Answers to Quick Quizzes
857
Here the first symbol e represents Euler’s number, the base
of natural logarithms. The second e is the charge on the
electron. The k
B
stands for Boltzmann’s constant, and T is
the absolute temperature. Set up a spreadsheet to
calculate I and R " !V/I for !V " 0.400 V to 0.600 V in
increments of 0.005 V. Assume I
0
"
1.00 nA. Plot R versus
!
V for T " 280 K, 300 K, and 320 K.
75.
Review problem. A parallel-plate capacitor consists of
square plates of edge length ! that are separated by a
distance d, where d 22 !. A potential difference !V is
maintained between the plates. A material of dielectric
constant 4 fills half of the space between the plates. The
dielectric slab is now withdrawn from the capacitor, as
shown in Figure P27.75. (a) Find the capacitance when
the left edge of the dielectric is at a distance x from the
center of the capacitor. (b) If the dielectric is removed at a
constant speed v, what is the current in the circuit as the
dielectric is being withdrawn?
the junction. This is indicative of scalar addition. Even
though we can assign a direction to a current, it is not a
vector. This suggests a deeper meaning for vectors
besides that of a quantity with magnitude and direction.
27.3 (a). The current in each section of the wire is the same
even though the wire constricts. As the cross-sectional
area A decreases, the drift velocity must increase in order
for the constant current to be maintained, in accordance
with Equation 27.4. As A decreases, Equation 27.11 tells
us that R increases.
27.4 (b). The doubling of the radius causes the area A to be
four times as large, so Equation 27.11 tells us that the
resistance decreases.
27.5 (b). The slope of the tangent to the graph line at a point
is the reciprocal of the resistance at that point. Because
the slope is increasing, the resistance is decreasing.
27.6 (a). When the filament is at room temperature, its
resistance is low, and hence the current is relatively large.
As the filament warms up, its resistance increases, and
the current decreases. Older lightbulbs often fail just as
they are turned on because this large initial current
“spike” produces rapid temperature increase and
mechanical stress on the filament, causing it to break.
27.7 (c). Because the potential difference !V is the same
across the two bulbs and because the power delivered to
a conductor is " " I !V, the 60-W bulb, with its higher
power rating, must carry the greater current. The 30-W
bulb has the higher resistance because it draws less
current at the same potential difference.
27.8 I
a
"
I
b
6
I
c
"
I
d
6
I
e
"
I
f
. The current I
a
leaves the
positive terminal of the battery and then splits to flow
through the two bulbs; thus, I
a
"
I
c
,
I
e
. From Quick
Quiz 27.7, we know that the current in the 60-W bulb is
greater than that in the 30-W bulb. Because charge does
not build up in the bulbs, we know that the same amount
of charge flowing into a bulb from the left must flow out
on the right; consequently, I
c
"
I
d
and I
e
"
I
f
. The two
currents leaving the bulbs recombine to form the current
back into the battery, I
f
,
I
d
"
I
b
.
Figure P27.75
!
!
x
d
v
∆V
Answers to Quick Quizzes
27.1 d, b " c, a. The current in part (d) is equivalent to two
positive charges moving to the left. Parts (b) and (c)
each represent four positive charges moving in the same
direction because negative charges moving to the left are
equivalent to positive charges moving to the right. The
current in part (a) is equivalent to five positive charges
moving to the right.
27.2 (b). The currents in the two paths add numerically to
equal the current coming into the junction, without
regard for the directions of the two wires coming out of