationship for reactance
 AC in an RC circuit
 AC in an RL circuit
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All mains electricity in New Zealand is made in hydroelectric dams. These are huge alternators that turn giant magnets within a nest of metal coils. The result of this is that the poles of the alternators alternate at 50 times per second, 50Hz. This produces Alternating Current or AC.
The equation for changing voltage is:
Voltage changes as a coil rotates
+
V = V max sin t
3T/4
V
t
T/4

‘I’ In
T
T/2
‘I’ Out
N
S
Coil rotations at different stages of the time period, T.
When this voltage is applied to a resistive circuit, the resulting current will also be alternating and sinusoidal:
I = V/R
I = (Vmax sint) /R
Thus:
I = Imax sint
Remember that in rotational motion, which a coil is doing, that = t
V = Vmax sin
and
I = Imax sin
AC is widely used in industry and homes as the power supply for electrical appliances because:
Produced by generators
The maximum voltage can b changed easily with a transformer.
It can be controlled by a wide range of components; R; C & L’s.
The frequency of AC is essential for timing
Most mains electricity is used for heating effect. The power dissipated in a resistance is given by P = VI. We can also write this as P = I2R substituting in V= IR into the above equation. As current varies with time so must power.
Voltage and current in an AC circuit produces a cyclic power output.
Peak power output
Power
(Watt)
Average power output
t (s)
One cycle (T)
WHY ARE ALL THE PEAKS POSITIVE?
ANS: when power is calculated by P = IV when I and V are both negative when mutiplied together they always produce a positive
This stands for root mean square. This is a way of “averaging out” the effect of the AC as the voltage and current are always changing.
The RMS is the DC value which produces the same power output.
The RMS values of voltage and current may be used directly in power formulae as stated earlier.
P = VRMS IRMS
P = I2RMSR
For sinusoidal alternating current, it can be shown that:
IRMS = IMAX
VRMS = VMAX
&
2
2
RMS values are so useful that any values quoted or measured will be the RMS values rather than the maximum values.
340V
240V
Vrms
t(s)
V
0.01
0.02
240V
340V
c. The ‘wattage’ of the lamp refers to its power output.
P = Vrms x Irms [rms values must be used in the power formula]
Irms = P / Vrms
= 100 / 240 = 0.42A
IPEAK = Irms x 2 = 0.42 x 2 = 0.59A
In a DC circuit containing a capacitor the current will only flow until the capacitor is full. In an AC circuit the flow alternates thus the capacitor charges and discharges 50 times a second. The lamp continually glows.
C
C
Lamp will not glow when capacitor is fully charged
Lamp will continue to glow as capacitor is continually charging and discharging
Readings of the capacitor voltage, VC, can be made for different values of current, I. This shows that the VC is proportional to the current, I.
VC(V)
VC I
There must therefore be a constant joining the two variables. The constant is called the reactance, XC.
AC current, I (A)
XC = gradient
VC= IXC
OR
XC= VC/I
Reactance is basically the resistance that the capacitor actually has to the AC current.
FACTORS AFFECTING THE REACTANCE OF THE CAPACITOR:
The larger the capacitance the smaller the reactance as larger capacitors store more charge. Reactance is inversely proportional to the capacitance.
The greater the frequency, the smaller the reactance. When the frequency, f, high, charge is flowing on and off the capacitor plates at a high rate.
XC 1/C
XC 1/f
XC 1/Cf
The proportionality constant in this relationship is:
1/2
The reactance of a capacitor C, with supply of frequency F, is given by:
XC = 1/2fC
OR
XC = 1/C
(Since 2f = )
PHASE RELATIONSHIP FOR CAPACITANCE:
VS
In a DC circuit the voltages across the components connected in series with a supply add up to the supply voltage.
VS = VC + VR
In AC containing ‘C’ and ‘R’ this does not apply. Why?
ANS: the two voltages are out of phase
C
R
VC
VR
+Vnax
t (s)
0
Vnax
T/4
T/2
3T/4
T
Vnax
VOLTAGE vs TIME GRAPH
PHASOR DIAGRAM
Phasor diagram for t = 0 on resistor voltagetime graph
VR
t
VC
Phasor diagram for t = 0 on capacitor voltagetime graph
t
VC
The angle between the phasors is 90o. The phasor VR is ahead of the phasor VC. These phasors are vectors and thus must be treated as so. The supply voltage, VS, is thus the two phasor vectors added together.
