Physics 21 Fall, 2014 Solution to HW-9

Physics 21
Fall, 2014
Solution to HW-9
26-34 In the circuit shown in the figure, the 6.0 Ω resistor
is consuming energy at a rate of 25.0 J/s when the current
through it flows as shown. (a) Find the current through
the ammeter A. (b) What are the polarity and emf of the
battery E, assuming it has negligible internal resistance?
(a) Let’s first use the information about the power dissipated in the 6.0 Ω resistor. We can get both the current
through
P = I 2 R so
q and the voltage across2 this resistor.
√
W
V
25.0 W · 6.0 Ω =
I = 25.0
6.0 Ω = 2.04 A. P = R so V =
12.2 V.
This circuit can be greatly simplified by replacing the parallel
20.0 Ω resistors with the equivalent resistor, which is 10.0 Ω.
We could further simplify the circuit by replacing any series
resistors with their equivalent resistors, but that’s not going
to save as much work as the parallel resistor simplification.
We won’t replace the series resistors. Finally, the ammeter in
the circuit is assumed to be ideal, and will have no influence
on the currents or voltages in the circuit. We can simply
ignore it. Our simplified circuit looks like the figure below.
The current through each branch, including assumed direction, is indicated on the figure. Two loops are also shown in
the figure, which we will use later when we apply Kirchhoff’s
loop rule.
The current in the middle branch of the circuit, I1 , is known
to be 2.04 A. The other two currents and the emf E make 3
unknown quantities in this circuit. We will need a system of
3 equations to solve this. They are:
−3.0 (2.04) + 25 − 12.2 − 10.0I2 − 19I2 − I2 = 0
25 − 12.2 − 17I3 − E − 13I3 − 3.0 (2.04) = 0
2.04 = I2 + I3
(1)
(2)
(3)
where equation (1) is Kirchhoff’s loop rule for the loop labeled (1) in the figure, equation (2) is Kirchhoff’s loop rule
for the loop labeled (2) in the figure, and equation (3) is
Kirchhoff’s junction rule for the top junction. All known
values have been filled in. For clarity, we have left units off
of the known values.
The only unknown in equation (1) is I2 , so we may solve
for it directly, obtaining I2 = 0.221 A. This is the current
through the ammeter in the original circuit drawing.
(b) We can now substitute our value for I2 into equation
(3), and we find I3 = 1.82 A. Substituting this into equation
(2) and solving for E results in E = −48.0 V. Note that
the negative result means that the polarity of the battery as
drawn in the original figure is reversed.
26-42 A 12.8µF capacitor is connected through a 0.890 MΩ
resistor to a constant potential difference of 60.0 V. (a) Compute the charge on the capacitor at the following times after
the connections are made: 0 s, 5.0 s, 10.0 s, 20.0 s, and
100.0 s. (b) Compute the charging currents at the same
instants.
(a) As derived, the formula for the charge on a charging
capacitor as a function of time is:
q(t) = Qf 1 − e−t/RC ,
where the final charge Qf = CV = 7.68 × 10−4 C. The time
constant τ = RC = 11.392 s. The table below gives q(t) at
the times specified.
(b) The relationship between charge and current is i = dq/dt,
so we can determine the current as a function of time by
differentiating the expression for q(t) above:
dq
Qf −t/RC
i(t) =
=
e
= I0 e−t/RC
dt
RC
where we substituted Qf = CV and noted that the initial
current I0 that flows is the battery voltage V divided by the
resistance R. I0 is 6.74 × 10−5 A.
Here is a table of the charge and current at various times.
Note that 100 s is about nine times the time constant, and
at that point the capacitor is essentially fully charged, and
the current from the battery is essentially zero.
t (s)
0
5
10
20
100
q(t) (C)
0
2.73 × 10−4
4.49 × 10−4
6.35 × 10−4
7.68 × 10−4
i(t) (A)
6.74 × 10−5
4.35 × 10−5
2.80 × 10−5
1.16 × 10−5
1.00 × 10−8
September 26, 2014
26-47 In the circuit shown in the figure below each capacitor initially has a charge of magnitude 3.60 nC on its plates.
After the switch S is closed, what will be the current in the
circuit at the instant that the capacitors have lost 80.0% of
their initial stored energy?
