Download V(t)

alternating currents & electromagnetic
waves
PHY232
Remco Zegers
[email protected]
Room W109 – cyclotron building
http://www.nscl.msu.edu/~zegers/phy232.html
Alternating current circuits
R
I
R
I
V
V
 previously, we look at DC circuits (the
voltage delivered by the source was
constant).
 Now, we look at AC circuits, in which case
the source is sinusoidal. A
is
used in circuits to denote the difference
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A circuit with a resistor
IR(A)
R
I
V(t)=V0sint
V0=10 V
R=2 Ohm
=1 rad/s
 The voltage over the resistor is the same as the voltage delivered by
the source: VR(t)=V0sint=V0sin(2ft)
 The current through the resistor is: IR(t)= V0/R sint
 Since V(t) and I(t) have the same behavior as a function of time, they
are said to be ‘in phase’.
 V0 is the maximum voltage
 V(t) is the instantaneous voltage
  is the angular frequency; =2f f: frequency (Hz)
 SET YOUR CALCULATOR TO RADIANS WHERE NECESSARY
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lon-capa
 you should now do problem 1 from set 7.
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|IR|(A) |VR|(V)
IR(A)
rms currents/voltages
Vrms
Irms
 To understand energy
consumption by the circuit, it
doesn’t matter what the sign of
the current/voltage is. We need
the absolute average currents
and voltages (root-mean-square
values) :
 Vrms=Vmax/2
 Irms=Imax/2
 The following hold:
 Vrms=IrmsR
 Vmax=ImaxR
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P(W)
|IR|(A) |VR|(V)
power consumption by an AC circuit
Vrms
Irms
 We already saw (DC):
 P=VI=V2/R=I2R
 For AC circuits with a single
resistor:
 P(t)=V(t)*I(t)=V0I0sin2t
 The average power
consumption:
 Pave=Vrms*Irms=V2rms/R=I2rmsR
 Pave=(Vmax/2)( Imax/2)=
ImaxVmax/2
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vector representation
V0
=t
V
-V0
time (s)
The voltage or current as a function of time can be
described by the projection of a vector rotating with
constant angular velocity on one of the axes (x or y).
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phasors
IR(A)
I(t), V(t) are in phase, so
point in the same direction
=t
I(t) V(t)
t
The instantaneous current and voltage over R are the projections on
the t-axis (horizontal axis) of vectors rotating with ang. frequency .
The length of the vectors indicate the maximum current or voltage.
I(t)=V(t)=0
t
I(t)=5A V(t)=10 V t
V(t)=-10V I(t)=-5A
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question
V(t) I(t)
t
Given a phasor diagram for a single resistor in circuit.
If the voltage scale is V and current scale Ampere,
then the resistor has a resistance
a) < 1 Ohm
b) > 1 Ohm
c) 1 Ohm
Since the V-vector is shorter than the R vector, R=V/I<1 Ohm
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A circuit with a single capacitor
C
IC(A)
I
V(t)=V0sint
Vc= V0sint
Qc=CVc=CV0sint
Ic=Qc/t= CV0cost= CV0sin(t+/2)
So, the current peaks ahead in time (earlier) of the voltage
There is a difference in phase of /2 (900)
why? When there is not much charge on the capacitor it readily accepts more
and current easily flows. However, the E-field and potential between the plates
increase and consequently it becomes more difficult for current to flow and
the current decreases. If the potential over C is maximum, the current is zero.
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IC(A)
phasor diagram for capacitive circuit
=t
I(t)
V(t)
t
Note: Imax= CV0
For a resistor we have I=V0/R so ‘1/C’ is similar to ‘R’
And we write: I=V/Xc with Xc= 1/C the capacitive reactance
Units of Xc are Ohms. The capacitive reactance acts as a resistance
in this circuit.
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power consumption in a capacitive circuit
There is no power consumption in a purely capacitive circuit:
Energy (1/2CV2) gets stored when the (absolute) voltage over the
capacitor is increasing, and released when it is decreasing.
Pave = 0 for a purely capacitive circuit
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question

