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Physics 9 Fall 2009
Homework 6 - Solutions
1. Chapter 32 - Exercise 8.
How much power is dissipated by each resistor in
the figure?
————————————————————————————————————
Solution
First, let’s figure out the current in the circuit. Since the two resistors are in series, we
can replace them by an equivalent resistance, Reqv = R1 + R2 using Kirchhoff’s loop
law,
X
∆Vi = E − IR1 − IR2 = 0.
12
12
E
= 12+18
= 30
= 25 = 0.40 amp. now, the power
This gives for the current, I = R1 +R
2
dissipated by each resistor, Ri , is Pi = I∆Vi = I 2 Ri . So, we find for each resistor
P1 = I 2 R1 = (0.4)2 12 = 1.92 W
P2 = I 2 R2 = (0.4)2 18 = 2.9 W.
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2. Chapter 32 - Exercise 27.
Determine the value of the potential at points a
to d in the figure.
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Solution
Because the wire at point d is connected to a ground, the potential at point d is at
0 V, by definition. In going from point d to point a, e pass through a battery, which
bumps up the voltage by ∆V = E = 9 V. So, the potential at point a is 9 V. To get the
potentials at points b and c we will need the current in the circuit. From the loop law,
∆Vloop = E1 − IR1 − E2 − IR2 = 0, where E1 = 9 V, R1 = 2Ω, E2 = 6 V, and R2 = 1Ω
(note that ∆V across E2 is −E2 because we’re going from positive to negative). Thus,
solving for the current, we find
I=
E1 − E2
= 1 A.
R1 + R2
So, the drop across the first resistor is ∆V = −IR1 = −2 V, which means that the
potential at point b is 9 - 2 = 7 V. Next, the battery connected backwards fights the 7
V, dropping it to 7 - 6 = 1 V. So, the potential at point c is 1 V. Finally, we cross the
last resistor which drops the potential by ∆V = −IR2 = −1 V, giving an net potential
of 1 - 1 = 0 V at point d, which we know. So, altogether,
Point
Voltage
a
9V
2
b
7V
c
1V
d
0V
3. Chapter 32 - Exercise 34.
What value resistor will discharge a 1.0 µF capacitor to 10% of its initial charge in 2.0
ms?
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Solution
For a capacitor of initial charge Q0 connected to a resistor, R, the charge falls away
exponentially fast,
Q (t) = Q0 e−t/RC .
So, a capacitor will fall to 10% of it’s initial charge after a time t0 such that
Q (t0 ) = 0.1Q0 = Q0 e−t0 /RC ⇒ 0.1 = e−t0 /RC .
Taking the natural logs of both sides and rearranging (using the fact that − ln 0.1 =
ln 10) gives
t0
.
R=
C ln 10
Plugging in the numbers gives
2 × 10−3
t0
= −6
= 0.87kΩ.
R=
C ln 10
10 ln 10
3
4. Chapter 32 - Problem 37.
The figure shows six identical bulbs connected to
an ideal battery. All the bulbs are glowing. Rank
in order, from brightest to dimmest, the brightness of bulbs A to F. Explain.
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Solution
The brightness of the bulbs depends on the power dissipated in the bulb, which depends
on the current. So, whichever bulb has the highest current has the highest power
dissipated, and so is the brightest. Now, because bulb A is in parallel with the rest of
the bulbs, the current from the battery is split between bulb A, and all the rest of the
bulbs! So, A is brightest.
The current next passes through bulb D before splitting up to the rest of the bulbs,
and so D is next brightest. Next, the current passes through the next set of bulbs.
Bulbs C and E are in parallel with each other, so they are equally bright. The current
through F is the sum of the currents through C, E, and B, and so F is brighter than
C and E. Now, B is in parallel with C, E, and F. But, since the CEF combination has
a higher resistance than B, more current passes through B than the others since they
are all at the same potential. So, we have for brightness, A > D > B > F > C = E.
4
5. Chapter 32 - Problem 41.
You have three 12 Ω resistors. Draw diagrams showing how you could arrange all three
so that their equivalent resistance is (a) 4.0 Ω, (b) 8.0 Ω, (c) 18 Ω, and (c) 36 Ω.
————————————————————————————————————
Solution
The four diagrams are seen in the figure below.
(a) In the first diagram, all three resistors are in parallel. So, the equivalent resistance
is
1
1
1
1
1
1
1
1
=
+
+
=
+
+
= ,
Reqv
R1 R2 R3
12 12 12
4
which gives Reqv = 4 Ω.
(b) In this case the top two resistors are in series and are both in parallel with the
bottom resistor. The top resistors add to give R1 + R2 = 12 + 12 = 24. This
equivalent resistance combines with the resistor on the bottom to give
1
1
1
3
1
=
+
=
= ,
Reqv
24 12
24
8
and so Reqv = 8 Ω.
