Before you begin: Use black pencil. Write and bubble your NAME on top left and your SBU ID Number at bottom left.
Fill bubbles fully and erase cleanly if you wish to change! 24 Questions, each question is 10 points, except when noted.
Each question has at most one correct answer: marking 2 or more bubbles will automatically give you 0 points.
1. My exam version is: (0 points)
A
B
C
D
q
2. Two unlike-sign point charges are located 3.0 cm apart; Q=6.0 C. The magnitude of the net force acting
on a third point charge q=+2.0 C located at a distance of 3.0 cm from either one of them is closest to …
Solution: The force from +Q on q has magnitude k Qq r 2 120 N with direction 60° up from right.
Likewise, the force from –Q on q is 120 N, with direction 60° down from right. The NET force is
the vector sum and equals 120 N (forms the side of an equilateral triangle), with direction Right.
+Q
B 120 N
C 240 N
D none of these
A 3.6 N
–Q
3. The direction of the force on charge q is … (5 points)
A Up
B Down
C Right
D Left
E none of these
4. A tiny plastic bead with mass m = 4.9×10 9 kg is charged by the addition of excess electrons. It is kept suspended in
an electric field of magnitude E=3.0×103 V/cm. g=9.80 m/s2. The number of excess electrons is closest to …
Solution: The force from gravity is down: w=mg.
The electric force must balance the weight w and point up: FC=qE=–Ne e E. Because the excess charge is negative, the
electric field must be directed DOWN, and the number of excess electrons Ne=mg/(eE)=0.999×106.
A 1.0×106
B 1.0×107
C 1.6×107
D 1.0×108
E none of these
5. In the previous problem: the direction of the electric field is: (5 points)
A Up
B Down
C Right
D Left
E none of these
6. A positive point charge q=2.0 C is released from a point d=1.0 cm to the right of a fixed positive point charge
Q=5.0 C. Only the electric force from Q is acting on q. As q moves, it will move with … (5 points) Q
d
Solution: The electric force by Q diminishes with the square of the distance between q and Q, and thus the
acceleration decreases as q moves farther to the right.
The kinetic energy it gains as it is pushed by the electric force, equals the decrease in its potential energy: K=– U=
U(1 cm) – U(9 cm)=q[V(1 cm) – V(9 cm)]= kqQ[1/0.01 – 1/0.09]=8.00 J.
A constant
acceleration
B increasing
acceleration
C decreasing
acceleration
D zero acceleration
q
E none of these
7. In the previous problem: the kinetic energy of q, when it is 9.0 cm away from Q, is closest to …
A 8.0 J
B 11 J
C 800 J
D 900 J
8. The work done by the electric force when a q=2.0 C charge moves from A to B is closest to …
Solution: Because the static electric force is a conservative force, the electric work done is
q times the potential difference between A and B, independent of the path taken. VAB=0 and
the electric work is therefore zero.
A 0.0 J
B 2.0 V
C 4.0 J
B 4.0 J
C –4.0 J
B
C
2V
–3 V
–2 V
–1 V
1V
D depends on the
path taken
9. The work done by the electric force when a q=2.0 C charge moves from D to B is closest to …
Solution: Because the static electric force is a conservative force, the electric work done is
q times the potential difference between D and B, independent of the path taken.
VDB=VB –VD=2 V=2 J/C and the electric work is therefore q VDB=4.0 J
A 2.0 V
E none of these
D
0V
A
D –2.0 V
10. Comparing the magnitudes of the fields at A and B, the electric field magnitude at point A is …
Solution: Because the field strength is largest when the voltage gradient is steepest, the field in A is smaller in
magnitude than the field in B where the equipotential lines are closer together.
A greater than that at B smaller than that at C equal to that at
point B
point B
point B
D not enough
information
