March 31, 2009
PHY2054 Discussion-Spring ‘09
Quiz 7 (Chapter 21.1-22.4)
Name:
UFID:
*1. (2.5pts) An RCL series combination of R = 50 Ω, C = 40 μC and L = 200 mH is connected across
an AC generator with ΔVrms = 120 V and f = 60.0 Hz. Calculate the rms voltage across the capacitor.
The inductive reactance and the capacitive reactance are
XL = 2πfL = 2π×60×0.2 = 75.4 Ω
XC = 1//(2πfC) = 1//(2π×60×40×10-6) = 66.3 Ω
The impedance of the combination is
Z = √[R2+(XL-XC)2] = √[502+(75.4-66.3)2] = 50.8 Ω
The rms current in the circuit is the same as that in the capacitor. Thus we have
IC,rms = Irms = ΔVrms /Z = 120/50.8 = 2.36 A
The voltage across capacitor is given by
ΔVC,rms = IC,rms XC = 2.36×66.3 = 156 V
***2. (2.5pts) An ideal transformer has a 240-turn primary coil and 80-turn secondary coil. The
primary coil is connected across an AC source with an rms voltage of 120 V and an internal
resistance of 20.0 Ω. What is the average power consumed in the secondary coil if it is connected
across a 40-Ω resistor?
Since the power in R2 is proportional to the current in the secondary coil
I2, we express I2 in terms of the supplied voltage. Applying Kirchhoff’s
loop rule to the primary coil and the secondary coil, we have
ΔV-I1R1-ΔV1 = 0 & ΔV2-I2R2 = 0
Since the transformer is ideal, the currents and the voltages in the
primary and secondary coils are related as follows:
ΔV1/N1 = ΔV2/N2 ⇒ ΔV1 = 3 ΔV2 , I1N1 = I2N2 ⇒ I1 = I2/3
Using these relationships, we solve Kirchhoff’s equations for I2.
ΔV-(I2/3)R1-(3 ΔV2) = 0 & ΔV2-I2R2 = 0 ⇒I2 = ΔV/(R1/3 + 3R2)
This relation is valid for rms values. Therefore, the average power in R2 is
Pav = R2 (I2,rms)2 = R2(ΔVrms)2/(R1/3 + 3R2)2 = 40×1202/(20/3+3×40) = 35.9 W
Constants & Formulas
Impedance of RLC Series: Z = √[R2+(XL-XC)2]
Force delivered by electromagnetic waves:
where XL = 2πfL & XC = 1/(2πfC)
F = 2IA/c (Reflection)
F = IA/c (Absorption), where c = 3.00×108 m/s
Ideal Transformer: ΔV1/N1 = ΔV2/N2, I1N1 = I2N2
Kirchhoff’s Rules: Iin = Iout & ΣΔV = 0
Kinematics Eqns: v = at, Δx = (1/2)at ,v = 2aΔx
Snell’s Law: n1sinθ1 = n2sinθ2
2
2
**3. (2.5pts) As a model of a spaceship, we consider a perfectly reflecting sheet exposed to sunlight.
Sunlight with intensity of 1 340 W/m2 is incident normally on a 1000-kg large square sheet 200 m on
each side. How long does it take the spaceship to move 1.00 km if it starts from rest?
When electromagnetic waves strike a surface, force is delivered to the surface. The force on the
reflecting surface is
F = 2IA/c = 2×1340×2002/(3×108)= 0.357 N
The acceleration of the sheet is
a = F/m = 0.357/1000 = 3.57×10-4 m/s2
Using one of the kinematics equations, the interval is calculated as
Δx = (1/2)at2 ⇒ t = √(2Δx/a) = √(2×1000/3.57×10-4) = 2.37×103 s
**4. (2.5pts) A ray of light strikes a flint glass block at an angle of incidence of 25.0º. The block has
a thickness of 8.00 cm and an index of refraction of 1.66. When the ray passes through the block, it
shifts laterally by a distance d. Find the value of d.
Applying Snell’s law, the angle of refraction is
sinθ1 = 1.66sinθ2 ⇒ θ2 = sin-1(sin25º/1.66) = 14.7º
The angle between the extension of the incident ray and the refracted ray is
α = θ2 - θ1 = 25-14.7 = 10.3º
Letting the path length of the beam inside of the block be L, the lateral shift
d is given by
d = Lsinα = (t/cosθ2)sinα = (8/cos14.7º)×sin10.3º = 1.50 cm