$&&LUFXLW7KHRU\3DUW,,
7.
Power in an AC Circuit
Earlier the result was had that
ref: SerwayCollege Physics 6edn' Sec 21.5
P
2
av
= Irms ⋅ R
in an AC circuit with one series resistor.
This is the rate at which energy is dissipated in a resistor (realised as heat).
What about energy dissipation in the capacitor and inductor of an RLC circuit ?
From DC electric theory studied earlier it is known already that
E =
c
1
2
⋅ C ( ∆V)
2
where Ec is the energy stored in the capacitor and ∆V is the voltage drop across the capacitor.
In an AC circuit, with only a generator and a capacitor in series, at some instant when the
voltage drop is maximum, ∆Vm, then there must be a corresponding maximum energy
stored in the capacitor, ECm, given at that instant by:
1
2
E
= ⋅ C ∆Vm
Cm
2
(
)
But when the current reverses in the cycle, charge will be drawn away from the plates of the
cap', and the energy content of the capacitor must then decreases since ∆V across the cap'
decreases as the charge Q decreases.
The energy must go somewhere, and it is returned to the AC generator.
There is thus a continual exchange of energy between the AC source and the capacitor, with
no loss of energy from this system (unlike the resistor which 'pours out' energy from a circuit
in the form of heat..requiring continual supply of energy to the circuit).
Earlier it was also derived that (in study of inductance) that
1 2
E = ⋅ LI
L
2
where an inductor is placed in series with a DC source, and EL is the energy stored in the
magnetic field of the inductor (coil loop).
This energy must be supplied by the battery in such a DC circuit.
Accordingly, in an AC circuit. containing one inductor in series with an AC source, there will be
some value of maximum current Ιm which means that, at that instant, there is a maximum
amount of energy stored in the magnetic field of an inductor and is given by:
E
Lm
=
1
2
2
⋅ L⋅ Im
When the currrent Ι decreases in the cycle, the energy content of the magnetic field must then
decrease. The energy contained in the magnetic field is returned to the AC source.
[ See that this agree with Lenz's law since the circuit loops (inductor) now attempts to maintain
the original current to maintain also the magnetic flux value...thus energy is GIVEN by the
inductor TO the source in an attempt to 'keep the original current value']
There is therefore a continual exchange of energy between inductor and capacitor in an AC
circuit, with no net loss of energy from the system (circuit).
page 1 of 10
!"$#&%(')+**
,-./01233
Hence, the only LOSS of energy from an AC circuit must be only as a result of the energy lost
as HEAT through the resistor !!
So, the equation for power loss for the series resistor is also the equation for power
loss from an RLC circuit.
P
Equation for Power Loss from an AC circuit with
R, L and C in series
2
av
= Irms ⋅ R
∆VRm
Ohms law lets us calculate R as: R =
Ιm
(this equation is from section 6 of these notes that just uses the max' volt drop across
the resistor at the instant of max current to calculate Ι..straight forward use of Ohms law
using two specific values of current and voltage drop at one instant of time)
Can now use the 'trick' used earlier to express things in terms of rms values, and simply
divide both numerator and denominator by the same value of 2 .
The numerator then becomes
∆VRm
2
Ι Rm
The denominator becomes:
2
Therefore also can write
R=
=V
Rrms
= Ι rms
∆VRrms
Irms
Substituting this into the expression for average power loss in the RLC circuit:
P
2 ∆VRrms
av
= Ι rms ⋅
Irms
Pav = Irms⋅ ∆VRrms
= Irms⋅ ∆VRrms
Equation for average power loss in RLC
circuit
(rms current and voltage drop as measured
by meters)
It is also possible to express ∆VRm in terms of the voltage of the source Vm.
The below figure is replicated from the previous section.
page 2 of 10
4567
283098-!:2;0$<&=(/0)3+,,
>?@ABCDEE
.