(c) Next, we have a two resistors in parallel in series with a third resistor. The two in
1
1
1
= 12
+ 12
, giving an equivalent resistance of 6 Ω for
parallel combine to give Reqv
that branch of the circuit. Adding this in series with the other resistance gives
Reqv = 6 + 12 = 18 Ω.
(d) Finally, we just have three resistors in series, which all add together to give
Reqv = R1 + R2 + R3 = 12 + 12 + 12 = 36 Ω.
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6. Chapter 32 - Problem 42.
What is the equivalent resistance between points
a and b in the figure?
————————————————————————————————————
Solution
We can form an
equivalent circuit in
the following steps.
In the first step, we break up the three parallel resistors into a single equivalent resistor.
1
6
Since they are in parallel, R1eq = 16 + 12
+ 14 = 12
, and so Req = 2 Ω.
Next, the element that contains the 10 Ω resistor in parallel with the wire has a net
resistance of zero! Why? The current will take the “path of least resistance,” bypassing
the resistor altogether. (The current doesn’t take a single path, in general - it splits
between the two paths depending on the load resistance, as we’ll see). This circuit is
a short circuit. Can we see this? Think of the wire as a resistor with R = 0. Then, for
1
+ 10 = ∞. So, Req = 0.
the resistor and wire in parallel, R1eq = R11 + R12 = 10
So, we can just ignore the 10 Ω resistor to get the three resistors in series in the third
diagram. This gives a net resistance of Req = 2 + 2 + 3 = 7 Ω.
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7. Chapter 32 - Problem 49.
(a) Load resistor R is attached to a battery of emf E and internal resistance r. For
what value of the resistance R, in terms of E and r, will the power dissipated by
the load resistor be a maximum?
(b) What is the maximum power that the load can dissipate if the battery has E =
9.0 V and r = 1.0 Ω?
(c) Why should the power dissipated by the load have a maximum value? Explain.
————————————————————————————————————
Solution
(a) When the battery is attached to the load resistor, R, the internal resistance, r,
of the battery is in series with the external resistor. From our usual loop law,
E
. The power dissipated in the load resistor is
the current in the wire is I = R+r
E 2R
2
P = I R = (R+r)2 . The power dissipated is a maximum when dP
= 0. Taking
dR
the derivative gives
!
E 2R
(R + r)2 − 2R (R + r)
d
2
=E
=0
dR (R + r)2
(R + r)4
Solving gives (R + r)2 − 2R (R + r) = R2 − r2 = 0, and so the power dissipated
is a maximum when R = r.
(b) The power is maximized when R = r, which gives P =
in the numbers gives P =
9.02
4×1
E 2R
(R+r)2
=
E2
.
4r
So, plugging
= 20.25 W.
(c) Suppose R is very small. Then the current is big, I ≈ E/r. But, P = I 2 R, which
is forced to be small by the factor of R. So, the smaller R gets, the smaller the
power goes. What if R is big? In this case, the current is also small, I ≈ E/R.
Then, the power becomes P = I 2 R ≈ E 2 /R, which is driven to zero as R gets
big. So, in either case the power is driven down for either extreme, R → 0, or
R → ∞. There has to be some intermediate value of R that maximizes the power
dissipated, and this is what we found in part a.
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8. Chapter 32 - Problem 63.
For the circuit shown in the figure, find the current
through and the potential difference across each
resistor. Place your results in a table for ease of
reading.
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Solution
First, we need to determine the current being supplied by the battery. We begin by
simplifying the circuit as in the following diagram, applying usual rules for combining
resistors in parallel and in series.
To move from the first to the second diagram, we combine the 4 Ω and 12 Ω resistors
in parallel for a net resistance of 3 Ω. Then we combine the 3 and 5 Ω resistors in
series to get the 8 Ω resistors in the third diagram. Next, combine the 24 and 8 Ω
resistors in parallel to get the 6 Ω resistor in the fourth diagram. Finally, we combine
the 6 and 3 Ω resistors in series to get the equivalent 9 Ω resistor in the last circuit
Applying Kirchhoff’s loop law to the equivalent circuit gives E1 − IR − E2 = 0, where
the voltage drops moving across the second battery since the current is moving from
2
positive to negative. This gives a current, I = E1 −E
= 12−3
= 1 amp.
R
9
So, now we need to work backwards, reconstructing the circuit as seen in the following
diagram.
8
We begin with step 1. Because current is conserved in the last diagram, we know that
1 A flows through both the 3 and 6 Ω resistors. The drop in voltage across the 3 Ω
resistor is thus ∆V = IR = 3.0 V. The change in voltage across the equivalent 6 Ω
resistor is ∆V = IR = 6 V.
