Concept Questions
A wire, initially carrying no current, has a radius
that starts decreasing at t = 0. As it shrinks, which
way does current begin to flow in the loop?
A) Clockwise B) Counter-clockwise C) No current
D) Insufficient information
•Calculate flux downwards – we get EMF clockwise
dB
E
•Flux is decreasing
dt
•Derivative of flux is negative
•EMF is positive clockwise
•Current will flow clockwise
•Right hand rule – B-field
The current that flows will then create a
downwards
magnetic field, which inside the loop, will
•Reinforces magnetic field
A) Strengthen it
B) Weaken it
•Tries to keep the B-flux constant
C) No change D) Insufficient information
Concept Question
What happens as I drop the magnet into
the copper tube?
A) Falls as usual
B) Falls slower
C) Falls faster
D) Floats constant
E) Pops back up and out
•As magnet falls, some places have magnetic fields
that diminish
•Current appears, replacing magnetic field
•This acts like a magnet, pulling it back up
•At bottom end, current appears to oppose change
•This repels the magnet, slowing it down
•Current is only caused by motion of magnet
•If motion stops, resistance stops current
•If motion is small, opposition will be small
•It doesn’t stop, it goes slowly
S
N
S
N
N
S
What if we used a
superconductor?
Concept Question
What is Kirchoff’s law for the loop shown?
A) E + L (dI /dt) = 0 B) E – L (dI /dt) = 0
C) None of the above
D) I don’t know Kirchoff’s law for switches
I
dI
0  E L
dt
dI E

dt L
E
I t
L
•The voltage change for an inductor is L (dI/dt)
•Negative if with the current
•Positive if against the current
L
+
E–
6A
6A
6A
R2 = 4 
L
+
–
In the steady state, with the switch closed,
how much current flows through R2? How
much current flows through R2 the moment
after we open the switch?
A) 0 A
B) 6 A
C) 3 A
D) 2 A
E) None of the above
R1 = 2 
Concept Question
E = 12 V
•In the steady state, the inductor is like a wire
•Both ends of R2 are at the same potential: no current through R2
•The remaining structure had current I = E/R1 = 6 A running through it
I = E /R1 = 6 A
•Now open the switch – what happens?
•Inductors resist changes in current, so the current
instantaneously is unchanged in inductor
I=6A
•It must pass through R2
I=1A
R2 = 1 k
L
+
–
The circuit at right is in a steady state.
What will the voltmeter read as soon as
the switch is opened?
A) 0.l V
B) 1 V
C) 10 V
D) 100 V
E) 1000 V
R1 = 10 
Concept Question
E = 10 V
•The current remains constant at 1 A
•It must pass through resistor R2
V  IR  1 A 1000  
•The voltage is given by V = IR
V  1000 V
•Note that inductors can produce very high voltages
•Inductance causes sparks to jump when you turn a
switch off
+
–
Loop has unintended inductance
V
Concept Question
A capacitor with charge on it has energy U =
Q2/2C, but Q is constantly changing. Where does
the energy go?
A) It is lost in the resistance of the wire
B) It is stored as kinetic energy of the electrons
C) It is stored in the inductor
D) Hollywood!
Q  Q0 cos t 
I

1
Q
C
CL
L
•Let’s find the energy in the capacitor and the inductor
Q02
UC 
cos 2 t 
2C
dQ
 Q0 sin t 
I 
dt
U L  12 LI 2  12 LQ02 2 sin 2 t 
Q02
UL 
sin 2 t 
2C
Q02
UC  U L 
2C
Energy sloshes
back and forth
Concept Question
If the voltage from a source looks like the graph
below, about what voltage should it be labeled?
A) 0 V
B) 170 V
C) 120 V
D) 85 V
E) It should be labeled some other way
•Average voltage is zero, but that doesn’t tell us anything
•Maximum voltage 170 V is an overstatement
•Power is usually proportional to voltage squared
Concept Question
A 60 W light bulb is plugged into a
standard outlet (Vrms = 120 V). What
is the resistance of the bulb?
A) 15 
B) 30 
C) 60 
D) 120 
E) 240 
R  240 
2

V
2
P  I rms
R  rms
R
2
R
V
R
pP
2
rms
120 V 


60 W
2
Capacitors and Resistors Combined
•Capacitors and resistors both limit the current – they both have
impedance
I max  Vmax R
•Resistors: same impedance at all frequencies
I max  Vmax X C
•Capacitors: more impedance at low frequencies
X C  1 C
Concept Question
The circuit at right might be designed to:
(A) Let low frequencies through, but block high frequencies
(B) Let high frequencies through, but block low frequencies
(C) Let small currents through, but not big currents
(D) Let big currents through, but not small currents
Impedance Table
Resistor
Impedance
R
Phase
0
Vector
Direction
right
Inductors are good for
(A) Blocking low frequencies
(B) Blocking high frequencies
(C) Blocking large currents
(D) Blocking small currents
Capacitor
XC 
1
C
Inductor
X L  L

