Electricity and Magnetism
Transformers and Alternating Current
Lana Sheridan
De Anza College
Nov 30, 2015
Last time
• mutual inductance
• oscillations in RLC circuits
Overview
• alternating current
• transformers
• Maxwell’s equations
Mutual Inductance Applications
If there is a changing current in one coil, an emf can be induced in
the other coil.
The current can be transferred to a whole different circuit that is
no directly connected.
This can be used for wireless charging.
It is also used in transformers: devices that change the voltage and
current of a power supply.
For either of those applications to work, there must be a
constantly changing current.
Alternating Current (AC)
Alternating current (AC) power supplies are the alternative to
direct current (DC) power supplies.
the current
leadsfrom
the driving
emf.
nergy
is thus
the gener-
Alternating Current (AC)
sin θ
+1
vailable at WileyPLUS
pated in the resistor can be
0 are the alternative to
Alternating current (AC) power supplies
π
2π
3π
0
direct current (DC) power supplies.
in2(vd t " f).
θ
(31-68)
–1
In however,
an alternating
current
resistor,
is the
aver-supply, the voltage and current vary
(a)
sinusoidally
withof
time:
e, the average
value
sin u,
1
e average value of sin2 u is 2
sin θ
sin2 θ
d areas under
the curve but
+1
+1
the unshaded spaces below
.
0
0
π
2π
θ
3π
(31-69)
–1
(a)
0
0
∆v = ∆Vmax sin(ωt)
or rms, value of the current i:
t).
+ –12
π
2π
3π
θ
(b)
2
2
θ
31-17as P
(a)=APplot
sin u versus u.
Thesinpower
delivered to a load Fig.
fluctuates
(ωt).
0 sinof
+1
The average value over one cycle is zero. (b)
(31-70)
+ –12
A plot of sin2 u versus u.The average value
Alternating Current (AC)
We describe the amount of current and potential difference across
the circuit by its root-mean-square (RMS) value.
The RMS voltage supplied is
∆Vrms =
∆Vmax
√
2
The RMS current supplied is
Imax
Irms = √
2
Average power:
1
P = Irms (∆Vrms ) = Pmax
2
Alternating Current (AC) Example 33.1
The voltage output of an AC source is given by the expression
∆v = 200 sin(ωt), where ∆v is in volts.
Find the rms current in the circuit when this source is connected to
a 100 Ω resistor.
Alternating Current (AC) Example 33.1
The voltage output of an AC source is given by the expression
∆v = 200 sin(ωt), where ∆v is in volts.
Find the rms current in the circuit when this source is connected to
a 100 Ω resistor.
200
∆Vrms = √ = 141 V
2
Alternating Current (AC) Example 33.1
The voltage output of an AC source is given by the expression
∆v = 200 sin(ωt), where ∆v is in volts.
Find the rms current in the circuit when this source is connected to
a 100 Ω resistor.
200
∆Vrms = √ = 141 V
2
Irms =
∆Vrms
= 1.41 A
R
AC and Types of circuits
The current and potential difference both change sinusoidally in
AC circuits.
However, they do not necessarily reach their peaks at the same
time across a particular circuit component.
It depends on the type of circuit.
across the
resistance
is equal to
the current
and
potential
AC in Resistance-only Circuits
write this as
difference are in phase.
vR , iR
IR
VR
R
iR vR
31-8 A resistor is connected across an
nating-current generator.
φ = definition
0° = 0 radof resistanc
From the
i
R
resistance
as
iR
vR
From Eq. 31-29, we can also wri
t
0
T/2
T
iR
where IR is the amplitude of t
31-31 and 31-32, we see that for
Instants
The voltage and current are in phase.
(a)
represented in (b)
(Definition of resistance! R = ∆v
i )
Fig. 31-9
(a) The current iR and the potentia
AC in Inductance-only Circuits
L
Our concern, h
find the forme
iL vL
Fig. 31-12 An inductor is connected
across
alternating-current
Kirchoff’s loop law an
holds
at each instant generator.
in this circuit:
∆v − L
di
=0
dt
∆v = ∆Vmax sin(ωt) = L
This means
i=
di
dt
∆Vmax
sin(ωt − π/2)
ωL
We now m
notation, we
(31-49)
at the ratedifference
diL/dt as
potential
AC in Inductance-only Circuits
by 90º.
. The unit of the
ohm, just as it is
vL , iL
L
iL vL
itive direction.
. 31-12 An inductor is connected
oss an alternating-current generator.
as
If we combine Eqs. 31-45 and 31-46,
φ = +90° = + π /2 rad
d sine:
(31-50)
VL
IL
vL
diL
dt
vL
Our concern,
however, is with the cu
find the formeriLby integrating Eq. 31
0
T/2
iL !
!
diT
L !
tL
V
L
!
s
We now modify this equation in
notation, we introduce the quantit
Instants
represented in (b)
The voltage leads the current. Voltage peaks first.
(a)
The potential difference across
the inductor is large when the
∆i
change (31-51)
in current is large. (VL = −L ∆t
)
entiate Eq. 3
AC in Capacitance-only Circuits
C
iC vC
A capacitor is connected
across
an
alternating-current
Kirchoff’s loop law holds at each instantgenerator.
in this circuit:
Fig. 31-10
∆v −
i=
q
=0
C
d
dq
= C (∆Vmax ) sin(ωt)
dt
dt
i = ωC (∆Vmax ) sin(ωt + π/2)
We now
tion, we intro
defined as
current leads the potential
qC !