11. A parallel-plate capacitor is connected to a 100 V battery until it is fully charged. The distance between the plates is d
and the space between the plates is filled with air ( =1.0). Then, the battery is disconnected.
3d
d
If the distance between the plates is increased to 3d, the voltage of the capacitor will be
closest to …
Solution: Initially the battery has 100 V, and a certain charge Q. When the plates are moved
apart, the charge remains Q, but the capacitance C changes: the capacitance varies inversely
with the separation distance between the plates and thus decreases by a factor 3. Q=CV, and because the charge
remains constant in this problem, the voltage must INCREASE by a factor 3 to 300 V.
A 33 V
B 100 V
C 300 V
D 900 V
12. In the previous problem: if, instead of changing the distance d between the plates, a dielectric
with =3.0 is inserted between the plates such that it fully fills the space between the plates,
the voltage of the capacitor will be closest to …
Solution: Initially the battery has 100 V, and a certain charge Q. When a dielectric is inserted
between the plates, the capacitance increases by a factor 3. Q=CV, and because the charge
remains constant in this problem, the voltage must DECREASE by a factor 3 to 33 V.
A 33 V
B 100 V
C 300 V
d
d
before
after
D 900 V
13. In the circuit shown R1=2.0 and R2=3.0 . The power delivered by the battery is closest to … (5 points)
Solution: The two resistors are in series with the battery and act like a single R=5 resistor.
Therefore, the current that runs in the loop is I=V/R=2.0 A. The power delivered by the battery
equals the product of its voltage and its delivered current: P=V×I=20 W.
The power dissipated by the 3.0 resistor equals the product of the voltage across that resistor
and the current through it: PR=VR×I=IR×I=I2R=12 W.
A 6.0 W
B 12 W
C 18 W
R1
10V
R2
D 20 W
14. The power dissipated in resistor R2 is closest to … (5 points)
A 6.0 W
B 12 W
C 18 W
D 20 W
15. The equivalent resistance of the resistor configuration (R=90 ) between terminals a and b is
closest to …
a
Solution: The left three 90 resistors are equal in value and in parallel; there equivalent
resistance equals 90 /3=30 . This 30 is in series with another 90 resistor, giving a total
resistance of the network of 120 .
A 23
B 90
C 120
B 67 F
C 133 F
R
R
R
D 300 F
C
a
C b
C
17. If the potential difference across a 36 mF capacitor equals 220 V, its charge is closest to …
Solution: The charge, capacitance, and voltage of a capacitor are related as: Q=CV. Thus: Q=36 mF×220 V=7.9 C.
A 7.9 C
B 160 C
C 9 kC
b
D 360
16. The equivalent capacitance of the configuration (C=100 F) between terminals a and b is …
Solution: The left two 100 F capacitors are in parallel and simply act as a single larger-plate
capacitor of 200 F. This 200 F capacitance is in series with another 100 F capacitor. This
series capacitance acts like a single capacitor of [(200 F)–1 + (100 F)–1] –1=67 F.
A 25 F
R
D none of these
18. A dielectric-filled parallel-plate capacitor has plate area 279 cm2, plate separation 0.500 mm and
dielectric constant =4.00. The capacitor is connected to a V=9.00 V battery. The energy of this
capacitor is closest to …
Solution: The capacitance of the parallel plate capacitor equals C= 0A/d=1.975 nF. Its energy,
when charged to 9.0 V equals: U= ½ CV2=80 nJ.
Without dielectric, the capacitance would by four times smaller, and its energy would only be 20 nJ. The capacitor
with only half the space filled with dielectric is equivalent to half an air-filled capacitor (energy 10 nJ) in parallel with
half a dielectric-filled capacitor (energy 40 nJ), giving a total of 50 nJ: “none of these” is the correct answer!
A 8.9 nJ
B 80 nJ
C 8.0 mJ
D none of these
19. The dielectric material is now slowly pulled halfway out of the capacitor, which remains