The first thing to recognize is that this problem is dealing with discharging three capacitors connected in series as
shown in the figure above. Once the switch is thrown, the
initial charge on each capacitor will discharge through the
resistor that is also connected to the capacitors. So, we will
need the relationship that describes the charge on the capacitors as a function of time as they discharge. This is just
the equation for a discharging capacitor that was discussed
in class:
q(t) = Q0 exp(−t/RC),
where Q0 is the initial charge on the capacitor, and RC is
the time constant of the circuit. However, this relationship
describes one capacitor, and we have three. The value of C
that we must use is Ceff , the effective capacitance given by
1
1
1
1
=
+
+
Ceff
C1
C2
C3
1
1
1
1
=
+
+
=
10.0 pF 20.0 pF 15.0 pF
4.62 pF
⇒ Ceff = 4.62 pF
We can take into account all three capacitors by writing
q(t) = Q0 exp(−t/RCeff )
However, the problem has not given us information in terms
of initial and final charge on the capacitors. Instead, we have
been given information in terms of initial and final energy
that is stored in the capacitor. We must use the relation
(found on the equation sheet) between the energy stored in
a capacitor and the charge on the capacitor:
Ucap =
Q2
2C
We are asked to find the current in the circuit at the time
when the energy stored in the capacitors has lost 80% of its
initial value. We’ll do this in two steps: first, find the time
at which the energy has dropped 80%, and second, find the
current at the time we’ve found. For the first step, we can
write an equation that says the ratio of the energy ucap at
time t to the initial energy is 1.00 − 0.80 = 0.20:
2
1
q(t)2 /Ceff
ucap(t)
q(t)
−2t
=
0.2 =
= 21 2
= exp
ucap (0)
Q0
RCeff
2 Q0 /Ceff
Taking the natural log of both sides, we can solve for t:
ln 0.2 = −
2t
RCeff
⇒
t = − 21 RCeff ln 0.2
(4)
Now, we have an expression for the time at which the capacitors have lost 80% of their initial energy. However, we
need to find the current at this time. One way is to use the
relation i(t) = −dq/dt (as discussed in the lecture, we use
the minus sign since dq/dt is negative, and we want i(t) to
be positive). Then
d d
Q0 exp(−t/RCeff )
i(t) = − q(t) = −
dt
dt
Q0
t
=
(5)
exp −
RCeff
RCeff
Now one could solve Eq. (4) explicitly for the time t and
then substitute into Eq. (5). One can also proceed formally,
− 21 RCeff ln 0.20
Q0
Q0 √
i(t) =
0.20
=
exp −
RCeff
RCeff
RCeff
Finally, plugging in the values for Q0 , Ceff , and R gives
√
3.60 × 10−9 C
0.20
(25.0 Ω)(4.62 × 10−12 F)
= 13.9 A
i(t) =
26-43 In the circuit shown in the figure both capacitors are
initially charged to 45.0 V. (a) How long after closing the
switch S will the potential across each capacitor be reduced
to 15.0 V? (b) What will be the current at that time?
(a) We can replace the two capacitors by an equivalent one
with C = 35 µF, and the two resistors by R = 80.0 Ω. This
change won’t affect the current or the potential differences
in the circuit; the voltage across the equivalent capacitor is
the same as across either of the two real capacitors.
From the text and from the lectures we know that the charge
q(t) of the discharging capacitor is given by
q(t) = Q0 e−t/RC .
26-76 A 7.00 µF capacitor that is initially uncharged is
connected in series with a 5.00 Ω resistor and an emf source
with E = 58.0 V and negligible internal resistance. At the
instant when the resistor is dissipating electrical energy at
a rate of 336 W, how much energy has been stored in the
capacitor?
This circuit is of the type shown in Figure 26.20b in the text:
the current i flows through the resistor R and charges the
capacitor C. We can start by solving for the current at the
time when the resistor is dissipating energy at the rate given:
r
336 W
2
= 8.20 A
P = i R = 336 W ⇒ i =
5Ω
Now we can use the loop equation for this circuit (discussed
in the lecture or given in the textbook) to find the charge
q on the capacitor when the current has the value we just
found:
V − iR − q/C = 0
q = 7 × 10
which can be solved for the time t for the charge to drop to
a fraction q(t)/Q0 of its initial value:
q(t)
Q0
t = −RC ln
.
= RC ln
Q0
q(t)
The general relation between the voltage across any capacitor and its charge is
v(t) = q(t)/C.
The key point is that the voltage v(t) is proportional to q(t).
Hence the time at which the voltage drops from 45 V to
15 V is the same as the time at which the charge drops to
45/15 = 1/3 of its original value. We can find this time from
the equation for t given above:
t = RC ln 3 = 80.0 Ω (35 × 10−6 F) ln 3 = 3.08 ms
(b) At any time, the circuit satisfies v(t) = i(t)R, so at the
given value of v(t),
i(t) =
15.0 V
v(t)
=
= 0.188 A.
R
80 Ω
⇒
q = C(V − iR)
F (58.0 V − 8.20 A × 5.00 Ω) = 119 µC
For this charge on the capacitor, the energy stored is
We can rearrange this equation to obtain
q(t)
= e−t/RC ,
Q0
−6
Ucap =
1
2
q2
=
C
1
2
(119 × 10−6 C)2
= 1.01 mJ
7 × 10−6 F

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