ID: A
UCSD Physics 2B
Answer Section
Unit Exam 3A
Current, Resistance, Circuits
MULTIPLE CHOICE
1. ANS: B
Each differential area element dA of a circle (cross section) is a ring of thickness dr and length
(circumference) 2π r. The area of this ring is the product, dA = 2π r dr
∫
I = J • dA =
= 2π K
∫
R
0
J (r)2π r dr =
∫
R
0
ÊÁ 2 ˆ˜
Á Kr ˜ 2π r dr = 2π K
Ë
¯
∫
R
r 3 dr
0
ÊÁ
R4 1
A ˆ˜ Ê
ˆ4
= (3.14) ÁÁÁÁ 122 4 ˜˜˜˜ ÁÁ 166 × 10 −2 m˜˜ = 1.45 kA
4
2
¯
m ¯Ë
Ë
TOP: CURRENT DENSITY + INTEGRATION
2. ANS: D
This is a series combination. When all resistors are identical
R SER = NR = 3 × 33Ω = 99Ω
TOP: SERIES
3. ANS: D
This is basically a unit conversion problem if you think about it:
energy
⇒ energy = power × time
time
ÊÁ
ÊÁ 60 s ˆ˜
J ˆ˜ ÊÁ
C ˆ˜
˜˜
E = P × t = (V × I ) × t = ÁÁÁÁ 12.0 ˜˜˜˜ ÁÁÁÁ 5.00 ˜˜˜˜ (3.00 min) ÁÁÁÁ
˜˜
C
s
min
Ë
Ë
¯Ë
¯
¯
Power =
= 1.08 × 10 4 J = 10.8kJ
TOP: ENERGY
4. ANS: B
Important to keep track of direction here. Since the protons and electrons are moving in the same
direction, you must subtract the electron contribution to the flux since the charge is negative.
Calculate the number of particles times the charge/particle and then divide by time
+
+
−
n p e + − n e e + ÊÁË n p − n e ˆ˜¯ e
Q n p e + n ee
I=
=
=
=
t
t
t
t
ÊÁ
ˆÊ
ˆ
Á 6.5 × 10 17 − 3.5 × 10 17 ˜˜ ÁÁ 1.6 × 10−19 C ˜˜
Ë
¯Ë
¯
=
= 0.048A = 48mA
1s
TOP: CURRENT
5
ID: A
5. ANS: B
Each parallel pair has resistance R/2, so the total resistance of the circuit is
R TOTAL = 3R + 3(R / 2) = 4.5R = 225Ω . Since the resistance between A and B (going around clockwise)
is R AB = 2R + 2 (R / 2) = 3R, the ratio of the resistances is
Hence the voltage V AB =
3R
2
= .
4.5R 3
2
(180V ) = 120V
3
Note: once you see that there are three identical sub-networks in series, the actual resistance of each
element is immaterial. Just the ratio matters. You must careful: if you go the other way around, you
must include the battery voltage.
6. ANS: A
Two ways to do this:
1. This is a compound circuit, so we must find the current around the entire loop, and therefore the
equivalent resistance of the entire circuit. First, reduce the parallel leg to its equivalent. Since
R 50Ω
=
= 25Ω . Now this is in series with R3, so
N
2
= R 12 + R 3 = 25.0Ω + 50.0Ω = 75.0Ω . The total circuit current is thus
V
120V
I=
=
= 1.60A
R Series 75.0Ω
R 1 = R 2 , R par =
R Circuit
Since the entire circuit current must pass through R 3
P 3 = I 2 R 3 = (1.60A) (50.0Ω ) = 128W
2
2. Think “voltage divider”. Since R 3 = 2 R 12 , V3 =
Now apply the power formula locally to R3
V 23
2
(80V )
P3 =
=
= 128W
R3
50Ω
TOP: NET + POWER
2
2
2
V = (120V ) = 80V
3
3
ID: A
7. ANS: B
Let's think of this device as being made up of many coaxial shells (layers) of length L , "width" w = 2π r and
thickness dr. Since the current flows directly outward, each layer can be treated as a parallel plate resistor
d
whose resistance is R = ρ . The differential element for this layer is
A
ÊÁ d ˆ˜
dr
dr
dr
dR = d ÁÁÁÁ ρ ˜˜˜˜ = ρ
=ρ
=ρ
π
rL
A
A
(
r
)
Lw
2
Ë
¯
Since each layer acts as one element in a series network (why?), the total resistance is the sum of these layers,
that is, an integral:
∫
R = dR =
ÊÁ b ˆ˜
b
ρ
ρ
dr
=
ln ÁÁÁÁ ˜˜˜˜
∫
2π L a r
2π L Ë a ¯
Plugging in the numers, we get
ÊÁ b ˆ˜
ÊÁ 2.33 ˆ˜
ρ
22Ω ⋅ m
˜˜ = 125 mΩ
R=
ln ÁÁÁÁ ˜˜˜˜ =
lnÁÁÁ
2π L Ë a ¯ 2π (1.36m) ÁË 2.22 ˜˜¯