Relationship of Vm to ∆VRm
Vm
3
∆VLm-∆
∆VC
∆VRm
The vector ∆VRm is clearly the projection of Vm onto the horizontal (x axis) and is thus given
by:
∆VRm =Vm.cos(φ)
Divide both sides by
∆VRm
2
So
=
2 to convert this equation into rms values:
Vm
2
⋅ cos ( φ )
∆VRrms = Vrms⋅ cos ( φ )
Then using
Pav = Irms⋅ ∆VRrms this gives:
Pav = Irms⋅ Vrms⋅ cos ( φ )
The quantity cos(φ ) is called the power factor.
Pav = Irms⋅ Vrms⋅ cos ( φ )
Equation for average power loss in an AC
circuit
in terms of rms current through circuit and
the AC source rms voltage
Study : Example 21.5
Serway College Physics 6th edn'.
Practice Problems:
29, 31
Serway College Physics 6th edn
page 3 of 10
FGHIJKLMNKO!PJQM$R&S(TM)L+UU
VWXYZ[\]]
8.
Resonance in a Series RLC circuit
Physics 6edn'
ref Serway College
From earlier (section 6):
Irms =
Vrms
Z
Vrms
=
2
(
)
2
R + XL − XC
Now if frequency is altered, then so too do the reactances XL and XC alter.
Recall these reactance values are given by:
X = ω ⋅ L = 2⋅ π ⋅ f ⋅ L
&
L
X =
C
1
ω⋅C
=
1
2⋅ π ⋅ f ⋅ C
Now inspecting the denominator of eqn (*#) above means the denominator will be the
smallest value when XL = XC. This condition will produce the largest possible value of Ιrms
in the circuit.
The particular frequency at which this occurs, denoted by fo, is called the resonant
frequency.
[ The phenomenon of systems having a maximum response at some particular frequency is
called resonance, and in this case it is the rms current (as well as the max current of course
Ιm)
that is our measured 'response'].
Thus it is the condition
X L = XC
X L = XC
that must be met for resonance.
Resonance Condition for AC Series RLC circuit
Using this the resonant frequency can be found:
Since
X L = XC
when f = fo, then:
1
2⋅ π ⋅ fo⋅ L =
1
2
fo =
fo =
fo =
2⋅ π ⋅ fo⋅ C
2
2
2 ⋅ π ⋅ L⋅ C
1
2
2
2 ⋅ π ⋅ L⋅ C
1
2⋅ π ⋅ L⋅ C
=
1
2⋅ π ⋅ L⋅ C
Equation for resonant requency
in terms of L and C
page 4 of 10
^_`a
\b]ZcbW!d\eZ$f&g(YZ)]+VV
hijklmnoo
Resonance in RLC Circuit
rms current (A)
2
1
0
0
20
40
60
80
100
f (Hz)
Plot example of resonance for a 10V rms voltage, 5 Ω resistor,
100µF capacitor and 0.2 H Inductance.
The radiowaves of radiostations intercepted by aerials on radios produce an alternating
(small) voltage across the aerial. This 'source of AC' is the signal source for the radio.
By varying the value of a capacitor (usually) in the radio, the condition for resonance for a
particular radiowave frequency occurs. The ensuing current is maximum with which to drive the
speakers and the station is heard!
Resonance of radiowaves is also important in magnetic resonance imaging, which 'tunes' into
radiowaves emitted by rotating magnetic vectors (formed from multiple nuclear magnetic dipole
moments in tissue) situated near receiving coils (aerials). Different slices of the patient are
made to have different frequencies of magnetic vectors and so produce different frequency
electromagnetic waves (all at radio frequencies).
Study :
Example 21.6
Practice Problems:
33, 35
Serway College Physics 6th edn'.
Serway College Physics 6th edn'
page 5 of 10
pqrstuvwxuy!zt{w$|&}(~w)v+
€‚ƒ„…†‡‡
9.