AC in Capacitance-only Circuits
difference
byhowever,
90º. is with the
Our concern,
vC , iC
IC
VC
C
iC vC
A capacitor is connected
s an alternating-current generator.
31-10
entiate Eq. 31-37 to find
φ = –90° = – π /2 rad
iC !
d
d
We
vC now modify Eq. 31-38 in t
tion, we introduce the quantity X
defined as
t
0
T/2
T
1
iC
XC !
"
Instants
represented in (b)
The current leads the voltage. Current peaks first.
(a) the capacitor
The potential across the capacitor is only high once
has built up some charge.
Transformers
which is alm
Transformers change ∆Vrms and Irms simultaneously, while keeping
rule: Transm
the average power Pavg = Irms ∆Vrms constant (conservation of
energy).
The Ideal
ΦB
The transm
for efficient
Np
and consum
Vp
Vs
R
lower (for u
Ns
sentially con
by Faraday’
Primary
Secondary
The ide
Fig. 31-18 An ideal transformer (two
bers of turn
This workscoils
via mutual
If theincurrent
wound inductance.
on an iron core)
a basicin the first coil
In use, the
did not constantly
change
(AC)An
thisacwould
not work.
transformer
circuit.
generator
progenerator w
duces current in the coil at the left (the priS
Ns
∆vp (the secondary)
mary). The coil ∆v
at the
s = right
Np
is connected to the resistive load R when
Transformers
The reason for the voltage relation is that the iron core ideally
contains all the magnetic flux lines produced.
Then the emf per turn Et = − dΦ
dt is the same in both solenoids.
∆vp = −Np
dΦ
dΦ
and ∆vs = −Ns
dt
dt
∆vs = ∆vp
Ns
Np
Imagine wh
age. Energy wou
as previously, bu
Transformers
Let ∆V now refer to the rms voltage.
which
If ∆Vs > ∆Vp (and therefore Ns > Np ), the transformer is
calledis aalmost
rule:
Transmit at
step-up transformer.
This is a step up transformer:
The Ideal Tran
ΦB
The transmission
for efficient high
Np
and consumption
Vp
Vs
R
lower (for use) th
Ns
sentially constan
by Faraday’s law
Primary
Secondary
The ideal tra
Fig. 31-18 An ideal transformer (two
bers of turns, wo
coils wound on an iron core) in a basic
Incalled
use, the
If ∆Vs < ∆Vptransformer
(and therefore
N
<
N
),
the
transformer
is
a prim
s ac generator
p
circuit. An
progenerator
whose
step-down transformer.
duces current in the coil at the left (the priS
mary). The coil at the right (the secondary)
Example from Lecture 19
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
P = I ∆V = (5 A)(1.2 × 106 V) = 6 MW
Example from Lecture 19
A power station supplies current I = 5 A and potential difference
∆V = 1200 kV to a particular installation along the electric grid.
How much power is supplied to the installation?
P = I ∆V = (5 A)(1.2 × 106 V) = 6 MW
Suppose the power station is 1000 km from the installation and
delivers the power over copper wires. Assume the resistivity of
copper is 1.69 × 10−8 Ω m and the radius of the high tension wire
is 2 cm. What is the resistance of the wire delivering the electricty?
R=
ρL
= 13.4 Ω
A
Example from Lecture 19
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
P = I 2 R = (5 A)2 (13.4 Ω) = 336 W
Example from Lecture 19
How much power is dissipated as heat in the transmission lines to
the installation (current I = 5 A and potential difference
∆V = 1200 kV are supplied to the station)?
P = I 2 R = (5 A)2 (13.4 Ω) = 336 W
How much power would be dissipated as heat in the transmission
lines to the installation if instead the station supplied 6 MW of
power with current I = 500 A and potential difference
∆V = 12 kV?
P = I 2 R = (500 A)2 (13.4 Ω) = 3.36 MW
Much more loss!
round a common iron core that is found inside all
Transformers
The circuit diagram for a transformer looks like:
I1
Circuit diaormer.
I2
RL
!v1
N1
!v2
N2
RL is the load resistance.
The equivalent resistance, as seen by the generator on the primary
side (Req = ∆Vp /Ip is:
2
Np
Req =
RL
Ns
Transformers
Np 2
RL
Req =
Ns
Since the “effective resistance” is different from the actual load
resistance RL , transformers are also used for load matching.
33.8 The Transformer and Power Transmission
101
Maximum power is delivered when the emf source’s internal
Figure 33.20 Electronic devic
primary winding in this transformer is
are often powered by AC adaptor
resistance, r = RL . The
attached to the prongs of the plug, whereas
containing transformers such as
this one. These adaptors alter the
AC voltage. In many applications
the adaptors also convert alterna
ing current to direct current.
the secondary winding is connected to the
power cord on the right.
nsformer is smaller than the one in
ning photograph of this chapter. In
, it is a step-down transformer. It
. Cengage Learning/George Semple
. Cengage Learning/George Semple
Sometimes, this is not possible, but using a transformer we can
make Req = r .
Summary
• transformers
• alternating current
• Maxwell’s equations
Collected Homework 4!
due tomorrow.
Homework
Serway & Jewett:
• PREV: Ch 32, Probs: 49
• NEW: Ch 33, onward from page 1021. Obj. Qs: 12, 13;
Conc. Qs.: 8; Probs: 1, 3, 5, 49, 51, 57