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Physics 202
Tuesday, April 27, 1999
Announcements:
Recitation and Lab Grades:
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Recitation overall score (HW and quizzes) will be normalized to an average
of 75% for each instructor. The lowest quiz and homework assignment will
be dropped.
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Lab reports will be normalized to an average of 85%
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Lab quizzes will not be normalized. The lowest lab quiz and report will
be dropped.
Review Sessions:
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Eric Wertz:
May 2 1:00 - 2:30
117 Osmond
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Seiji Takemore May 2
2:30 - 4:00 101 Osmond
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Yuri Zadorozhny May 2
6:00 - 7:30 117 Osmond
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Steve Minnick May 3
10:30 - 12:00 117 Osmond
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Karl Nelson
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Extended Office Hours:
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Professor Samarth: May 3, 1-3
pm and May 4, 1-3 pm
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Professor Chan
May 4, 10 am-12 pm and 3-5 pm
Info about the Final:
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When? Tuesday, May 4, 6:50 to 8:40 pm
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Where?
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Sections 1,2,5,16,12,13,15
111 Forum
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Sections 3,4,7,11,17,20
108 Forum
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Sections 6,8,19,9,10,14,18
105 Forum
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The exam will have 25 multiple choice questions of which 15 will be on
new material
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You may bring a 8.5 x 11" sheet of paper will writing on both sides.
Lecture notes:
Continuing with RLC circuits...
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Let's measure the voltage across R in an RLC circuit; also across L; across
C
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Note: x = vR + vL + vC
at all times
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What are the phase relations between vR and i? They are
in phase.
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between vL and i? vL leads i by 90o
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between vC and i? vC lags behind i by 90o
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See figure 33-11 on page 824 of the text for pictures of the phasor diagrams.
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Note: xm2 = VR2
+ (VL -VC)2 = (IR)2 + (IXL
- IXC)2 = I2[R2 + (XL
- XC)2]
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or similarly I = xm / [R2
+ (XL - XC)2]1/2
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Impedance of RLC circuit Z = [R2 + (XL - XC)2]1/2
= [R2 + (wL - (1/wC)2]1/2
, so I= xm / Z
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If wL = 1/wC, then
I has a maximum. For a fixed L and C, we can change w
so that wL = 1/wC,
therefore the current is a maximum. This is called tuning the RLC circuit
to resonant condition. w2 = 1/LC
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See figure 33-13 on page 826 of the text for a picture of Resonance curves
Power in a RLC, alternating current circuit
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x = xm sin wt,
then i = Isin (wt - phase constant)
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In an RLC circuit, energy supplied by a generator is stored in electric
field in C and in magnetic field in L. Average energy stored in C and L
is constant. The transfer of energy is from the generator G to the resistor.
(from electromagnetic energy to heat).
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P = i2R = [ I sin (wt - phase constant)]2R
= I2Rsin2 (wt - phase
constant)
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P is never negative
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Paverage = I2R/ 2 = (I/(2)1/2)2R
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See figure 33-34 on page 827 of the text.
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I/(2)1/2 = the root mean square or rms value of i
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I/(2)1/2 = Irms or Paverage = Irms
2R
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Similarly Vrms = V/ 21/2, xrms
= xm / 21/2 , then Irms
= xrms /Z = xrms
/[R2 + (XL - XC)2]1/2
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V = 21/2 Vrms , where V is the amplitude of the voltage
and I = 21/2 Irms
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Paverage = Irms 2R = (xrms
/ Z) Irms R = xrms
Irms R/Z = xrms
Irms cosq where the cosq
is the power factor
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In order to maximize the the rate energy is supplied to a resistive load,
need to keep cosq close to 1 or q
= 0
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How? cosq = R /[R2 + (XL
- XC)2]1/2 ; match XL to XC
Transformers
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In transmission of electrical energy from a power
station to towns or cities, there must be current flowing in wires (with
resistance); in order to minimize loss in the transmission cables, current
must be kept low. but since total power supplied by the generator is P
= xI, x must
be kept high.
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Typically, the voltage of a long distance transmission
line is kept at about 500 kV and the power generator runs at 200 megawatts.
Let's assume this power station is 1000 km from a city and the resistance
of the cable is 0.2 W / km.
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P = 300 MW = xI,
I = 600 amps
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The power lost in the transmission = I2R
= 72 MW
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What if the transmission line voltage is kept at
250 kV?
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xI= 300 MW and I = 1200
amps
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Power lost in the transmission is 288 MW.
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Therefore, potential needs to be kept high in transmission,
but high voltage is dangerous for household appliances. There are transformers
to step down the voltage just before power is fed into houses.
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