VR
VR
VS
VC
VS
VC
VS
SOLUTION:
a. VS = 6.0 + 8.0
= 6.02 + 8.02
= 10V
b. = tan1 (8.0/6.0)
= 53o
C
6.0V
R
VS
VC
VR
8.0V
C = 100F
R = 25Ω
VS
50Hz
C = 500F
R=5.0Ω
VC
2V
a. Current, I, is calculated from the voltage across the resistor, VR.
I = V/R
= 2.0/5.0
= 0.40A
b. VS2 = VC2 + VR2
= (2.52 + 2.02)
= 3.2V
c. As the resistor voltage is in phase with the current, is the phase difference.
= tan1 (2.5/2.0)
= 51O
VR = 2.0V
VC = 2.5V
VS
Resistance and reactance both impede the flow of current in a circuit. The combined effect of resistance and reactance is called impedance, symbol Z, unit Ohm (Ω).
Just as VR = IR for resistor
And VC = IXC for a capacitor,
VS = IZ for the whole circuit.
R
Z = R2 + XC2
Thus:
Z
XC
10V
50Hz
10V
50Hz
10V
50Hz
C
0.50A
0.25A
C
A
A
A
R
R
d. The impedance of the circuit containing both R & C.
e. The current in the circuit containing both R & C.
e. I = VS /Z
= 10 / 44.7
= 0.22A
Inductors always oppose a change in current, so in an AC circuit where the current is continually changing, the inductor will act to limit the amount of current.
A voltage is induced across the inductor whenever the current (and hence the flux) in it is changing. When ‘L’ is linked to AC the current is always changing thus there is a continual voltage, VL across the inductor. The voltage will act against the current according to Lenz’s law. Using the rheostat the relationship between current and inductor voltage can be investigated
R
A
L
VL
Readings of the inductor voltage, VL, can be made for different values of current, I. This shows that the VL is proportional to the current, I.
VL(V)
VL I
There must therefore be a constant joining the two variables. The constant is called the reactance, XL.
AC current, I (A)
XL = gradient
VL= IXL
OR
XL= VL/I
Reactance is basically the resistance that the inductor actually has to the AC current. Real inductors do have resistance but at this level, unless otherwise stated, the resistance is said to be negligible.
FACTORS AFFECTING THE REACTANCE OF THE INDUCTOR:
The larger the inductance the larger the reactance this is due to greater inductance causing a greater opposing voltage and therefore a smaller current. Reactance is proportional to the inductance.
The greater the frequency, the smaller the current. When the frequency, f, high, the opposing voltage will be greater causing a smaller current. Reactance is proportional to the frequency.
XL L
XL f
XL Lf
The proportionality constant in this relationship is:
2
The reactance of a inductor L, with supply of frequency f, is given by:
XL= 2fL
OR
XL = L
(Since 2f = )
A 0.50H inductor is connected to a 10V 50Hz AC supply. Assuming the inductor has negligible resistance, what is:
a. The reactance of the inductance?
b. The current in the circuit?
PHASE RELATIONSHIP FOR INDUCTANCE:
VS
In a DC circuit the voltages across the components connected in series with a supply add up to the supply voltage.
VS = VL + VR
In AC containing ‘L’ and ‘R’ this does not apply. Why?
ANS: the two voltages are out of phase
L
R
VL
VR
+Vnax
t (s)
0
Vnax
T/4
T/2
3T/4
T
Vnax
VOLTAGE vs TIME GRAPH
PHASOR DIAGRAM
Phasor diagram for t = 0 on resistor voltagetime graph
VR
t
VL
Phasor diagram for t = 0 on inductor voltagetime graph
VL
t
The angle between the phasors is 90o. The phasor VR is behind of the phasor VL. These phasors are vectors and thus must be treated as so. The supply voltage, VS, is thus the two phasor vectors added together.
VL
VS
VR
VS
VR
VL
SOLUTION:
a. VR = I x R
= 0.0032 x 940
= 3.0V
b. VL = I x 2fL
= 0.0032 x 2 x 100 x 2.0
= 4.0V
= 42 + 32
= 25
= 5.0V
d. The current is in phase with the resistor voltage, and so is the phase difference:
= tan1 (4.0/3.0)
= 53O
The supply voltage leads the current by 53O.
VS
VL = 4.0V
VR = 3.0V
Resistance and reactance both impede the flow of current in a circuit. The combined effect of resistance and reactance is called impedance, symbol Z, unit Ohm (Ω).
