ID: A
UCSD Physics 2B
Answer Section
Unit Exam 5A
AC Circuits
MULTIPLE CHOICE
1. ANS: A
Calculate and compare the AC reactance for each of the three load components:
X R = R = 470Ω
XC =
1
ωC
1
= Ê
= 546Ω
5
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
2. ANS: C
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
3. ANS: C
Flux through a solenoid is the field times cross-section area times the number of turns:
ˆ˜ Ê 2 ˆ μ0 N 2 π R 2 I
ÊÁ
N
˜˜ ÁÁ π R ˜˜ =
Á
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
ˆ˜
ÊÁ
−6
Á 1.26 × 10 H / m˜ (3.14)
¯
Ë
Answer is rounded to 3 sig figs
TOP: SELF-INDUCTANCE
3
ID: A
4. ANS: B
Power is conserved, so P = V house I house = V line I line or
I line
ˆ
ÊÁ V
ÊÁ
ÁÁ house ˜˜˜
˜˜ = (52.3A) ÁÁÁ 117V
Á
= I house ÁÁ
Á 550,000V
Á V line ˜˜
Ë
¯
Ë
ˆ˜
˜˜ = 1.11 × 10 −2 A = 11.1mA
˜˜
¯
TOP: POWER TRANSMISSION
5. ANS: C
This circuit can decay, but a circuit must have both an inductor L and capacitor C in order to
oscillate.
TOP: RC CIRCUIT
6. 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
7. ANS: A
Apply the transformer equation - the ratio of voltages = ratio of turns:
ÊÁ 18,000 ˆ˜
VP
Ns
NP
˜˜ = 50,000V
=
⇒ VS = VP
= (7500V ) ÁÁÁÁ
˜
ÁË 2700 ˜˜¯
VS
NS
Np
TOP: TRANSFORMER
8. 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
9. ANS: D
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
Á 2π f ˜ˆ NB (2) (3.14) (60Hz) (300) (0.025T )
Ë
¯
TOP: GENERATOR
2
ID: A
10. ANS: D
1
V rms =
V peak = (0.707) (736V ) = 520V
2
TOP: RMS
11. ANS: A
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
Ô
˜ˆ˜
rad
˜˜ (1.54s) Ô˝Ô
sinÌÔ ÁÁÁ 377
˜
s ¯
ÔÔÓ Ë
ÔÔ˛
Remember to set your calculator to radian mode, or convert to degrees!
TOP: AC VOLTAGE
12. ANS: D
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: A
cycle ˆ˜˜˜
rad ÊÁÁÁ
rad
ω = 2π f = 2 (3.14)
ÁÁ 440
˜ = 2.76 × 10 3
sec ˜˜
cycle Á
sec
Ë
¯
TOP: ANGULAR FREQUENCY
14. ANS: B
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
15. ANS: A
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
3