=t
I(t)
V(t)
t
The angle  between the current vector and voltage
vector in a phasor diagram for a capacitive circuit is
a) 00
b) 450
c) 900
d) 1800
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A circuit with a single inductor
IL(A)
L
I
V(t)=V0sint
VL= V0sint=LI/t
IL=-V0/(L)cost= V0 /(L )sin(t-/2)
(no proof here: you need calculus…)
So, the current peaks later in time than the voltage
There is a difference in phase of /2 (900)
why? As the potential over the inductor rises, the magnetic flux produces a
current that opposes the original current. The voltage across the inductor
peaks when the current is just beginning to rise, due to this tug of war.
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IL(A)
phasor diagram for inductive circuit
=t
I(t)
V(t)
t
Note: Imax= V0/(L)
For a resistor we have I=V0/R so ‘L’ is similar to ‘R’
And we write: I=V/XL with XL= L the inductive reactance
Units of XL are Ohms. The inductive reactance acts as a resistance
in this circuit.
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power consumption in an inductive circuit
There is no power consumption in a purely inductive circuit:
Energy (1/2LI2) gets stored when the (absolute) current through the
inductor is increasing, and released when it is decreasing.
Pave = 0 for a purely inductive circuit
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question
 The inductive reactance (and capacitive reactance) are
just like the resistance of a normal resistor, I.e. if I know the
inductive reactance, I can calculate the current at any
time given the voltage using I=V/XL.
 a) True
 b) False
answer: False; it tells something about maximum currents
and voltages, but ignores the phases involved.
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Combining the three: the LRC circuit
L
C
R
I
V(t)=V0sint
 Things to keep in mind when analyzing this system:
 1) The current in the system has the same value
everywhere I=I0sin(t-)
 2) The voltage over all three components is equal to the
source voltage at any point in time: V(t)=V0sin(t)
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An LRC circuit
I
VR
VC
VR
I
VL

Vtot
=t
t
VL
VC




For the resistor: VR=IRR and VR and IR=I are in phase
For the capacitor: Vc=IXc and Vc lags Ic=I by 900
For the inductor: VL=IXL and VL leads IL=I by 900
at any instant: VL+Vc+VR=V0sin(t), that is the total voltage
Vtot is the vector addition of the three individual
components
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impedance
VR
VR
VL
Vtot
VL
tvector sum V +V
L
C
VC
 Vtot= VL+Vc+VR (vectors)
t
VC
 Vtot=[VR2+(|VL|-|VC|)2]=
[ (IR)2+(IXL-IXC)2]=I[R2+(XL-Xc)2]
 define X=XL-Xc : reactance of an RLC circuit
 define Z=[R2+(XL-Xc)2]= [R2+X2] : impedance of RLC circ.
 Vtot=IZ & I=Vtot/Z looks like Ohms law
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phase angle
VR
I
VL

VR
Vtot
VL
=t
t
VC

Vtot
t
vector sum VL+VC
VC
 The current I and the voltage Vtot are out of phase by an
angle . This angle can be calculated with:
 tan=opposite/adjacent=(|VL| -|Vc| )/VR=X/R
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question
 If the maximum voltage over the capacitor equals the
maximum voltage over the inductor, the difference in
phase between the voltage over the whole circuit and the
voltage over the resistor is:
 a) 00
VR
 b)450
Vtot
0
 c)90
 d)1800
V
L
vector sum VL+VC=0
t
VC
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power consumption by an LRC circuit
 Even though the capacitor and inductor do not consume
energy on the average, they affect the power
consumption since the phase between current and
voltage is modified.
VR
P=I2rmsR=IrmsVR

 VR=Vrmscos (since cos=VR/Vtot)
VL
 So: P=VrmsIrmscos
 cos: power factor of a circuit VL+VC

