Physics 1B Electricity & Magne4sm Frank Wuerthwein (Prof) Edward Ronan (TA) UCSD Some Details on Final Exam •  20 Ques4ons •  Will make available a prac4ce final late night on Monday, March 19th on website. •  Most of the ques4ons for final will be taken from: –  Homework –  Quiz 1,2,3,4 –  Prac4ce Final –  Examples worked in class Outline of today •  Addi4onal material on AC currents •  Chapter 24: Electromagne4c waves "
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AC Circuits with Ind. Consider a simple circuit containing an inductor and an AC source. Ini4ally the current in hindered by the inductor s induc4on. Even though the initial value of
current will be a maximum, the
voltage drop will oppose it.
When the time rate of change
of the current is maximum
then the voltage drop across
the inductor is a maximum.
AC Circuits with Ind. "
"
"
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Also, when the time rate of change of the current is
zero then the voltage drop across the inductor is also
zero.
Giving us the following:
As time moves on, we would
first see a maximum voltage
drop, then a maximum
current.
The voltage across the
inductor, ΔVL, leads the
current by 90o.
AC Circuits with Ind. "
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"
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What determines how much current the inductor gets? The inductance, L, of the inductor and the frequency, f,
of the AC current.
Because of this we define a new variable which basically
measures how much the inductor resists current for a
given AC source.
This variable is called inductive reactance, XL, and is
given by:
X L = 2"fL = #L
"
where f is measured in Hz and L is in H, this means that
XL will be in Ω.
AC Circuits with Ind. "
Since inductive reactance measures how an inductor
resists current in an AC circuit, we can say that Ohm s
Law becomes: "VL,rms = Irms X L
"
The generator in a purely inductive AC circuit has an
angular frequency of 120π rad/s. If Vmax = 140V and L =
0.100H,
!what is the rms current in the circuit?
Imax
"Vmax "Vmax
=
=
XL
#L
Imax
140V
=
= 3.71A
120" rad s (0.100H)
(
)
Imax 3.71A
Irms =
=
= 2.63A
2
2
RLC Circuit "
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When we examine the different graphs of the voltage
across circuit elements we compare them to the
current graph. The instantaneous voltage
across resistor is in phase
with the current.
The instantaneous voltage
across the inductor leads the
current by 90o.
The instantaneous voltage
across the capacitor lags the
current by 90o.
RLC Circuit "
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When you want to find the sum of the voltage drops
(ΔVmax) across the RLC circuit you simply cannot add
the separate maximum voltage values (they are not all in
phase).
You have to add them as vectors (called phasors).
Note that ΔVL and ΔVC
are on the same line (180o
difference) and that ΔVR
is 90o from both of them.
(ΔVL – ΔVC) comes
from 180o difference.
φ
RLC Circuit "
In order to find ΔVmax, use Pythagorean theorem:
2
R
"Vmax = "V + ("VL # "VC )
"
"
2
We define the angle between ΔVmax and the current as
the phase angle, φ.
! phase angle is given
The
by:
#VL $ #VC )
(
tan " =
#VR
φ
"
RLC Circuit "
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We can also define an overall RLC circuit resistance to
current flow that is based on the resistor, inductor, and
capacitor values.
We define this term to be impedance, Z, (which will
have units of Ω) in terms of R, XL and XC.
We can also define the
phase angle this way:
RLC Circuit "
For an RLC circuit, Ohm s Law becomes: "Vmax = Imax Z
"
"
No power losses are associated with pure capacitors
and pure inductors in an AC circuit.
Thus, !
the average power delivered to the generator is
converted to internal energy in the resistor.
Pavg = Irms"VR = Irms"Vrms cos #
"
cos
is called the power factor of the circuit.
"
Phase shifts can be used to maximize power outputs.
RLC Circuit "
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"
"
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Resonance occurs where you have the maximum
current value for a given circuit.
This will maximize the power output as well (need
cos to be 1).
You need to minimize
impedance for this to occur.
This occurs when XC=XL.