Note that the diameters a and b are so close that you can get an approximate answer by treating the whole
device as a rolled-up plate resistor with d = b − a and A = 2π aL. Then
ÊÁ 2.33
ˆ˜
ÁÁ
˜
ÁÁ 2.22 − 1 ˜˜˜
Ë
d
b−a
b/a−1
¯
R≈ρ =ρ
=ρ
= (22Ω ⋅ m)
= 128mΩ
A
2π aL
2π L
2π (1.36m)
TOP: RESISTIVITY + INTEGRATION
8. ANS: A
Calculate the "time constant" and plug it into the "discharging" formula -
Ê
ˆÊ
ˆ
τ = RC = ÁÁ 250 × 10 3 Ω ˜˜ ÁÁ 20.0 × 10 −6 F ˜˜ = 5.00s
Ë
¯Ë
¯
ÊÁ t ˆ˜
ÊÁ 3.00s ˆ˜
˜˜ = 82.3V
V (t) = V 0 exp ÁÁÁ − ˜˜˜˜ = (150V ) exp ÁÁÁÁ −
˜˜
ÁË τ ¯
5.00s
¯
Ë
TOP: RC TIME
9. ANS: D
Apply Ohm’s Law: R =
V
220V
=
= 25.1Ω
I
8.75A
TOP: OHM'S LAW
10. ANS: C
The resistors are in parallel, so they’re at the same potential (voltage) and therefore independent of
each other. That means you apply Ohm’s Law just to resistor R2
I2 =
V2
V
117V
=
=
= 6.80A
R 2 R 2 17.2Ω
TOP: POWER
3
ID: A
11. ANS: B
Apply the basic formula for a resistor in cartesian (x,y,z) coordinates:
R=ρ
L
s
=ρ 2
A
s
Ê
ˆ
⇒ ρ = Rs = (1330Ω ) ÁÁ 12.7 × 10 −2 m˜˜ = 169 Ω • m
Ë
¯
TOP: RESISITIVITY
12. ANS: B
Begin with the definition of current density and apply the formula for drift velocity:
I = JA = ÊÁË nqv d ˆ˜¯ A
To find q for each particle, add up the number of charges. Neutrons have no charge, but the one
proton has charge e+, Now invert the formula & solve for drift velocity
vd =
I
1.1 × 10 −4 A
m
= Ê
= 7.29 × 10 7
ˆ
ˆ
ˆ
Ê
Ê
12
−3
−19
−8
2
nqA ÁÁ 41 × 10 m ˜˜ ÁÁ 1 × 1.60 × 10 C ˜˜ ÁÁ 23 × 10 m ˜˜
s
¯Ë
¯
¯Ë
Ë
TOP: DRIFT VELOCITY
13. ANS: B
If you remembered that Q/L is charge density and speed is L/t, then
ΔQ Q L
Ê
ˆ
I=
= × = λ × v = ÁÁ 2.88 × 10 −3 C / m˜˜ (26.7m / s) = 76.9 mA
¯
Δt
L t
Ë
Even if you didn't, the only combination of the given quantities resulting in the correct units for current is the
product λ × v = (linear charge density) times ( velocity)
TOP: CURRENT CONCEPT
14. ANS: A
Resistance is directly proportional to length. Write an equation for each wire:
RA = ρ
LA
LB
and R B = ρ
A
A
Now divide the two equations to get the ratio formula:
LA
RA
LA
A
=
=
RB
LB
LB
ρ
A
ρ
⇒ R A = RB
LA
LB
Since wire A is half as long as wire B,
LA =
1
L
2 B
⇒
ÊÁ 1 ˆ˜ R B
LA 1
LA
=
= R B ÁÁÁÁ ˜˜˜˜ =
so that R A = R B
LB
LB
2
2
Ë2¯
TOP: RESISTIVITY
15. ANS: D
Definition of average current I =
Q 125 C
=
= 7.58 A
t
16.5 s
TOP: CURRENT
4
ID: A
16. ANS: A
The middle lamp is "shorted out" and therefore has no current through it since the potential difference across
it is zero. At the same time, the total resistance of the circuit is lowered so more current flows through the
remaining bulbs. You can also see this by recognizing the circuit as a voltage divider. When the middle bulb
is shorted, the battery voltage is divided between only two remaining bulbs, so they are brighter than when
the voltage was divided by 3.
TOP: DEMO
17. ANS: D
There’s only one way to do this without making yourself crazy. Make a string of coversion factors
necessary to get the units right. One bulb = 30 Watts, etc etc. You have to know that June has 30
days (“30 days has September... and some months I don’t remember”)
N=
ÊÁ bulb ˆ˜ ÊÁ 6800cents ˆ˜ ÊÁ 1000W • hr ˆ˜ ÁÊÁ day ˜ˆ˜ ÊÁ month ˆ˜
ÁÁ
˜Á
˜Á
˜
Á
˜
ÁÁ 30W ˜˜˜ ÁÁÁ month ˜˜˜ ÁÁÁ 12cents ˜˜˜ ÁÁÁÁ 24hr ˜˜˜˜ ÁÁÁ 30days ˜˜˜ = 26.2 bulbs
Ë
¯Ë
¯Ë
¯Ë
¯
¯Ë
TOP: POWER & ENERGY
5