Tranformers
ref Serway College Physics 6th edn'
Sec 21.7
By placing one coil of wire with a source of AC -called the 'primary coil'- near another coil of wire
which does not have an AC source-the so called ' secondary coil'- (see below figure) it is
possible to induce an alternating electric potential across the secondary coil.
[ This phenomenon is not really new, it is of course electromagnetic induction, and a secondary
coil potential is produced with a DC source attached to the primary coil, but only at the very
short time intervals of current increase and decrease when closing and opening the switch in
such a DC source circuit (see notes on EM induction).]
Because the current is continually changing with an AC source, then always a potential is
produced at the secondary coil.
.
AC potentials produced
at secondary coil if AC potentials
applied at primary coil.
Magnetic Field lines from the
primary coil are shown
intercepting the secondary coil
at one instant in time.
If an average voltage ∆V1 is applied to the primary coil (between instants of time t1 and t2)
then Faradays law of Induction relates the average magnetic flux change in that coil, ∆Φ1 , in
the time ∆t = t2-t1. Faradays Law states that:
∆Φ 1
∆V1 = −N1⋅
∆t
...where N1 is the number of loops in the primary coil.
IF (this is an inportant 'if') ALL the flux appearing in the primary coil was also produced in the
secondary coil, at any instant, then it would be also true that the flux change ∆Φ2 would equal
the flux change ∆Φ1 in the same time interval ∆t.
So, in the case of 'identical fluxes' we have
∆Φ1 = ∆Φ2 = ∆Φ , say
Now, for the secondary coil we have in general (when fluxes are not identical) that the
voltage drop induced across the coil is given by:
∆Φ 2
∆V2 = −N2⋅
∆t
page 6 of 10
∆
−
⋅
 ∆Φ 
⋅
⋅
 ∆ 
ˆ‰Š‹
†Œ‡„Œ!Ž†„$&‘(ƒ„)‡+€€
∆
−
⋅
∆Φ
’“”•–—˜™™
∆
Now divide one equation by the other (here the second by the first:)
 ∆Φ 2  1  ∆t 
⋅
⋅
= −N2⋅ 

∆V1
 ∆t  −N1  ∆Φ 1 
∆V2
or,
 ∆Φ 2 
N2  ∆t 

⋅
=
N1  ∆Φ 1 


 ∆t 
In the case where the average flux change is equal in any chosen time interval ∆t, then
∆Φ1 = ∆Φ2 = ∆Φ and so:
∆V2
∆V1
=
N2
N1
..which would occur in the case of perfect flux transfer.
Such an arrangement of coils which would make this possible is called a perfect
transformer.
An arrangement of primary and scondary coils as in the above diagram is simply called an
AC transformer.
Taking the least complicated case of the ideal transfromer, then where N2 > N1, ∆V2 >∆V1
and this is called a step up transformer, since the voltage appearing in the secondary coil is
'stepped up' from the value of voltage across the primary coil.
The converse case is called a step down transformer.
( In a non ideal case, to obtain the same secondary voltage, then the value of N2 must be
 ∆Φ 2 


∆t 

rather greater that for a perfect transformer since the ratio of
will be < 1, and clearly
 ∆Φ 1 


 ∆t 
the number of turns must be increased in the non ideal case by the inverse of this ratio).
How can perfect flux linkage be achieved ?
The true answer is that it cannot, but it can be approached. With the arrangement in the
above diagram where the magnetic field lines of the primary coil intercepts the secondary coil
some distance away, through air, then linkage is poor (you can see most of the magnetic
vectors will not occur within the secondary coil loops, but will be spread out in space).
If now a common magnetic material is used as a common core for both coils, what occurs is
that
the vast majority of magnetic field lines are confined to the core rather than spread in space.
page 7 of 10
š›œ
˜ž™–Ÿž“! ˜¡–$¢&£(•–)™+’’
¤¥¦§¨©ª««
.very few magnetic field vectors (small magnitude)
appear outside the core
Magnetic Field Linkage
(and so magnetic flux linkage)
by use of common magnetic
(eg iron) core
S
AC source
R
Whilst it is true that greater voltages can be obtained than those at the primary, this does not
constitute a breakdown of the conservation of energy law.
