Prof. Anchordoqui
Problems set # 6
Physics 169
March 24, 2015
1. Find the current I in the circuit shown Fig. 1.
4Ω
8Ω
1Ω
16 Ω
16 Ω
I
1Ω
3Ω
20 V
Figure 1: Problem 1.
2. In the circuit shown in Fig. 2, the power produced by bulb1 and bulb2 is 1 kW and 50 W,
respectively. Which light has the higher resistance? (Assume the resitance of the light bulb remains
constant with time.)
bulb1
100 V
i1
bulb2
i2
Figure 2: Problem 2.
3. Consider the circuit shown in Fig. 3, with the start up switch T1 open (for a long time). Now,
close the switch and wait for a while. What is the change in the total charge of the capacitor?
4. A regular tetrahedron is a pyramid with a triangular base. Six R = 10.0 Ω resistors are
placed along its six edges, with junctions at its four vertices, as shown in Fig. 4. A 12.0-V battery
is connected to any two of the vertices. Find (i) the equivalent resistance of the tetrahedron between
these vertices and (ii) the current in the battery.
T1
V1
V2
VC
C
R2
R1
Figure 3: Problem 3.
R
R
R
R
R
R
Figure 4: Problem 4.
5. Find the equivalent resistance in the limit n → ∞ for the circuits shown in Figs. 5 and 6.
A
1Ω
2Ω
4Ω
8Ω
Rn−1
Rn
B
Figure 5: Problem 5 (i).
6. The circuit in Fig. 7 has been connected for a long time. (i) What is the voltage across the
capacitor? (ii) If the battery is disconnected, how long does it take the capacitor to discharge to
one tenth of its initial voltage?
7. Find the voltage across A and B (i.e. VAB ) as a function of time in the circuit shown in
Fig. 8
8. The switch in Fig. 9(a) closes when ∆Vc > 2∆V /3 and opens when ∆Vc < ∆V /3. The
voltmeter reads a voltage as plotted in Fig. 9(b). What is the period T of the waveform in terms
37. The circuit in Figure P28.37 has been connected for a long
A
time. (a) What
is the voltage across the capacitor? (b) If the
R take
R the capacitor
battery is disconnected,
1Ω
2 Ωhow
3 Ωlong
4 Ω does it
to dischargeB to one tenth of its initial voltage?
n−1
n
Figure 6: Problem 5 (ii).
1.00 Ω
8.00 Ω
1.00 µF
10.0 V
4.00 Ω
2.00 Ω
Figure
P28.37
Figure 7: Problem 6.
R
A
38. In places such as a hospital operating
room and a factory
C
C
for electronic circuit
boards,
electric
sparks must be
V
C
avoided. A person standing on
a grounded floor and
B
touching nothing else can typically have a body capacitance of 150 pF, in parallel
with a foot capacitance of
Figure 8: Problem 7.
80.0 pF produced by the dielectric soles of his or her
of R , R , and C?
shoes.9. This
The
problemperson
illustrates howacquires
a digital voltmeterstatic
affects the electric
voltage across a charge
capacitor in an from
RC circuit. A digital voltmeter of internal resistance r is used to measure the voltage across a
interactions
with furniture, clothing, equipment, packagcapacitor after the switch in Figure P28.76 is closed. Because the meter has finite resistance, part
the current supplied by
the battery
passes through the
meter. (i) Apply Kirchhoff’s
to this static
ingof materials,
and
essentially
everything
else. rules
The
charge is conducted to ground through the equivalent
resistance of the two shoe soles in parallel with each other.
A pair of rubber-soled street shoes can present an equiva0
1
2
Problems
ted in paral-
f 40.0 V and
charge two
ving an emf
0 !. If the
t additional
hat rate does
he batteries,
at rate does
each other.
termine the
vision of the
sults in less
any other
t in a direct
er delivered to
891
R1
Voltage–
controlled
switch
∆V
R2
V ∆Vc
C
(a)
∆Vc(t)
∆V
2∆V
3
∆V
3
T
t
(b)
Figure P28.66
Figure 9: Problem 8.
Three
lightbulbs
are toconnected
circuit, 67.
and use
the fact60.0-W,
that iC = 120-V
dq/dt to show
that this leads
the differential across
equation a
d using the
120-V power source,dq as q shown
in Figure P28.67. Find
r
P28.63. The
Req +
=
E,
(a) the total powerdt delivered
C
r + R to the three bulbs and
oltmeter has
voltage
that the
resistance
of
where Req = (b)
rR/(rthe
+ R).
(ii) Showacross
that theeach.
solutionAssume
to this differential
equation
is
e of actual
in reality the resiseach bulb is constant
(even
though
r
q=
CE 1 − e−t/(R C)
ect to within
r+R
tance might increase
markedly with current).
ircuit shown
eq
and that the voltage across the capacitor as a function of time is
VC =
r
E(1 − e−t/(Req C) ).
r+R
(iii) If the capacitor is fully charged, and the switch is then opened, how does the voltage across
the capacitor behave in this case?
R 1semiconductors are put in contact, the relative affinities
10. When two slabs of n-type and p-type
of the materials cause electrons to migrate out of the n-type material across the junction to the
R
R3
120 V
p-type material. This leaves behind a volume in2 the n-type material
that is positively charged and
creates a negatively charged volume in the p-type material. Let us model this as two infinite slabs
of charge, both of thickness a with the junction lying on the plane z = 0. The n-type material lies
Figure P28.67
68. Switch S has been closed for a long time, and the electric
0 equation is
differential
increas
r
q#
C!0 (1 ( e (t/R
#+
0 < eqz C<)a
r & R%
%
28.8 (c). A c
There
branch
!(x, y, z) = !(z) = $" !0
" a< z< 0
%
Voltmeterz > a
&0
28.9 (b), (d
charge
r
Charge
a) Find the electric field everywhere.
R
pacitor
b) Find the potential difference between the pointsC P1 and P2. . The point P1. is locatedthe re
on a plane parallel to the slab a distance z1 > a from the center of the slab. Thecharge
point P is located on plane parallel to the slab a distance z < !a from the centertery vo
2.
2
28.10 (c), (i)
on the
circuit
Solution:
ε
S
half of
Figure P28.76of two slabs of opposite charge density.
paralle
n this problem, the electric field is a superposition
Figure 10: Problem 9.
time, t
6
the rig
After Joseph Priest, "Meter Resistance: Don't Forget It!" The Physics
exists o
Teacher, January 2003, p. 40.
of the slab.
!
Figure
11: E
Problem
10.
Outside both slabs, the field of a positive
slab
P (due to the p -type semi-conductor ) is
!
n -type
constant andin points
away
and
the
field
of
a
negative
slab
to the
E
the range 0 < z < a and has uniform charge density +ρ0 . The adjacent
material
lies insemiN (duep-type
< z < 0 and and
has uniform
charge
density −ρthe
Fig. 11. so
Hence:
0 ; see slab,
conductor) the
is range
also−aconstant
points
towards
when we add! both

contributions we find that the electric field is
+ρ0 outside
0 < z < athe slabs. The fields E P are
zero
ρ(x, y, z) = ρ(z) =
−ρ0 −a < z < 0 ! .
shown on the figure below. The superposition 
of 0these
fields ET is shown on the top line

|z| > a
n the figure.
(i) Find the electric field everywhere. (ii) Find the potential difference between the points P1 and
P2 . The point P1 is located on a plane parallel to the slab a distance z1 > a from the center of the
slab. The point P2 is located on plane parallel to the slab a distance z2 < −a from the center of
the slab.