Name: ________________________________________ Code # _________
UCSD Physics 2B
Practice Exam 5
Summer Session 2008
ID: X
Inductance & AC Circuits
READ ME !!!
Helpful Hints:
Do this exam a few times... Once without the answers,
Once after seeing the solutions & Once again with your formula sheet
This is a good time to add formulas/constants to your sheet.
Remember, this is a timed exam... work swiftly so you have time to check your answers.
1. How much does the maximum emf produced by a generator (a rotating coil) change if the period of
rotation is doubled?
A.
B.
C.
D.
E.
It is the same
It is doubled
It is quadrupled
It is halved
It is quartered
2. A 200-turn coil is rotating in a magnetic field of magnitude B = 0.250T at a frequency of f = 60.0 Hz.
The area of the coil A = 5.00 × 10−2 m2 . What is the maximum emf in the coil?
A.
B.
C.
D.
150 V
450 V
942 V
240 V
3. The rms value of house voltage in much of Europe is 220 V. What is the peak voltage?
A.
B.
C.
D.
E.
440 V
310 V
220 V
160 V
120 V
4. If you plug a 1200W heater and a 600W toaster into 110V electrical outlets, what total rms current
would these two appliances draw if the wiring is in parallel? (it always is!)
A.
B.
C.
D.
E.
5.45 A
10.9 A
11.6 A
16.4 A
23.1 A
1
Name: ________________________
ID: A
5. If you double the rms voltage in an AC circuit, the peak voltage
A. increases by a factor of 2
B. decreases by a factor of 2
C. increases by a factor of 2
D. decreases by a factor of
E. does not change
2
6. Your house is wired for 220 V and you want to draw no more than 17 kW. What circuit breaker do
you need in order not to shut off the power? Note: If circuit breakers are rated in 10A increments,
e.g., 10A, 20A, 30A, etc., you must choose one just above the value you calculate.
A.
B.
C.
D.
E.
20 A
30 A
40 A
80 A
150 A
7. A country house draws 52.3 A with all its appliances running. A single high-voltage line delivers
power at 10,500 V to a transformer on the pole which steps the voltage down to 117 V for the house.
What is the current in the high-voltage power line?
Hint: voltage and current may change, but power remains fixed.
A.
B.
C.
D.
E.
52.3 A
583 mA
11.1 mA
4.69 kA
none of these
8. The generators at a power plant operate at 1250 V which is stepped up to 25,000 V for transmission
. If the primary of the transformer has 500 windings, how many windings must the secondary
have?
A. 25
B. 10,000
C. 62,500
2
D. 250
Name: ________________________
ID: A
9. In the circuit below, C = 0.13 mF and L = 270 mH. What is the natural oscillation frequency of the
circuit?
A.
B.
C.
D.
E.
2.85 x 10 4 rad/sec
1.69 x 10 2 rad/sec
5.93 x 10-3 rad/sec
3.51 x 10-5 rad/sec
none of these
10. In the circuit below, L = 55 mH and R = 160 Ohms. What is the natural oscillation frequency of
the circuit?
A.
B.
C.
D.
E.
8.80 Hz
1.14 Hz
2.91 kHz
0.344 mHz
zero
3
Name: ________________________
ID: A
11. The LR circuit shown below has a resistance R = 25 Ω , an inductance L = 5.4 mH , and a battery of
emf V = 9.0 V. After the switch S is closed, how much energy is stored in the inductance of this
circuit when a steady current is achieved?
A.
B.
C.
D.
E.
zero
0.15 J
0.35 mJ
0.70 mJ
0.97 mJ
12. What is the time constant of an RL circuit with resistance R = 25Ω and inductance L = 5.4 mH?
A.
B.
C.
D.
E.
7.4 s
4.6 s
0.14 s
0.22 ms
1.5 ms
13. In the circuit below R = 1.50kΩ , L = 7.20mH and V = 75.0V . From the time switch S is closed, how
long does it take the current through the inductor to decay to one-third its maximum value?
A. 10.8ms
B. 96.4 μ s
C. 5.40 s
4
D. 5.27 μ s
Name: ________________________
ID: A
14. In other parts of the world, not only is the house line voltage different, but the line frequency is as
well. If the frequency in England f = 50 Hz, what is the angular frequency ω in rad/s?
A. 3.19 × 10 −3
B. 314
rad
s
rad
s
rad
s
rad
D. 0.216
s
C. 7.96
15. A solenoid with rectangular cross-section has N = 3250 turns, length l = 52.0 cm and area A = 110 cm2 .
If it is carrying a current I = 53.0 A, what is the energy density inside?
A. 1.66
J
m3
B. 3.09 × 10 −4
C. 6.91 × 104
J
m3
D. 13.8
J
m3
J
m3
16. Three circuits shown below each have an identical AC voltage source driving a single load
component: a resistor R = 187 Ω , an inductor L = 33.3 μ H and a capacitor C = 220 pF . If the driving
source has frequency ω = 57.2 × 105 rad/s, which component draws the least current?
A.
B.
C.
D.
