Name _________________
Solutions to Test 3
April 16, 2008
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
RL Circuits
Possibly useful formulas:
Constant
μ0 = 4π ×10−7 T ⋅ m/A
τ =L R
LC Circuits
ω = 1 LC
Q = Q0 cos (ωt )
EM waves
E0 = cB0
Impedance
X C = 1 Cω , X L = Lω
1
Z = R + ( X L − XC )
S=
μ0
(E × B)
S = cB02 2μ0
P= S c
2
2
AC Circuits
ΔV = ΔVmax sin (ωt )
I = I 0 e −t τ
2
φ = tan −1 ⎡⎣( X L − X C ) R ⎤⎦
ω0 = 1 LC
I=
E
1 − e −t τ )
(
R
I = I max sin (ωt − φ )
RLC Circuits
2
1
R
− 2
LC 4 L
Q = Q0 e − Rt 2 L cos (ωt )
ω=
RMS values
ΔV
ΔVrms = max
2
I
I rms = max
2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. In an RLC circuit, if the resistance R is very small, the energy in the circuit will
A) Disappear quickly
B) Disappear slowly
R
C) Remain indefinitely
L C
D) Increase slowly
E) Increase quickly
2. A prism turns white light into a rainbow by
A) Having a different index of refraction n for different colors, so you get
different refraction
B) Absorbing certain colors in different directions, so one direction removes all but
red, etc.
C) Having atoms that are spaced the same distance apart as the different colors, so
only certain colors can pass through at different angles
D) Bouncing different frequencies off of electrons that are orbiting at the correct
frequencies, but in different orientations
E) Shifting the frequencies by different amounts as the light bounces off of moving
atoms
3. The Poynting vector points in the direction of
A) Motion of the EM wave
B) Electric field
C) Magnetic field
D) Frequency
E) None of the above
4. Inductors resist changes in
A) Resistance B) Voltage
C) Current
D) Frequency
E) Charge
5. For an ideal transformer, if the voltage is stepped up from 120 V to 240 V, what will
happen to the current I?
A) It will double
B) It will quadruple
C) It will cut in half
D) It will be cut by ¼
E) It will remain the same
6. If a wave has a wavelength of 2 m, the wave number k must be
B) 1/π m-1
C) 4π m-1
D) π m-1
A) 4/π m-1
E) 2π m-1
7. The ray approximation, or geometric optics, should only be applied when
A) The frequency of light is much higher than any other frequency in the problem
B) The frequency of light is much lower than any other frequency in the problem
C) The wavelength is much longer than any other distance in the problem
D) The wavelength is much shorter than any other distance in the problem
E) None of the above
8. What does a diode do in a power supply?
A) It stores charge from cycle to cycle, so the voltage can be kept
uniform
B) It blocks sudden changes in current caused by external power surges
C) It converts the high voltage AC from the outlet into low voltage AC for a device
D) It reduces the resistance in subsequent components by increasing the voltage
E) It only allows current to flow one way, helping to turn AC into DC
9. To make a circuit where there is a lot of power fed to a resistor at low frequencies,
but very little at high frequencies, you should put the resistor in series with
A) a capacitor
B) an inductor
?
C) a capacitor and an inductor
D) another resistor
E) none of these will work
10. If a light beam is moving at 1.00×108 m/s, the index of refraction n must be about
A) 1/3
B) 1/2
C) 1
D) 2
E) 3
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences).
11. Place the following in order from shortest wavelength to
longest: gamma rays, infrared, microwaves, radio waves,
ultraviolet, visible light, X-rays.
The correct order appears in the box at right.
12. At t = 0, the capacitor at right has a charge on it, and there
is no current flowing. Explain qualitatively the behavior of
the circuit. You should include at least one equation.
Gamma rays
X-rays
Ultraviolet
Visible light
Infrared
Microwaves
Radio Waves
C
The charge will begin to flow out of the capacitor and
L
through the inductor, eventually charging the capacitor in the other
direction, and then the process will reverse, and so oscillate indefinitely. The charge on
the capacitor will take the form Q = Q0 cos (ωt ) , where Q0 is the initial charge, and the
angular frequency is given by ω = 1
capacitance respectively.
LC , where L and C are the inductance and
13. As a light wave moves from air to glass, which of the following quantities will
increase, decrease, or stay the same: velocity, frequency, angular frequency,
wavelength, wave number.
The frequency f and angular frequency ω will stay the same, but the speed of light
v (velocity) decreases. By the formulas ω k = f λ = v , it is clear that the wave number k
must increase while the wavelength λ decreases.
Part III: Calculation: [60 points]
Choose three of the following four questions and perform the indicated
calculations (20 points each)
14. In the circuit sketched at right, the switch is initially left closed for a long time
(a) Describe the steady state circuit. In particular, find the current in the
inductor in the steady state, and explain where else,
if anywhere the current is flowing.
+
12 V
This current also flows through the 12 V battery and 2 Ω resistor.
(b) At t = 0, the switch is suddenly opened. At this moment, explain which way
the current flows, if any. At t = 0, find the voltage across the 8 Ω resistor, if
any.
The inductor resists any changes in current, so at t = 0, the current in the inductor
must now still be 6 A. Therefore, the resistor on the right must have 6 A running through
it, and using Ohm’s Law again, we have
ΔV = IR = ( 6 A )( 8 Ω ) = 48 V
(c) Find an expression for the current at all subsequent times t > 0.
The characteristic time for an RL circuit is just given by
τ = L R = ( 4 H ) ( 8 Ω ) = 0.5 s
We now plug this into our formula for the decay of the current:
I = I 0 e −t τ = ( 6 A ) e −t ( 0.5 s ) = ( 6 A ) e −2t ,
where in the last formula, we assume t is in seconds.
