Recitation Week 5
Chapters 24 and 25
Problem 24.53. Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In
1
s with an average light power output of 2.70 · 105 W. (a) If the conversion of electrical
one such unit, the flash last for 675
energy to light is 95% efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor
for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value
calculated in (a). What is the capacitance?
(a) The total light energy released is
Ulight = Plight · ∆t .
(1)
That is 95% of the total energy released from the capacitor, so
0.95UC = Ulight
(2)
5
UC =
Plight · ∆t
2.70 · 10 ·
Ulight
=
=
J = 421 J .
0.95
0.95
0.95 · 675
(3)
(b) The capacitance is given by
1
CV 2
2
2UC
2 · 421 J
C=
=
= 53.9 mF .
V2
(125 V)2
UC =
(4)
(5)
Problem 24.56. In Fig. 24.9a, let C1 = 9.0 µF, C2 = 4.0 µF, and Vab = 28 V. Suppose the charged capacitors are
disconnected from the source and from each other, and then reconnected to each other with plates of opposite sign together.
By how much does the energy of the system decrease?
a-----+---------+
+|+
+|+
Vab C1 === Q1 C2 === Q2
-|-|b-----+---------+
The initial energy is
Ui =
1
2
(C1 + C2 )Vab
= 5.10 mJ .
2
(6)
To find the final energy, we need to find the final voltage or charge. The initial charges are
Q1 = C1 Vab = 252 µC
(7)
Q2 = C2 Vab = 112 µC .
(8)
Flipping C2 around gives a total charge in the top wire of
Qf = Q1 − Q2 = 140 µC .
(9)
The charge in the bottom wire is −Q1 + Q2 = −Qf , as you’d expect since capacitors should have equal and opposite charges
on either plate.
This yields a final energy of
Uf =
Q2f
(140 µC)2
=
= 700 µJ ,
2(C1 + C2 )
2 · 14.0 µF
(10)
for a net decrease of
−∆U = Ui − Uf = 4.40 mJ .
(11)
Problem 24.57. For the capacitor network show in Fig. 24.31, the potential difference across ab is 12.0 V. Find (a) the
total energy stored in this network and (b) the energy stored in the 4.80 µF capacitor.
6.20 11.8
+--||--||--+
8.60 4.80 |
|
a--||--||--+
+--b
|
|
+----||----+
3.50
(all capacitances are given in µF)
(a) Bundling down to a single equivalent capacitance, we see that the 8.60 and 4.80 µF capacitors are in series, as are the
6.20 and 11.8 µF capacitory. The circuit is thus equivalent to
4.06
+--||--+
3.08 |
|
a--||--+
+--b
|
|
+--||--+
3.50
where
CL =
CT =
1
1
+
8.60 4.80
−1
1
1
+
6.20 11.8
−1
µF = 3.08 µF
(12)
µF = 4.06 µF .
(13)
Then we see that the 4.06 and 3.50 capacitors are in parallel, so the circuit is equivalent to
3.08 7.56
a--||----||--b
where
CR = (4.06 + 3.50) µF = 7.56 µF .
These last two capacitors are themselves in series, so we can move down to a single equivalent capacitance
−1
1
1
C=
+
µF = 2.19 µF .
3.08 7.56
(14)
(15)
We use the capacitor energy equation on this effective capacitance to find the total energy
U=
1
1
CV 2 = · 10.6 µF · (12.0 V)2 = 158 µJ .
2 ab
2
(16)
(b) Stepping back out into our more detailed picture, we remember that capacitors in series share the same charge (because
the wire between them remains neutral as a whole), so both the CL = 3.08 and CR = 7.65 µF capacitors have the same
charge Q as the single equivalent capacitance CT . Similarly, the two series capacitors that make up CL also have the same
charge Q as CL (which is the same Q as CT ). Therefore, the charge on the 4.80 µF capacitor is the same as the charge on
CT .
Because the charge is the same on both CT and C4.8 , we’d like to find a formula for energy that involves charge. Plugging
the capacitor formula Q = CV into our usual energy formula yields
2
1
Q
Q2
1
=
.
(17)
U = CV 2 = C
2
2
C
2C
Isolating the constant Q and applying the formula to both CT and C4.8 yields
Q2 = 2CT UT = 2C4.8 U4.8
CT
2.19 µF
2.19
U4.8 =
UT =
158 µJ =
158 µJ = 71.9 µJ
C4.8
4.80 µF
4.80
(18)
(19)
Problem 25.17. In household wiring, copper wire 2.05 mm in diameter is often used. Find the resistance of a 24.0 m length
of this wire.
We find the resistivity for copper ρ = 1.72·10−8 Ω·m in Table 25.1 (page 851). The resistance is then given by Equation 25.10
(on page 853):
(1.72 · 10−8 Ω·m) · (24.0 m)
ρL
ρL
= 125 mΩ .