Resistor
Capacitor
Inductor
Insufficient Information
17. An AC line source has voltage given by V (t) = V 0 sin (ω t) where V 0 = 256 V and ω = 27.3 rad / s. What
is the instantaneous voltage on the line at time t = 12.6 s ?
A.
B.
C.
D.
11.7 V
88.1 kV
-70.6 V
-256 V
5
Name: ________________________
ID: A
18. A town center requires P = 1.16 MW of power during peak usage hours and a small generating plant supplies
that power via a high-voltage transmission line at V = 37.5 kV . How much power is lost on each leg between
towers if the resistance R = 3.89 × 10 −1 Ω ?
A.
B.
C.
D.
30.93 W
372 W
1.20 W
4.35 x 1010 W
19. A coil with a self-inductance of 6.5 H carries a current that is changing at the rate of 50 A/s. What
is the induced emf in the coil?
A.
B.
C.
D.
E.
0.13 V
7.7 V
32 V
65 V
0.33 kV
20. If you want to build a filter to prevent high frequencies passing from an AC voltage source to a load, which
of the following components would you insert in series?
A.
B.
C.
D.
Resistor
Capacitor
Inductor
None of these
6
ID: A
UCSD Physics 2B
Answer Section
Practice Exam 5
Inductance & AC Circuits
MULTIPLE CHOICE
1. ANS: D
dΦ
d
emf =
= NBA sin ω t = NBAω cos ω t
dt
dt
Since the max emf is proportional to frequency, it is inversely proportional to the period.
TOP: GENERATOR
2. ANS: C
First write Faraday’s Law and make sure to express the angular velocity properly:
dΦ
d
d
= NBA cos (ω t) = NBA cos ÊÁË 2π f t ˆ˜¯ = NBA ÊÁË −2π f ˆ˜¯ sin ÊÁË 2π f t ˆ˜¯
dt
dt
dt
Now use the fact that the extremes of the sine function are ±1. Then
ÊÁ
1 ˆ˜
Ê
ˆ
maxÁÊË emf ˜ˆ¯ = NBA ÁÊË 2π f ˜ˆ¯ = (200) (0.25T ) ÁÁ 5.00 × 10 −2 m2 ˜˜ (6.28) ÁÁÁÁ 60.0 ˜˜˜˜ = 942V
Ë
¯
s¯
Ë
emf =
TOP: GENERATOR
3. ANS: B
V peak = 2V rms = 2 (220V ) = 311V
TOP: RMS
4. ANS: D
Total power used is 1200 W + 600 W = 1800 W.
P = IV ⇒ I =
P 1800W
=
= 16.4A
V
110V
TOP: RMS POWER
5. ANS: A
V peak = 2 V rms which is directly proportional by the factor
2
TOP: RMS
6. ANS: D
Use the power equation to find the current drawn at a particular voltage
P = IV ⇒ I =
P 17 × 10 3 W
=
= 77.3A ⇒ 80A
V
220V
TOP: CIRCUIT BREAKER
1
ID: A
7. ANS: B
Since Energy is conserved, Power must be also, so P = V house I house = V line I line
ˆ
ÊÁ V
Á house ˜˜˜
ÁÊ
˜˜ = (52.3A) ÁÁÁ 117V
I line = I house ÁÁÁÁ
˜
Á 10,500V
Á V line ˜
Ë
¯
Ë
˜ˆ˜
˜˜ = 0.583A
˜
¯
TOP: POWER TRANSMISSION
8. ANS: B
Apply the transformer equation
ÊÁ 25,000V
NP
VP
VS
=
⇒ NS = NP
= (500) ÁÁÁÁ
ÁË 1,250V
VS
NS
VP
ˆ˜
˜˜ = 10,000
˜˜
˜¯
TOP: TRANSFORMER
9. ANS: B
1
1
ω=
=
= 169 rad/sec
LC
ÁÊÁ 270 × 10 −3 H ˜ˆ˜ ÁÊÁ 0.130 × 10 −3 F ˜ˆ˜
Ë
¯Ë
¯
TOP: LC CIRCUIT
10. ANS: E
A circuit must have both an inductor L and capacitor C in order to oscillate.
TOP: RC CIRCUIT
11. ANS: C
When steady current is achieved, the magnetic field in the inductor is no longer changing, so the
inductor becomes "invisible" to the current flow. Hence, the current I = V/R and the stored energy is
thus
U=
2
1 2 1 ÊÁÁÁ V ˆ˜˜˜
1Ê
ˆ ÊÁ 9.0V
L I = L ÁÁ ˜˜ = ÁÁ 5.4 × 10 −3 H ˜˜ ÁÁÁÁ
2
2 ËR¯
2Ë
¯ Ë 25Ω
ˆ˜ 2
˜˜ = 3.5 × 10 −4 J
˜˜
¯
Another way to see this is to look at the Inductor Equation (AC Ohm’s Law) V = −L
dI
.
dt
When current finally becomes steady, the voltage drop across the inductor is zero, then we use only
the resistor to complete & analyze the circuit for current.