connected to the battery, until it covers only half of the plates. The energy of the capacitor is now
closest to …
A 10 nJ
B 20 nJ
C 30 nJ
D none of these
20. A 550 W heater heats 0.50×10–3 m3 of water. Water has a specific heat of 4190 J/kg/°C and density of 1000 kg/m3.
Ignoring the small heat taken by the container, the time it takes to raise the water temperature from 21 °C to boiling is
closest to … (assume standard atmospheric pressure)
Solution: The water heater raises the temperature of 0.50 L or m=0.50 kg of water by T=79 °C (water boils at 100
°C). Power is energy (in this case in the form of heat Q) delivered per time period t: P=Q/ t. Thus t=Q/P. The
heat required to heat this amount of water is Q=mc T=1.66×105 J. Thus, t=P/Q=301 seconds, which is close to 5
minutes.
A 30 seconds
B 1.0 minute
C 5.0 minutes
D 30 minutes
E none of these
S
21. The C=20.0 F capacitor is initially uncharged. Then, switch S is closed. Immediately after
the switch is closed, the current through the R=300 resistor is closest to …
R
Solution: The capacitor is initially uncharged and will initially have a 0 V potential difference
C
across its plates. Thus, initially, the resistor sees the full 16 V voltage drop of the battery and
16V
–2
the initial current is 16 V/ 300 =5.33×10 A=53.3 mA.
After 12 ms, the capacitor will be partially charged and now has a voltage drop across its
plates and the current will be decreased from its initial value. The characteristic time of this exponential decrease is
RC=6.0 ms, i.e. after 6 ms the current will be 1/e of its initial value, and after another 6 ms it is down by another
factor 1/e. Thus, after 12 ms, the current will be 53.3 mA / e2 =7.21 mA.
A 53 mA
B 20 mA
C 7.2 mA
D 0 mA
22. After 12 ms, the current through the resistor will be closest to …
A 53 mA
B 20 mA
C 7.2 mA
D 0 mA
23. The nodes of Ranvier on a myelinated axon of radius 20 m are spaced 0.80 mm apart. The ionic resistance in the
axon between successive nodes is 20 M . The myelinated sections between successive nodes have a capacitance of
2.0 pF. The signal speed along this axon is closest to …
Solution: The characteristic time constant of a section of myelinated axon between two nodes is RC=20 M × 2.0
pF=40 s. This is the time it takes for the potential to rise to 1/e of its peak value at the next node. The rise is enough
to trigger another action potential at this next node. Thus the propagation is 0.80 mm / 40 s=20 m/s.
A 0.50 m/s
B 1.0 m/s
C 10 m/s
D 20 m/s
E 25 m/s
24. In the resting state of a cell, the ion pumps in the membrane … (5 points)
Solution: The ion pump pumps 3 sodium ions out of the cell for every 2 potassium ions it pumps in. The INSIDE
potential is –70 mV relative to the OUTSIDE. The pump is powered by the chemical energy from the reaction ATP
ADP + P, and it pumps AGAINST the electric field (and thus cannot use its energy) and does NOT use solar energy.
A maintain a
potential of about
–70 mV on the
outside relative to
the inside of the
cell
B pump 3 sodium
C pump 3 sodium
ions out of the cell
ions into the cell
for every 2
for every 2
potassium ions
potassium ions
they pump into the
they pump out of
cell
the cell
D are powered by the E use solar energy
strong electric field
for their pumping
inside the cell’s
action
membrane
Additional Midterm #1 Textbook Practice Problems Chapter 20: Problems #14, 26, 36, 44 Chapter 21: Problems #12, 18, 22, 35, 60, 77 Chapter 22: Conceptual/Multiple Choice #24, 26, 27; Problems #3, 26, 38, 40 Chapter 23: Conceptual/Multiple Choice #31, 32, 34; Problems #26, 39, 40, 44, 48, 50 Chapter 20 Solutions: Problems #14, 26, 36, 44 P20.14. Prepare: Please refer to Figure P20.14. Charges A, B, and C are point charges. Charge A experiences an