The frequency for this to
occur is, fo:
RLC Circuits •  Example •  In a series RLC circuit, suppose R = 300Ω, L = 60mH, C = 0.50μF, Vmax = 50V, and ω = 10,000rad/s. Find the impedance Z of the circuit and the phase angle, ϕ. Answer
"   First, we should calculate the reactances, XL and
XC, and then the impedance.
"
RLC Circuits "
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Answer
The reactances are:
X L = "L = (10,000
!
"
rad
The impedance will be given by:
2
Z = R + ( X L " XC ) =
(300#)
Z = 500"
"
H) = 600%
1
1
XC =
=
= 200%
$6
"C (10,000 rad sec)(0.50 #10 F)
2
!
)(60 #10
sec
$3
To find the phase angle use:
2
+ (600# " 200#)
2
"
"
RLC Circuits Answer
This gives us:
% 600$ # 200$ (
#1% 400$ (
" = tan '
* = tan '
*
&
)
& 300$ )
300$
#1
" = 53°
"
"
"
"
The
! positive angle means that the circuit is more
inductive than capacitive.
So, the voltage leads the current.
Now, from these pieces of information observable
! can be determined.
information
For example, what if the question also asked for the
current amplitude in the circuit and the voltage
amplitude across each circuit element?
"
"
RLC Circuits Answer
Since we know the impedance and the maximum
voltage we can easily compute:
Imax
"
Vmax
50V
=
=
= 0.10A
Z
500"
Which means that for the circuit elements:
"VR ,max = Imax R = (0.10A)( 300#) = 30V
!
"VL,max = Imax X L = (0.10A)(600#) = 60V
"VC ,max = Imax X C = (0.10A)(200#) = 20V
Chapter 24 Electromagne4c Waves Plane Electromagnetic waves
- description E & B fields are the
same everywhere
within the plane.
E-field
B-field
Wave fronts (infinite & parallel)
E & B fields always
orthogonal.
E & B fields always
orthogonal to wave
velocity.
E & B fields vs x,t E(x,t) = E sin(kx - ωt) in y-direction
B(x,t) = B sin(kx - ωt) in z-direction
E & B are in phase and orthogonal in space.
Speed of Electromagne4c wave "
1
=
=c
k
#0µ0
Maxwell unified E&M&O by realizing that
his equations allow for waves that propagate
at the speed of light!
Relationship of E and B
"
E p = B p = cBp
k
!
The Electromagnetic wave
spectrum
Polarization
The orientation of E in the y-z plane is arbitrary.
Polarized EM waves have well defined orientation.
However, once E is chosen, the orientation of B is fixed.
Polarized light sources
- examples (typically) polarized:
radio, TV, radar, ... EM from simple antennas.
Lasers.
(typically) unpolarized:
sun light
light from light bulbs, or other hot sources (e.g. the glow of fire).
Polarizers
Law of Malus 2
I = I0 cos "
E.g.: Intensity,I, of unpolarized light drops
by
factor 2 when going through a polarizer
because “unpolarized” = random
polarization
and the average of the square of a cos is
Q to ponder: •  You have unpolarized light that traverses a sequence of two polarizers whose polariza4on axes form and angle of 60 degrees. •  What frac4on of the intensity of the unpolarized light makes it through both polarizers? Answer: •  Unpolarized light drops by factor 2 when going through first polarizer. •  Cos260o = 0.52= 0.25 •  The remaining intensity ager both polarizers is 12.5% of the ini4al intensity of the unpolarized light. Q to ponder: •  Is it possible to rotate the polariza4on axis of light by 90 degrees ? •  If yes, how would one do it? Answer •  Yes. •  Imagine subjec4ng polarized light with a sequence of polarizers, all of which make a small angle with each other. •  Each polarizer reduces the intensity by cos2 •  However, each polarizer also forces the light to be polarized along the polarizers axis. •  Imagine 900 polarizers, each with 0.1 degree from the previous: cos18000.1o = 99.7% •  In contrast: cos290o = 0. •  It is thus in principle possible to rotate light with consecu3ve polarizers at small angles, and with only small losses in intensity. For Next Time (FNT) "  Start preparing for final exam