Vtot
t
VC
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lon-capa
 you should now do problem 4 from LON-CAPA 7
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Given:
R=250 Ohm
L=0.6 H
C=3.5 F
f=60 Hz
V0=150 V
example
L
C
R
I
 questions:
V(t)=V0sint
 a) what is the angular frequency of the system?
 b) what are the inductive and capacitive reactances?
 c) what is the impedance, what is the phase angle 
 d) what is the maximum current and peak voltages over each element
 Compare the algebraic sum of peak voltages with V0. Does this make
sense?
 e) make the phasor diagram. Include I,VL,VC,VR,Vtot, . Assume VR is in the
first quadrant.
 f) what are the instantaneous voltages and rms voltages over each
element. Consider Vtot to have zero phase.
 g) power consumed by each element and total power consumption
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answers
 a) angular frequency  of the system?
 =2f=260=377 rad/s
 b) Reactances?
 XC=1/C=1/(377 x 3.5x10-6)=758 Ohm
 XL= L=377x0.6=226 Ohm
 c) Impedance and phase angle
Given:
R=250 Ohm
L=0.6 H
C=3.5 F
f=60 Hz
V0=150 V
 Z=[R2+(XL-Xc)2]=[2502+(226-758)2]=588 Ohm
 =tan-1[(XL-XC)/R)=tan-1[(226-758)/250]=-64.80 (or –1.13 rad)
 d) Maximum current and maximum component voltages:
 Imax=Vmax/Z=150/588=0.255 A
 VR=ImaxR=0.255x250=63.8 V
 VC=ImaxXC=0.255x758=193 V
 VL=ImaxXL=0.255x266=57.6 V
 Sum: VR+VC+VL=314 V. This is larger than the maximum voltage delivered
by the source (150 V). This makes sense because the relevant sum is not
algebraic: each of the voltages are vectors with different phases.
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e) the phasor diagram
Imax=Vmax/Z=0.255 A
VR
VR=ImaxR=63.8 V
VC=ImaxXC=193 V
VL=ImaxXL=57.6 V
=-64.80 (or –1.13 rad)
Vtot=150 V
VL
I

t
Vtot
=t
VL+VC
start with VR (in first quadrant). The I vector
V
is in the same direction. VC vector is perpendicular to VR, C
900 later in time (counter-clockwise). VL is perpendicular
to VR, 900 faster in time (clockwise). Add VL and VC as vectors, and
add the results to VR as vectors to make Vtot. The angle between the
horizontal x-axis and Vtot is the angle t. The angle between I (or VR)
and Vtot is the phase angle .
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Imax=Vmax/Z=0.255 A
VR=ImaxR=63.8 V
answers
VC=ImaxXC=193 V
VL=ImaxXL=57.6 V
=-64.80 (or –1.13 rad)
Vtot=150 V
 f) instantaneous voltages over each element (Vtot has 0 phase)?
 start with the driving voltage V=V0sint=Vtot
 VR(t)=63.8sin(t+1.13) (note the phase relative to Vtot)
 VC(t)=193sin(t-0.44) phase angle : 1.13-/2=-0.44
 VL(t)=57.6sin(t+2.7) phase angle : 1.13+/2=2.7
 rms voltages over each element?
 VR,rms=63.8/2=45.1 V
 VC,rms=193/2=136 V
 VL,rms=57.6/2=40.7 V
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answers
 g) power consumed by each element and total power consumed?
 PC=PL=0 no energy is consumed by the capacitor or inductor
 PR=Irms2R=(Imax/2)2R=0.2552R/2=0.2552*250/2)=8.13 W
 or: PR=Vrms2/R=(45.1)2/250=8.13 W (don’t use Vrms=V0/2!!)
 or: PR=VrmsIrmscos=(150/2)(0.255/2)cos(-64.80)=8.13 W
 total power consumed=power consumed by resistor!
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lon-capa
 you should now try problem 6 of lon-capa set 7, except
for the last part
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LRC circuits: an overview
Reactance of capacitor: Xc= 1/C
VL
Reactance of inductor: XL= L
Current through circuit: same for all components
‘Ohms’ law for LRC circuit: Vtot=IZ
vector sum
VL+VC
 Impedance: Z=[R2+(XL-Xc)2]




VR
I

Vtot
=t
t
VC
 phase angle between current and source voltage:
tan=(|VL| -|Vc| )/VR=(XL-Xc)/R
 Power consumed (by resistor only): P=I2rmsR=IrmsVR
P=VrmsIrmscos
 VR=ImaxR in phase with current I, out of phase by  with Vtot
 VC=ImaxXC behind by 900 relative to I (and VR)
 VL=ImaxXL ahead of 900 relative to I (and VR)
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quiz (extra credit)
 The sum of maximum voltages over the resistor, capacitor
and inductor in an LRC circuit cannot be higher than the
maximum voltage delivered by the source since it
violates Kirchhoff’s 2nd rule (sum of voltage gains equals
the sum of voltage drops).
 a) true
 b) false
answer: false
The maximum voltages in each component are
not achieved at the same time!
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Resonances in an RLC circuit
 If we chance the (angular) frequency the reactances will change
since:
 Reactance of capacitor: Xc= 1/C
 Reactance of inductor: XL= L
 Consequently, the impedance Z=[R2+(XL-Xc)2] changes
 Since I=Vtot/Z, the current through the circuit changes
 If XL=XC (I.e. 1/C= L or 2=1/LC), Z is minimal, I is maximum)
 = (1/LC) is the resonance angular frequency
 At the resonance frequency =0 (see question on slide 23)
Z
I
0