electric force FB on A due to charge B and an electric force FC on A due to charge C. The force FB on A is directed to the right,

and the force FC on A is directed to the left.
Solve: Coulomb’s law yields:
FB on A =
(9 ×109 N ⋅ m2 /C2 )(1.0 ×10−9 C)(1.0 ×10−9 C)
= 9.0 ×10−5 N
(1.0 ×10−2 m)2
FC on A =
(9 ×10 N ⋅ m2 /C2 )(1.0 ×10−9 C)(4.0 ×10−9 C)
= 9.0 ×10−5 N
(2.0 × 10−2 m)2
9
The net force on A is



Fon A = FB on A + FC on A = (9.0 × 10−5 N, + x-direction) + (9.0 × 10−5 N, − x-direction) = 0 N
Assess: The force on A by C is the same (but in the opposite direction) as that of B on C because C has four times the
charge and is twice the distance away compared to B. Check this statement against Coulomb’s law!
P20.26. Prepare: A field is the agent that exerts an electric force on a charge. Because the weight of the plastic ball
acts downward, the electric force must act upward.



Solve: Newton’s second law on the plastic ball is Σ( Fnet ) y = Fon q − w. To balance the weight with the electric force,
Fon q = w ⇒ |q| E = mg ⇒ E =
mg (1.0 × 10−3 kg)(9.8 N/ kg)
=
= 3.3 × 106 N/C
|q|
3.0 × 10−9 C
Because Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball must be

pointing downward. Thus E = (3.3 × 106 N/C, downward).






Assess: F = q E means the sign of the charge q determines the direction of F or E . For positive q, E and F are pointing in


the same direction. But E and F point in opposite directions when q is negative.