For any time interval ∆t, conservation of energy occurs, and with a perfect flux linkage this
would mean that
∆E1 = ∆E2
in any time interval ∆t
Divide each side by ∆t gives:
∆E1 ∆E2
=
∆t
∆t
Or
Pav1 = Pav2
So:
I1rms⋅ ∆V1rms⋅ cos φ 1 = I2rms⋅ ∆V2rms⋅ cos φ 2
( )
( )
With no capacitance nor inductance in either primary or secondary circuit, then φ1 = φ2 = 0
So
I1rms⋅ ∆V1rms = I2rms⋅ ∆V2rms
This result is much more general though, since the argument is true for any interval ∆t, it
must also be true for the case where ∆t shrinks to zero.
Accordingly
or,
dE1
=
dE2
dt
dt
P 1 = P2
....equality
of instantaneous power values
I1⋅ ∆V1 = I2⋅ ∆V2
...where each current and voltage drop are instantaneous values of current and voltage drop
for primary side and secondary side of the transformer.
Then
In reality, there is not perfect energy exchange for transformers, and eddy currents in the iron
ore of a transformer will induce small loops of currents within the iron, causing Ι2R losses
within the core, which will heat up to some extent.
Consequently power efficiency is never 100% but can be in the range 90% to 99%, typically,
with iron cores.
page 8 of 10
¬­®¯
ª°«¨±°¥!²ª³¨$´&µ(§¨)«+¤¤
¶·¸¹º»¼½½
Transformers are used in power supply of electricity, with step up transformers used after initial
generation of electricity at the power station.
This is because if a high current is used (with corresponding small voltage) to transmit electricity,
then large Ι2R losses occur in the supply cables.
A step down transformer is used at local distribution stations, with a further step down transformer
at substations near actual supply outlets, giving either finally 220Vrms to 240Vrms (eg Europe,
Australia) or 110 to 120Vrms (eg. US, Japan) for use in a domestic power point socket.
Hospitals use partcular local step down transformers to get a range of output voltage supplies, a
say used in X ray facilities (called 'line voltage' and ~ 415V Europe/Australia).
Autotransformers are a particular arrangement of secondary and primary coil.
In thisthere is no physical or electrical seperation of secondary and primary coil, rather the
'secondary' coil is a 'tapped portion' of a single coil, wound upon a common iron core.
The magnetic field is common to both coils and the flux through the secondary is some factor of
the flux through the primary coil.
.
a
a
b
b
AC in
Vs
AC in
Autotransformer
Vs
c
c
d
d
The principles of the transformer apply as usual and
∆Vs Ns
=
∆Vp Np
where the subscripts s and p denote secondary and primary.
In the above figure(on the left), if there are 500 turns between a and d, and 250
between
b and c then Ns = 2Np and so ∆Vs = 2∆Vp.
With the autotransformer arrangement on the right hand of the figure then Ns = 250 and
Np = 500. Thus Vs = Vp / 2.
Autotransformers allow for 'mobile pickoff points' for the secondary part. They are used
in X ray circuits routinely (this autotransformer allows setting of a variable voltage that
after use of another -step up-transformer consequently allows a variable choice of
kilovoltage applied across the X ray tube).
page 9 of 10
¾¿ÀÁ
¼Â½ºÃ·!ļź$Æ&Ç(¹º)½+¶¶
ÈÉÊËÌÍÎÏÏ
Study
Example 21.7
Practice Problems:
39 , 41
Serway College Physics 6th edn'
Serway College Physics 6th edn'
page 10 of 10
ÐÑÒÓ
ÎÔÏÌÕÔÉ!ÖÎ×Ì$Ø&Ù(ËÌ)Ï+ÈÈ