TOP: INDUCTOR ENERGY + CONCEPT
12. ANS: D
The time constant (see text p.840) for and RL circuit is
τ=
L 5.4 × 10 −3 H
=
= 2.2 × 10 −4 s = 0.22ms
R
25Ω
TOP: LR CIRCUIT
2
ID: A
13. ANS: D
The RL circuit time constant is τ RL = L R so the discharching or “unloading” formula is
ÊÁ R ˆ˜
I (t) = I 0 exp ÁÁÁ − t ˜˜˜˜
ÁË L ¯
ÊÁ R ˆ˜ 1
I (t)
1
= exp ÁÁÁÁ − t ˜˜˜˜ = . Take natural log of both sides
When current reaches initial value,
3
I0
Ë L ¯ 3
ÈÍ
˘
ÊÁ R ˆ˜ ˙˙˙
ÊÁ 1 ˆ˜
ÍÍ
R
ln ÍÍÍ exp ÁÁÁÁ − t ˜˜˜˜ ˙˙˙ = ln ÁÁÁÁ ˜˜˜˜ ⇒
t = ln 3
L
ÍÍÎ
Ë L ¯ ˙˙˚
Ë3¯
⇒ t=
7.20 × 10 −3 H
L
ln3 =
ln3 = 5.27 μ s
R
1.50 × 10 3 Ω
TOP: LR CIRCUIT
14. ANS: B
ÍÈÍ
ÍÎ
Conversion problem: ω = 2π f = ÍÍÍÍ 2(3.14)
˘
rad
rad ˙˙˙˙ ÊÁÁÁ cycle ˆ˜˜˜
˜˜ = 314
ÁÁ 50
˙
˙
s ˜
s
cycle ˙˚ Á
¯
Ë
TOP: ANGULAR FREQUNCY
15. ANS: C
N
B = μ0 n I = μ0 I
l
μ0 I 2 N 2 μ0 ÊÁÁ IN ˆ˜˜ 2
B2
Á ˜
⇒ uL =
=
=
2μ0
2 ÁÁË l ˜˜¯
2l 2
È
˘2
ÁÊÁ 1.26 × 10 −6 H / m˜ˆ˜ ÍÍÍÍ (53.0A) ÁÊÁ 3.25 × 10 3 ˜ˆ˜ ˙˙˙˙
J
Ë
¯ ˙˙˙
Ë
¯ ÍÍÍ
ÍÍ
˙˙ = 6.91 × 10 4 3
=
Ê
ˆ
2˜
ÍÍ
˙
2
Á
m
Á 0.520m ˜
ÍÍ
˙˙˙
Ë
¯
ÍÎÍ
˙˚˙
ALTERNATE SOLUTION (if the inductance is given)
UL =
1 2
LI
2
uL =
UL
LI 2
=
VOLUME 2(l × A)
TOP: ENERGY DENSITY SOLENOID
3
ID: A
16. ANS: B
Look at the AC reactance for each circuit since, for AC circuits, Ohm’s law becomes
I=
V
X
X R = R = 187Ω
XC =
1
1
= Ê
= 795Ω
5
ωC ÁÁ 57.2 × 10 rad / s ˆ˜˜ ÊÁÁ 220 × 10−12 F ˆ˜˜
¯Ë
¯
Ë
Ê
ˆÊ
ˆ
X C = ωL = ÁÁ 57.2 × 10 5 rad / s ˜˜ ÁÁ 33.3 × 10 −6 F ˜˜ = 190Ω
Ë
¯Ë
¯
The capacitor circuit has the highest reactance, and therefore draws the least current
TOP: REACTANCE
17. ANS: D
Since we’re given the voltage function, just plug & chug:
ÔÏ
Ô¸
V (t) = V 0 sin (ω t) = (256V ) sinÌÔ (27.3rad / s) (12.6s) ˝Ô = −256V
Ó
˛
REMEMBER to set your calculator to RADIAN mode for these problems!
TOP: AC VOLTAGE
18. ANS: B
First find the current through the line given the power and voltage
P 1.16 × 10 6 W
P = IV ⇒ I =
=
= 30.93A
V
37.5 × 10 3 V
Then use the current form of the power equation
ˆ
2Ê
P = I 2 R = (30.93A) ÁÁ 3.89 × 10 −1 Ω ˜˜ = 372W
Ë
¯
V2
NOTE: A very common mistake is to use either P = IV or P =
. But this only works when you know the
R
voltage across the circuit element in question (in this case one leg). The voltage we’re given is like the
“battery voltage” - not the same thing at all!
TOP: POWER TRANSMISSION LOSS
19. ANS: E
ÊÁ
dI
A ˆ˜
emf = −L
= − (6.5H ) ÁÁÁÁ −50 ˜˜˜˜ = 325V = 0.33 kV
dt
s¯
Ë
TOP: SELF-INDUCTANCE
20. ANS: C
R is out since the voltage drop across it is independent of frequency.
L has AC reactance X L = ωL which goes up with increasing frequency, that is, it's AC "resistance" goes up
and it therefore passes low frequencies and attenuates high ones.
C has AC reactance X C = 1 / ωC which goes down with increasing frequency, that is, it's AC "resistance"
goes down and it therefore favors passing high frequencies and attenuates low ones.
TOP: FILTER
4