(d) At what time t will the current through the inductor have dropped to 1 A?
We simply set the current to 1 A and solve for t:
1 A = ( 6 A ) e −2t ,
e −2t = 16 ,
−2t = ln ( 16 ) = −1.792,
t = 0.896 s
8Ω
–
4H
ΔV 12 V
I=
=
=6A
R
2Ω
2Ω
In the steady state, the inductor has no changing
current, and hence no EMF across it. Therefore, we can treat it
like a wire, so that the current will completely bypass the 8 Ω
resistor. The 12 V battery is resisted only by the 2 Ω resistor,
and hence will have a current
15. An AC source with ΔVmax = 90 V operating at 120 kHz is fed
through a 1.5 mH inductor, a 1 nF (= 10-9 F) capacitor, and
a 150 Ω resistor.
(a) What is the angular frequency ω?
1.5 mH
The angular frequency is simply given by
150 Ω
1.00 nF
ω = 2π f = 2π (1.20 ×105 s −1 ) = 7.54 ×105 s −1
(b) What is the impedance for the inductor alone? For the
capacitor alone?
ΔVmax = 90 V
f = 120 kHz
The impedances for each of these components separately is
given in the formulas above:
X L = Lω = ( 0.0015 H ) ( 7.54 ×105 s −1 ) = 1130 Ω
XC =
1
1
=
= 1330 Ω
−9
Cω (1.00 × 10 F )( 7.54 × 105 s −1 )
(c) What is the impedance for the whole circuit?
Again, the relevant formula is given above:
Z 2 = R 2 + ( X L − X C ) = (150 Ω ) + (1130 Ω − 1330 Ω ) = 6.06 ×104 Ω 2 ,
2
2
2
Z = 246 Ω
(d) Find the maximum voltage Imax and the average power dissipated in the
resistor
The maximum voltage is just given by
I max =
ΔVmax
90 V
=
= 0.365 A
Z
246 Ω
To find the power, we will use the formula
2
P = RI
2
rms
⎛I ⎞
2
= R ⎜ max ⎟ = 12 RI max
=
⎝ 2 ⎠
1
2
(150 Ω )( 0.365 A )
2
= 10.0 W
16. In the year 2100, the Space Ship Carlson will be launched. It will have mass of
1000 kg and will also have a 300 kg solar sail in the shape of a square of size 20.0
km × 20.0 km. It will be pushed by advanced lasers that can illuminate its entire
surface with an intensity of 10.0 kW/m2, which will impact perpendicular to the
solar sail. The solar sail blocks 1/3 of the light, reflects 1/3 of it back, and the
remaining 1/3 passes directly through it.
(a) Calculate the magnitude of the electric and magnetic fields from the
oscillating fields at this power.
The average value of the Poynting vector is the same as the intensity, so we have:
S =
cB02
2 μ0
2 ( 4π ×10−7 T ⋅ m/A )(1.00 ×104 W/m 2 )
2 μ0 S
=
= 8.38 ×10−11 T 2 , ,
8
3.00 ×10 m/s
c
−6
B0 = 9.15 ×10 T = 9.15 μ T
B02 =
The electric field is then given by
E0 = cB0 = ( 3.00 ×108 m/s )( 9.15 ×10−6 T ) = 2750 V/m
(b) Calculate the pressure on the solar sail due to the light that is absorbed and
reflected.
The pressure for light that is absorbed by the sail is P = S c ; however, only
one-third of the power is absorbed, so this must be modified to P = S 3c . For light
that is reflected, the formula is P = 2 S c , but again, only a third of this is reflected, so
this is modified to P = 2 S 3c . The transmitted light has no effect, so the net pressure
is
P=
S 2 S
S
1.00 ×104 W/m 2
+
=
=
= 3.34 ×10−5 N/m 2
8
3c
3c
c
3.00 ×10 m/s
(c) Calculate the total force and the acceleration of the rocket/solar sail.
To get the total force, we simply multiply by the area of the sail, so
F = Pσ = ( 3.34 ×10−5 N/m 2 )( 2.00 ×104 m ) = 1.33 ×104 N
2
We then us F = ma to get the acceleration, so
a=
F 1.33 ×104 N
=
= 10.3 m/s 2
1300 kg
m
17. A light beam enters a rectangular aquarium from the side filled with water (n =
1.333) at an angle of θ = 27° below level, as sketched below. At the back of the
tank is a mirror, which will reflect the light back into the water.
(a) Most of the light will refract and enter the water, as sketched below. At what
angle α will it enter the water?
We can find this angle simply by using Snell’s Law, which tells us
n1 sin17° = n2 sin α ,
sin α =
n1
sin17°
sin17° =
= 0.341
n2
1.333
α = 19.9°
(b) After it reflects off the back wall, it will reflect with an angle β compared to
normal. Find β.
I have added the angle γ, which by alternate interior angles, is equal to α. Now,
by the law of reflection, β = γ, so we have
β = 19.9°
(c) When the light reaches the top of the water, will it reflect, refract, or do a
combination of both? Sketch which direction(s) it goes, and calculate any
relevant angles.
The angle δ is contained in a right triangle with β, and hence is its complement,
given by
δ = 90° − β = 90° − 19.9° = 70.1°
The critical angle at which light will be totally internally reflected is given by
sin θ c = n1 n2 = 1 1.333 = 0.750,
θ c = sin −1 ( 0.750 ) = 48.6°
Since our incoming light is at an angle greater than this, it is totally internally reflected,
and we have all of it reflected. By the law of reflection, the angle is
θ = δ = 70.1°
water
α
27°
δ
γ
β
mirror
θ