(20)
= 2 =
R=
A
πr
π(2.05 mm/2)2
Problem 25.24. You apply a potential difference of 4.50 V between the ends of a wire that is 2.50 m in length and 0.654 mm
in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?
Juggling the resistivity/resistance relationship (Equation 25.10)
R=
ρL
A
(21)
and Ohm’s law (Equation 25.11)
V = IR ,
(22)
we get
ρ=
AV
πr2 V
π(0.654 mm)2 · (4.50 V)
AR
=
=
=
= 1.37 · 10−7 Ω·m .
L
LI
LI
(2.50 m) · (17.6 A)
(23)
This puts the resistivity between that of tungsten and steel in Table 25.1.
Problem 25.36. The circuit shown in Fig. 25.37 contains two batteries, each with an emf and an internal resistance, and
two resistors. Find (a) the current in the circuit (magnitude and direction); (b) the terminal voltage Vab of the 16.0 V
battery.; (c) the potential difference Vac of point a with respect to point c. (d) Using Fig. 25.21 as a model, graph the potential
rises and drops in this circuit.
1.6
16.0
+---a---/\/\/---|i---b---+
|
|
5.0 Z
Z 9.0
Z
Z
|
1.4
8.0
|
+-------/\/\/---|i---c---+
Resistances (ZZ and /\/\/) are in Ω. Battery (|i) voltages are in volts.
(a) We’re going to use Kirchoff’s loop rule: the sum of voltage changes around a loop is 0.
X
∆V = 0
(24)
loop
The book doesn’t name this rule until Chapter 26 (see Equation 26.6 on page 887), but we’ve been using since the capacitor
chapter (see page 821 “Capacitors in Parallel”), and this resistance chapter (see page 862, “Potential changes around a
circuit”). I think it’s best to just get the name out in the open :p.
Anyhow lets travel the loop along the path a → b → c → d → a, adding up voltage changes as we go. By Kirchoff’s loop rule,
the total voltage change must be 0. We’ll need to pick a direction for the current I to be flowing as well, so we know whether
the potential increases or decreases going across resistors. Let’s pick the same direction (clockwise) as the path we’re taking.
The total voltage change is then
0 = −1.6 ΩI − 16.0 V − 9.0 ΩI + 8.0 V − 1.4 ΩI − 5.0 ΩI
(25)
where we used Ohm’s law V = IR to find the voltage change across each resistor, and gave it a negative sign (drop) because
we were traveling in the same direction as I. Now it’s just algebra:
0 = −(1.6 + 9.0 + 1.4 + 5.0) ΩI + (−16.0 + 8.0) V
17.0 ΩI = −8.0 V
I = −470 mA .
(26)
(27)
(28)
The − sign shows us that the current actually flows in the opposite direction to the one we picked at the beginning, so there
is 470 mA of current flowing counterclockwise.
(b) This just an excerpt from (a). Starting from b and moving counterclockwise,
Vab = 16.0 V − 1.6 Ω · 470 mA = 15.2 V
(29)
(c) Another excerpt from (a). Starting from c and moving clockwise,
Vac = 8.0 V + 1.4 Ω · 470 mA + 5.0 Ω · 470 mA = 11 V
(30)
(d) Setting point Vb = 0 V, we have
Vb = 0 V
Vab/2 = 16 V
(31)
(32)
Va = 15.2 V
(33)
Vac/3 = 12.9 V
(34)
V2ac/3 = 12.2 V
(35)
Vc = 4.2 V
(36)
(37)
Problem 25.53. In the circuit in Fig. 25.39, find (a) the rate of conversion of internal (chemical) energy to electrical energy
within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of the electrical
enegy in the external resistor.
1.0
12.0
+---a---/\/\/---|i---d---+
|
|
|
|
+---b-------/\/\/----c---+
5.0
(a) First we need to find the current. Going counterclockwise around the loop from d and using only SI units
0 = 12.0 − 1.0I − 5.0I
12.0
I=
= 2.0 A
6.0
(38)
(39)
Power through any element is given by P = IV (Equation 25.18, page 864), and all the electrical energy from the battery is
being created by chemical energy creating the internal 12 V emf, so
Pchem = 2.0 A · 12.0 V = 24.0 W .
(40)
(b) The battery dissipates some of this energy into heat through it’s internal resisitance
Pi = IV = I(IR) = I 2 R = (2.0 A)2 · 1.0 Ω = 4.0 W .
(41)
(c) Similarly to (b), the external resistor dissipates
Pe = I 2 R = (2.0 A)2 · 5.0 Ω = 20.0 W .
(42)
Note that all 24.0 W of electrical power from the reaction in the battery is being dissipated into heat by the two resistors
(4 + 20 = 24 W).