Just as VR = IR for resistor
And VL = IXL for a inductor,
VS = IZ for the whole circuit.
Z = R2 + XL2
Thus:
XL
Z
R
A real 380mH inductor can be regarded as a pure inductance in series with its resistance, R.
10V DC
10V AC
50Hz
A
0.20A
A
0.20A
L=380H
L=380H
R
R
When connected to a 10V DC supply, a current of 0.20A flows. It is connected to a 10V 50Hz AC supply.
d. The current when connected to the AC supply.
LCR SERIES CIRCUIT:
When a capacitor, inductor and resistor are connected in series to an AC supply they form a very useful circuit with some important properties. The voltages and impedance in an LCR circuit will vary considerably as the frequency is changed.
The voltage across any of the components in the circuit will depend on the reactance/resistance of the component.
VR = IR R is constant & in phase with I
VC = IXC XC = 1 /2fC, 90o behind I
VL = IXL XL = 2fL, 90o ahead I
Although the resistance remains constant in the circuit the reactance’s depend upon the frequency.
In order to show how the reactance s vary is to consider a phasor type diagram for the impedance of the circuit.
XL = 2fL
R
Constant voltage
A
R
Phasor diagram of resistance and reactance’s
XC = 1/2fC
L
C
An LCR circuit
In the above phasor diagram the total reactance is the difference between the inductor reactance and the capacitor reactance.
XT = XC  XL
If the frequency of the supply is increased, the inductor reactance will increase but the capacitor reactance will decrease, as inversely proportional to the f.
In order to find the impedance of the circuit the total reactance is combined with the resistance:
R
XL  XC
R
XL  XC
Z
Z = R2 + (XL  XC)2
In order to find the supply voltage of the circuit the VC and VL is combined with VR:
VR
VR
VL  VC
VL  VC
VS
VS = VR2 + (VL  VC)2
VARIABLE FREQUENCY AC
Constant voltage
A
R
L=40mH
130 turns
C=100F
12Ω
c. Because the impedance diagram is mathematically similar to the voltage phasor diagram, the angle is the angle between the current and the resistor voltage.
= tan1 (19.264/12)
= 58O
The current leads the supply voltage by 58O.
I
R
XC  XL
Z
Resonance is a phenomenon that deals with changing frequency. We have already seen that reactance is effected by a change in frequency.
XL is proportional to frequency
XC is inversely proportional to frequency.
If we were to analysis the affect of changing the frequency on the change in current we get a graph shown below:
I (A)
fo = resonant frequency
f (Hz)
fo
The observations show that the current reaches a maximum at a specific frequency called the resonant frequency. After this as the frequency increases the current decreases.
When the resonant frequency has been reached we stated that the circuit has been tuned.
We know that current is inversely proportional to impedance as:
I = VS/Z
When I = max then Z = minimum.
We know that Z is dependent on R (which doesn’t change) & reactance, X (which changes with frequency).
From what we have learnt already:
At high frequencies:
XL > XC
XL < XC
At low frequencies:
XL = XC
fohas been reached when:
At this point XL XC = 0 & so:
Z = R
As Z is minimum this is therefore fo.
A 120Ω resistor, a 2.00H inductor and a 4.50F capacitor are connected in series to a 15V variable frequency AC supply.
Calculate the resonant frequency:
VARIABLE FREQUENCY AC
VAC = 15V
R
SOLUTION:
At resonance: XL = XC
fo = 1/2LC
fo = 1/[2(2.00 x 4.50x106)]
= 53.1Hz
C=4.50F
120Ω
L = 2.00H
WHAT HAPPENS TO THE VOLTAGE AT RESONANCE?
We know that at resonance XL = XC.
Thus:
VL/I > VC/I
As [VC = IXC & VL = IXL]
As I is a constant throughout a series circuit it can be cancelled out leaving:
VL = VC
Z = R
At resonance we have found that:
VL = VC
&
Vs = VR
As:
VL
VS = VR
Thus: VS = IZ or IR
VR = IR
VC
VS
A
0.020A
900Ω
VR
20V
VC
a. At resonance, VC = VL
so VC = 20V
b. VR = I x R
= 0.020 x 900
= 18V
c. At resonance, since VC = VL, the supply voltage is the same as VR.
VS = 18V
“RESONANCE & APPLICATIONS”
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&
COMPLETE
QUESTIONS 8
EXTENSION
“DC POWER SUPPLIES”
PAGE 164  165
S & C