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example
Using the same given parameters as the earlier problem,
what is the resonance frequency?
Given:
R=250 Ohm
L=0.6 H
C=3.5 F
f=60 Hz
V0=150 V
= (1/LC)=690 rad/s
f= /2=110 Hz
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question
 An LRC circuit has R=50 Ohm, L=0.5 H and C=5x10-3 F. An
AC source with Vmax=50V is used. If the resistance is
replaced with one that has R=100 Ohm and the Vmax of the
source is increased to 100V, the resonance frequency will:
 a) increase
 b)decrease
 c) remain the same
answer c) the resonance frequency only depends
on L and C
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loncapa
 You should now try question 6, part 7 and question 5 of
lon-capa set 7.
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transformers
transformers are used to convert
voltages to lower/higher levels
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transformers
primary circuit
with Np loops in Vp
coil
secondary
Vs circuit with Ns
loops in coil
iron core
If an AC current is applied to the primary circuit: Vp=-NpB/t
The magnetic flux is contained in the iron and the changing flux acts
in the secondary coil also: Vs=-NsB/t
Therefore: Vs=(Ns/Np)Vp if Ns<Np then Vs<Vp
A perfect transformer is a pure inductor (no resistance), so no power
loss: Pp=PS and VpIp=VsIs ; if Ns<Np then Vs<Vp and IS>Ip
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question
a transformer is used to bring down the high-voltage delivered
by a powerline (10 kV) to 120 V. If the primary coil has 10000
windings, a) how many are there in the secondary coil?
b) If the current in the powerline is 0.1 A, what is the maximum
current at 120 V?
a) Vs=(Ns/Np)Vp or Ns=(Vs/Vp)Np = 120 windings
b) VpIp=VsIs so Is=VpIp/Vs=8.33 A
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question
 Is it more economical to transmit power from the power
station to homes at high voltage or low voltage?
 a) high voltage
 b) low voltage
answer: high voltage
If the voltage is high, the current is low
If the current is low, the voltage drop over the power
line (with resistance R) is low, and thus the power
dissipated in the line ([V]2/R=I2R) also low
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electromagnetic waves
 James Maxwell formalized the basic equations governing
electricity and magnetism ~1870:
 Coulomb’s law
 Magnetic force
 Ampere’s Law (electric currents make magnetic fields)
 Faraday’s law (magnetic fields make electric currents)
 Since changing fields electric fields produce magnetic
fields and vice versa, he concluded:
 electricity and magnetism are two aspects of the same
phenomenon. They are unified under one set of laws: the
laws of electromagnetism
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electromagnetic waves
Maxwell found that electric and magnetic waves travel
together through space with a velocity of 1/(00)
v=1/(00)=1/(4x10-7 x 8.85x10-12)=2.998x108 m/s
which is just the speed of light (c)
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electromagnetic waves can be used to broadcast…
 Consider the experiment performed by Herz (1888)
I
Herz made an RLC circuit with L=2.5 nH, C=1.0nF
The resonance frequency is = (1/LC)=6.32x108 rad/s
f= /2=100 MHz.
Recall that the wavelength of waves =v/f=c/f=3x108/100x106=3.0 m
wavelength: =v/f
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He then constructed an antenna
dipole antenna
 charges and currents vary
sinusoidally in the primary and
secondary circuits. The charges in
the two branches also oscillate at
the same frequency f
I
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antenna
---------++++++
---------++++++
producing the electric field wave
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producing the magnetic field wave
E and B are in phase
and E=cB with
c: speed of light
---------++++++
The power/m2=0.5EmaxBmax/0
II
antenna
---------++++++
II
The energy in the wave is
shared between the
E-field and the B-field
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question
Can a single wire connected to the + and – poles of a
DC battery act as a transmitter of electromagnetic waves?
a) yes
b) no
answer: no: there is no varying current and hence no
wave can be made.
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c=f
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lon-capa
 now try questions 2 and 7 from set 7.
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quiz (extra credit)
R
L
I
V




At t=0, the switch is closed. Shortly after that:
a) the current slowly increases from I=0 to I=V/R
b) the current slowly decreases from I=V/R to I=0
c) the current is a constant I=V/R
The coil opposes the flow of current due to self-inductance, so the
current cannot immediately become the maximum I=V/R. It will
slowly rise to this value
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