Step 2 - We split up the equivalent resistor into the two original resistors. We know
that the change in voltage across each resistor is the same, and is equal to 6 V, but the
current, which splits between the two resistors, is not the same in each resistor. From
Ohm’s law, we know that ∆V = IR, and so I24 = 6/24 = 0.25 A, while I8 = 6/8 = .75
A. Notice that I24 + I8 = 1 A, which we need.
Step 3 - Now we break up the equivalent 8 Ω resistor into the two resistors in series.
Now the current is the same through each of these two resistors, but the voltage is not.
Again, going back to Ohm’s law, we can solve for the change in voltage through each
resistor. ∆V3 = 0.75 × 3 = 2.25 V, while ∆V5 = 0.75 × 5 = 3.75 V. Again, notice that
the voltages in these two resistors adds up to 6 V, which we knew was the case.
Step 4 - Finally, we break up the 3 Ω equivalent resistor into the 4 and 12 Ω resistors
in parallel. Now, the current through these two resistors aren’t equal, but the voltage
drop is. So, yet again, we go back to Ohm’s law. I4 = 2.25/4 = 0.56 A, while
I12 = 2.25/12 = 0.19 A. Again, the currents add up to the original 0.75 A, as it must.
So, now we have all our answers. Let’s put the all in a table
Resistor (Ω)
Voltage (V)
Current (A)
4
12
24
5
3
2.25 2.25 6.0 3.75 3.0
0.56 0.19 0.25 0.75 1.0
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9. Chapter 32 - Problem 77.
The capacitor in Figure 32.38a begins to charge after the switch closes at t = 0 s.
Analyze this circuit and show that Q = Qmax 1 − e−t/τ , where Qmax = CE.
————————————————————————————————————
Solution
When the switch closes, the battery begins to charge. Once it starts, Kirchhoff’s loop
law says that
q
E
q
E − IR − = 0 ⇒ I = −
.
C
R RC
But, I = dq
(the derivative is positive since the capacitor is charging). So,
dt
dq
dt
= − RC
, or
Separating and integrating gives q−CE
Z
0
Q
1
dq
=−
q − CE
RC
t
Z
dt ⇒ ln
0
Q − CE
CE
=−
dq
dt
=
q
E
− RC
.
R
t
.
RC
Solving for Q (t) gives
Q (t) = EC 1 − e−t/τ ,
where τ ≡ RC. So, we can write Q = Qmax 1 − e−t/τ , where Qmax ≡ EC. We know
that this is Qmax because it is the maximum charge that builds up after a long time
(t → ∞).
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10. Chapter 32 - Problem 78.
The switch in Figure 32.38a closes at t = 0 s and, after a very long time, the capacitor
is fully charged. Find expressions for (a) the total energy supplied by the battery as the
capacitor is being charged, (b) total energy dissipated by the resistor as the capacitor
is being charged, and (c) the energy stored in the capacitor when it is fully charged.
Your expressions will be in terms of E, R, and C. (d) Do your results for parts a to c
show that energy is conserved? Explain.
————————————————————————————————————
Solution
In problem 77 we found that Q (t) = CE 1 − e−t/RC for this circuit. We’ll use this in
what follows.
(a) The power being supplied by the battery as the capacitor is being charged is given
, or
by Pbatt = I∆Vbatt = IE. Now, as the capacitor is being charged, I = dQ
dt
I=
E
d CE 1 − e−t/RC = e−t/RC .
dt
R
2
So, Pbatt = IE = ER e−t/RC . Now, the power is how fast the energy is changing,
. So, we can integrate the power to get the total energy supplied by
Pbatt = dE
dt
the battery.
∞
Z
Z ∞
E2
E2 ∞
−t/RC
−t/RC dt e
= − RCe
= CE 2 ,
Ebatt =
Pbatt dt =
C 0
C
0
0
where we integrated to infinity because the capacitor has been charging for a long
time.
(b) Now, the power dissipated by the resistor is Pres = I∆Vres = I 2 R, using Ohm’s
2
law. So, plugging in the expression for the current gives Pres = ER e−2t/RC . But
again, Pres = dtd Eres . So we again integrate to find the total energy.
∞
Z ∞
Z
2
E2 ∞
E
C
1
Eres =
Pres dt =
dt e−2t/RC = −
e−2t/RC = CE 2 .
R 0
2
2
0
0
So, the total energy dissipated by a resistor is Eres = 21 CE 2 .
(c) The total energy stored by a capacitor when it has charged to Qmax = CE is
Ecap =
Q2max
(CE)2
1
=
= CE 2 .
2C
2
2
(d) The total energy supplied by the battery is E = CE, and is the only source
of energy. The energy that the capacitor stores up is 12 CE 2 , while the resistor
dissipates a total energy of 21 CE 2 . So, the total energy stored by the capacitor,
plus the energy dissipated by the resistor is 21 CE 2 + 12 CE 2 = CE 2 , which is precisely
the energy supplied by the battery. So, the total energy is conserved!
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