Download Notes 14

Notes for July 10, 2008
Prepared by Tim Ford
Quantitative Problem 7/9/08
1. An oscillating LC circuit consists of a 75.0mH inductor and a 3.60µF capacitor.
If the maximum charge on the capacitor is 2.90µC, (a) what is the total energy
in the circuit and (b) what is the maximum current?
L = 75.0mH
C = 3.60µF
PEL = ½ LImax2
PEC = ½ C(ΔVmax)2
ΔV = Q/C
(a) Total energy = PEL = PEC
PEC = ½ C(ΔVmax)2
= ½ C(Q/C)2
= ½ Q2/C
= ½ (2.90×10-6)2/ (3.6×10-6)
= 1.1681×10-6 J
There is no resistor in the LC
circuit, thus there is no power
dissipated across the circuit.
Therefore the system can be
considered a perpetual oscillation
system.
When the capacitor is fully charged
and the energy stored in the
capacitor is at its maximum, the
current across the circuit is zero.
When the energy is stored in the
inductor is maximal, the energy
stored in the capacitor is zero, and
current is a maximum.
Because energy is conserved, the
PEL = PEC
(b) Maximum current
PEL = PEC
½ LImax2 = ½ C(ΔVmax)2
½ LImax2 = ½ C(Q/C)2
½ LImax2 = ½ Q2/C
Imax = (Q2/CL)½
= ((2.90×10-6)2/ (3.6×10-6×75.0×10-3))½
= 5.581×10-3A
Quantitative Problem Chosen by Shusaku
(2) An AC source with a maximum voltage of 150V and f = 50.0Hz is connected
between points a and d. Calculate the rms voltages between points (a) a and
b, (b) b and c, (c) c and d
R = 40Ω
a
b
ΔV = 150V
L = 185mH
f = 50.0 Hz
d
c
C = 65.0µF
Reactances between points a and d are
XL= 2πfL = 2π(50.0)(185×10-3) = 58.12Ω
XC = 1/(2πfC) = 1/(2π(50.0)(65×10-6)) =
48.97Ω
Total Impedance between points a and d
Zad = (R2 + (XL – XC)2)½ = ((40.0)2 + (58.12 – 48.97)2)½ = 41.0Ω
(ΔVTotal)rms = (ΔVTotal)max / (2) ½ = 150/(2) ½ = 106.06V
(ITotal)rms = (ΔVTotal)rms / Z = 106.06 / 41.0 = 2.58A
(a) Between points a and b
ZR = (Rab2 + (XL,ab – XC,ab)2) ½ = (40.02 + (0 – 0)2) ½ = 40.0Ω
(ΔVR)rms = IrmsZab = (2.58)(40.0) = 103.2V
(b) Between points b and c
ZL = (Rbc2 + (XL,bc – XC,bc)2) ½ = (02 + (58.12 – 0)2) ½ = 58.12Ω
(ΔVL)rms = IrmsZbc = (2.58)(58.12) = 150.0V
(c) Between points c and d
ZC = (Rcd2 + (XL,cd – XC,cd)2) ½ = (02 + (0 – 48.97)2) ½ = 48.97Ω
(ΔVC)rms = IrmsZC = (2.58)(48.97) = 126.3V
VL
VR
Which comes first, the current or the total
voltage?
The total voltage comes before the current in this example.
VL > VC
VTotal
Vc
Conceptual Problems 7/10/08
(1) The color of an object is said to depend on the wavelengths the object
reflects. So, if you view the colored objects under water, in which the
wavelength of the light will be different, does the color change?
The color will not change because color depends on frequency of light, which
does not change under water.
(2) How is it possible that a complete circle of a rainbow can sometimes be seen
from an airplane?
Rainbow formation occurs when light passes through water molecules and
different wavelengths get bent at different angles due to refraction and
reflection. Due to the spherical shape of the water molecule, the shape of the
rainbow results in a ring. You only see half of this ring because there is a
horizon and you must be above this horizon to see the entire rainbow ring.
(3) If the number of turns on the transformer input is greater than the number on
the output, then the transformer:
(d) Outputs less voltage than at the input
VIN = (NIN/NOUT)VOUT
(4) When viewing your image in a hand-held mirror, if you move the mirror away
at a speed v, the image appears to:
(b) move away at 2v
For a flat mirror, |d0| = |d1|, so the distance from the object to the image is 2d0.
So, the rate at which the object distance increases is half the rate that the
distance between the object and image increases. (This is chapter 23)
Quantitative Problems 7/10/08
(1) Light is incident normal to a 1cm layer of water that lies on top of a flat Lucite
plate with a thickness of 0.5 cm. How much more time is required for light to
pass through this double layer than is required to traverse the same distance
in air? (n of Lucite = 1.59)
v = c/n
t = d/c = (1.5×10-2)/(3×108) = 5×1011s in air
twater, Lucite = (dwater / vwater + dlucite / vlucite)
= (dwater / (c/nwater) + dlucite / (c/nlucite))
= {(1×10-2 / (3×108/1.333)) + (0.5×10-2) / (3×108/1.59)} = 7.09×10-11s
(2) Light of wavelength 589nm in a vacuum passes through a piece of fused
quartz of index of refraction n = 1.458. (a) Find the speed of light in fused
quartz, (b) what is the wavelength of the light in fused quartz, and (c)what is
the frequency of light in fused quartz?
(a) v = c/n
3×108/1.458 = 2.06×108 m/s
(b) λn= λ0 / n = 589 / 1.458 = 404 nm
(c)
f = c / λ = 3×108 / 589×10-9 = 5.09×1014 Hz