ID: B
UCSD Physics 2B
Answer Section
Unit Exam 5B
AC Circuits
MULTIPLE CHOICE
1. ANS: D
Flux through a solenoid is the field times cross-section area times the number of turns:
2
2
˜ˆ˜ ÁÊ 2 ˜ˆ μ0 N π R I
ÁÊÁ
N
˜˜ Á π R ˜ =
Φ = NBA = N ÁÁÁ μ0 I
˜
length ¯ Ë
¯
length
Ë
2
2
Φ μ0 N π R
L length
=
⇒ N=
Self-inductance L =
I
length
R 2 μ0 π
N=
24.0 H
ÊÁ
ˆ2
Á 10.0 × 10−2 m˜˜
Ë
¯
25.0 × 10 −2 m
= 1.23 × 10 4
ÁÊÁ 1.26 × 10 −6 H / m˜ˆ˜ (3.14)
¯
Ë
Answer is rounded to 3 sig figs
TOP: SELF-INDUCTANCE
2. ANS: C
Calculate and compare the AC reactance for each of the three load components:
X R = R = 470Ω
XC =
1
1
= Ê
= 546Ω
5
ωC ÁÁ 57.2 × 10 rad / s ˆ˜˜ ÊÁÁ 320 × 10−12 F ˆ˜˜
¯Ë
¯
Ë
ˆÊ
ˆ
Ê
X L = ωL = ÁÁ 57.2 × 10 5 rad / s ˜˜ ÁÁ 127 × 10 −6 H ˜˜ = 726Ω
¯Ë
¯
Ë
The resistor has the lowest reactance, and therefore (by Ohm’s Law) draws the most current.
TOP: REACTANCE
3. ANS: A
dΦ
d
emf = −
= − NBA × cos (ω t) = NBA × ω sin (ω t)
dt
dt
V peak = max ÊÁË emf ˆ˜¯ = NBAω = NBA ÊÁË 2π f ˆ˜¯
V peak
(370V )
= 0.131m2
⇒ A= Ê
ˆ
Á 2π f ˜ NB (2) (3.14) (60Hz) (300) (0.025T )
Ë
¯
TOP: GENERATOR
1
ID: B
4. ANS: B
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
5. ANS: D
ω = 2π f = 2 (3.14)
rad
cycle
ˆ˜
ÁÊÁ
ÁÁ 440 cycle ˜˜˜ = 2.76 × 10 3 rad
ÁÁ
sec ˜˜
sec
¯
Ë
TOP: ANGULAR FREQUENCY
6. ANS: B
1
ω2 =
LC
⇒ L=
1
1
= È
˘2
Í
ω C ÍÍ ÊÁ
ˆÊ
ˆ ˙˙
ÍÍ ÁÁÁ 7500 cycles ˜˜˜˜ ÁÁÁ 2π rad ˜˜˜ ˙˙˙ ÊÁÁ 13.0 × 10 −3 F ˆ˜˜
ÍÍ ÁÁ
¯
sec ˜˜ ÁÁË cycle ˜˜¯ ˙˙˙˙ Ë
ÍÍÎ Ë
¯
˚
2
= 3.47 × 10 −8 H = 34.7 × 10−9 H = 34.7nH
TOP: LC CIRCUIT
7. ANS: C
Apply the transformer equation - the ratio of voltages = ratio of turns:
ÊÁ 18,000 ˆ˜
NP
VP
Ns
˜˜ = 50,000V
=
⇒ VS = VP
= (7500V ) ÁÁÁÁ
˜
VS
NS
Np
ÁË 2700 ˜˜¯
TOP: TRANSFORMER
8. ANS: D
As t → ∞, the field in the inductor becomes saturated and there is no "back emf". Hence, the inductor
acts like a short circuit and only R1 is left. We then have
I=
V
90V
=
= 1.5 × 10 −1 A = 150mA
R 1 600Ω
TOP: RC CIRCUIT CONCEPT
9. ANS: C
The time constant (see text p.840) for an RL circuit is
τ=
L
R
⇒ L = Rτ = (250Ω ) (3.3ms) = 825mH
TOP: LR CIRCUIT
2
ID: B
10. ANS: A
1
V rms =
V peak = (0.707) (736V ) = 520V
2
TOP: RMS
11. ANS: D
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 Ohm's Law to calculate the voltage drop
Ê
ˆ
V = IR = (30.93A) ÁÁ 3.89 × 10 −2 Ω ˜˜ = 1.20V
Ë
¯
2
V
Beware of using P =
since the tranmission voltage is not the voltage drop across the leg.
R
TOP: POWER TRANSMISSION LOSS
12. ANS: B
First solve for the current, then choose the next lowest available rating
P = IV ⇒ I =
P 6.4 × 103 W
=
= 55.7A ⇒ 55A
V
115V
TOP: CIRCUIT BREAKER
13. ANS: C
Power is conserved, so P = V house I house = V line I line or
ÁÊÁ V house ˆ˜˜
Ê
˜˜ = (52.3A) ÁÁÁ 117V
I line = I house ÁÁÁÁ
˜
ÁÁ 550,000V
Á V line ˜˜
Ë
¯
Ë
ˆ˜
˜˜ = 1.11 × 10 −2 A = 11.1mA
˜˜
¯
TOP: POWER TRANSMISSION
14. ANS: D
This circuit can decay, but a circuit must have both an inductor L and capacitor C in order to
oscillate.
TOP: RC CIRCUIT
15. ANS: C
Invert the given equation for instantaneous voltage to solve for the peak voltage:
V (t) = V P sin (ω t)
⇒ VP =
V (t)
=
sin (ω t)
122V
ÏÔ Ê
¸Ô = +212V
ÔÔ ÁÁ
ÔÔ
rad ˆ˜˜˜
Á
sinÌÔ ÁÁ 377
˜˜ (1.54s) ˝Ô
s
ÔÔÓ Ë
ÔÔ˛
¯
Remember to set your calculator to radian mode, or convert to degrees!
TOP: AC VOLTAGE
3