 12 
1
2
down
up
I max  Vmax R
I max  Vmax X C
I max  Vmax X L
Concept Question
In the mystery box at right, we can put a 2.0 F capacitor, a
4.0 H inductor, or both (in series). Which one will cause the
greatest current to flow through the circuit?
A) The capacitor
B) The inductor
C) both
D) Insufficient information
1.4 k
1.3 k
1.5 k
•We want to minimize impedance
X C  1.3 k
•Make the vector sum as short as possible
•Recall, capacitors point down, inductors up X L  1.5 k
•The sum is shorter than either separately
1.4 k
?
60 Hz
170 V
2.0 F
L= 4.0 H
Concept Question
P 
R
2

V
rms
Z2
Z  R   X L  XC 
2
L
C
R
f
2
Vmax
How will XL and XC compare at the frequency where the maximum power
is delivered to the resistor?
A) XL > XC B) XL < XC
C) XL = XC
D) Insufficient information
•Resonance happens when XL = XC.
•This makes Z the smallest
•It happens only at one frequency
•Same frequency we got for LC circuit
1
XC 
C
1
 0 L
0 C
X L  L
0 
1
LC
Concept Question
•In the example we just did, we found
only some frequencies get through
1 RC    R L
What happens if this is impossible to meet, because
1/RC > R/L?
A) The inequality gets reversed, R/L <  < 1/RC
B) Pretty much everything gets blocked
C) Only a very narrow frequency range gets through
L
C
R
f
Vmax = 5 V
Concept Question
When the voltage shown in
blue was passed through two
components in series, the
current shown in red resulted.
What two components might
they be?
A) Capacitor and Inductor
B) Inductor and Resistor
C) Capacitor and Resistor
 X L  XC 
  tan 

R


1
Voltage
Current
•The phase shift represents how the timing of
the current compares to the timing of the
voltage
•When it is positive, the current lags the
voltage
•It rises/falls/peaks later
•When it is negative, the current leads the
voltage
•It rises/falls/peaks earlier
10, 000 V 120 V

5, 000
N2
N2 
120 V
 60
2V
V2 = 120 V
N1 =5000
V1 = 10 kV
A transformer has 10,000 V AC going
into it, and it is supposed to produce
120 V AC, suitable for household use.
If the primary winding has 5,000 turns,
how many should the secondary have?
A) 120
B) 240
C) 60
D) None of the above
N2 =?
Concept Question
2V
E1 E2

N1 N 2
Concept Question
A wave has an electric field given by E  ˆjE0 sin  kz  t 
What does the magnetic field look like?
E0
ˆ
A) B  i sin  kz  t 
c
E0
ˆ
C) B   i sin  kz  t 
c
E0  cB0
E0
ˆ
B) B  k sin  kz  t 
c
E0
ˆ
B


k
sin  kz  t 
D)
c
•The magnitude of the wave is B0 = E0 / c
•The wave is traveling in the z-direction, because of sin(kz - t).
•The wave must be perpendicular to the E-field, so perpendicular to j
•The wave must be perpendicular to direction of motion, to k
•It must be in either +i direction or –i direction
•If in +i direction, then E  B would be in direction j  i = - k, wrong
•So it had better be in the –i direction
Concept Question
 Increasing
f Increasing
cf
Radio Waves
Microwaves
Infrared
Visible
Ultraviolet
X-rays
Gamma Rays
Red
Orange
Yellow
Green
Blue
Violet
Which of the
following waves
has the highest
speed in vacuum?
A) Infrared
B) Orange
C) Green
D) It’s a tie
E) Not enough info
Cross-Section
•To calculate the power falling on an object, all that matters is
the light that hits it
•Example, a rectangle parallel to the light feels no pressure
•Ask yourself: what area does the light see?
•This is called the cross section
P= S 
F = P
P S c
If light of intensity S hits an absorbing sphere
of radius a, what is the force on that sphere?
A) a2S/c B) 2a2S/c C) 4a2S/c
•As viewed from any side, a sphere looks
like a circle of radius a
•The cross section for a sphere, then, is a2
Equations for Test 3
Faraday’s Law
dB
E
dt
Impedance:
Power and
Pressure
P  I V
Inductors:
dI
E  L
dt
I max  Vmax Z
2
P  RI rms
P= S 
F = P
F = 2P
Transformers:
E1 E2

N1 N 2
I1V1  I 2 V2
Units:
Frequency,
Wavelength
1 H  1 V  s/A
  2 f
Speed of Light
f 1 T
f   k  c
k  2
c  3.00 108 m/s
End of material
for Test 3