P20.36. Prepare: Equation 20.8 tells us the force on a charged object in an electric field: F = q E.
on q
We are given q = e and E = 1.0 × 107 N/C.
Solve:
Fonq = qE = (1.6 × 10−19 C)(1.0 × 107 N/C) = 1.6 × 10−12 N
Assess: Notice the C’s cancel out leaving units of N. The answer is very small, but that is what we expect for such a
small charge.
P20.44.
Solve: (a) The electric field strength of q1 = + q = 1.0 nC is
E1 = K
|q1|
r
2
1
=
(9.0 ×109 N ⋅ m2 /C2 )(1 .0×10−9 C)
= 89,109 N/C
(0.0010 m)2 + (0.010 m)2
Similarly, the electric field strength of q2 = –q = 1.0 nC is
E2 = K
|q2|
r 22
=
(9.0 ×109 N ⋅ m2 /C2 )(1 .0×10−9 C)
= 89,109 N/C
(0.0010 m)2 + (0.010 m)2
We will now calculate the components of these electric fields. The electric field due to q1 is away from q1 in the fourth
quadrant and that due to q2 is toward q2 in the third quadrant. Their components are
E1x = E1 cosθ
E1y = −E1 sinθ
E2x = −E2 cosθ
E2 y = −E2 sinθ
The x and y components of the net electric field are:
(Enet ) x = E1x + E2 x = 0 N/C
(Enet ) y = E1y + E2 y = − E1 sin θ − E2 sin θ = −178,218 sin θ N/C = − 18,000 N/C
The electric field strength at (10 mm, 0 mm) is thus 18,000 N/C.
(b) The electric field strength of q1 = + q = 1.0 nC is
E1 = K
|q1|
r 21
=
(9.0 × 109 N ⋅ m2/C2 )(1 .0 × 10−9 C)
= 111,111 N/C
(0.0090 m)2
Similarly, the electric field strength of q2 = – q = 1.0 nC is
E2 = K
|q2|
r 22
=
(9.0 × 109 N ⋅ m2/C2 )(1 × 10−9 C)
= 74,380 N/C
(0.0110 m)2
The electric field strength at (0 mm, 10 mm) is thus 111,111 N/C – 74,380 N/C = 36,730 N/C ≈ 37,000 N/C. Chapter 21 Solutions: Problems #12, 18, 22, 35, 60, 77 P21.12. Prepare: The electric potential difference between the plates is determined by the uniform electric field in the
parallel-plate capacitor and is given by Equation 21.6.
Solve: (a) The potential difference ΔVC across a capacitor of spacing d is related to the electric field inside by Equation
21.6:
ΔVC
E=
⇒ ΔVC = Ed = (1.0 × 105 V/m)(0.002 m) = 200 V
d
(b) The electric field of a capacitor is related to the charge by Equation 20.7:
Q = ε 0 AE = (8.85 × 10−12 C2 /(N ⋅ m 2 ))(4.0 × 10−4 m 2 )(1.0 × 105 V/m) = 3.5 × 10−10 C
Assess: A charge of 0.35 nC on the positive plate and an equal negative charge on the negative plate create a significant
potential difference across the parallel.
P21.18. Prepare: Please refer to Figure P21.18. The net potential is the sum of the potentials due to each charge given
by Equation 21.10.
Solve: The potential at the dot is
V=
& 2.0 × 10−9 C 2.0 × 10−9 C 2.0 × 10−9 C )
1 q1
1 q2
1 q3
+
+
= (9.0 × 109 N ⋅ m 2 /C2 ) (
+
+
+ = +1400 V
4πε 0 r1 4πε 0 r2 4πε 0 r3
0.050 m
0.030 m *
' 0.040 m
Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
P21.22. Prepare: The magnitude of the electric field in terms of the potential difference ΔV between two points a
distance d apart is given by Equation 21.17.
Solve: When the electric field is uniform,
ΔV = Ex d = Ex Δx = (1000 V/m)(0.30 m − 0.10 m) = 200 V P21.35. Prepare: Equation 21.22 gives the capacitance of a parallel-plate capacitor with a dielectric (the paper).
C=
κε 0 A
d
We are given A = (0.35 m)2 = 0.1225 m 2 and d = 0.25 × 10−3 m. Assuming the science fair is at 20°C, we also look up the
dielectric constant of paper in Table 21.2: κ paper = 3.0.
Solve:
C=
κε 0 A (3.0)(8.85 × 10−12 F/m)(0.1225 m 2 )
=
= 13 nF
d
0.25 × 10−3 m
Assess: The answer could be expressed in scientific notation as 1.3 × 10−8 F but capacitances are usually given in pF
(sometimes pronounced “puff”), nF, or µ F. In our calculation the m2 cancels, leaving F.
P21.60. Prepare: Please refer to Figure P21.60. The proton at point A is at a potential of 30 V and its speed is 50,000
m/s. At point B, the proton is at a potential of −10 V and we are asked to find its speed. Clearly, the proton moves into a
lower potential region, so its speed will increase. Energy is conserved. Also note that the potential energy is determined by
the electric potential.
Solve: The conservation of energy equation Kf + Uf = Ki + Ui is
1 2
1
mv + (+e)(−10 V) = m(50,000 m/s)2 + (+e)(30 V)
2 f
2
⇒ vf = (50,000 m/s)2 +
2(1.60 × 10−19 C)(40 V)
= 1.0 × 105 m/s
1.67 × 10−27 kg
Assess: The speed of the proton is higher, as expected.
P21.77. Prepare: When the equipotential lines are widely spaced, as at A, it means the potential isn’t changing as
quickly. From Equation 21.17, E = (ΔV )/d, this means the field is weaker there.
Solve: (a) Even though they are on the same equipotential line, the electric field strength at A is smaller than the electric
field strength at B because the equipotential lines are farther apart at A.
(b) The same argument applies here. Since the equipotential lines are farther apart at C than at D, then the electric field
strength at point C is smaller than the field strength at D.
(c) We can tackle the direction first. We know that the electric field lines are perpendicular to the equipotential
lines and that they point in the direction of decreasing potential. So the field line at D points in the positive xdirection.
In order to get a feel for the spacing of the equipotential lines, imagine one more at 3 V going perpendicularly down
through the origin. Along the x-axis the equipotential lines are quite evenly spaced. Then we can calculate the potential
difference (by counting equipotential lines or by subtracting the last from the first) from the point (0 cm, 0 cm) to the point
(1 cm, 0 cm). It looks like about a 5.5 V potential difference between those two points, so E = 5.5 V/cm, in the positive xdirection.
Assess: In part (b) it does not matter that point C is not on a drawn equipotential line. You could estimate its potential
as −1.5 V, halfway between the adjacent equipotential lines, but even that is not needed to answer the question.


In part (c) a positive little test charge at point D would feel a force equal to F = q E, so it would feel a force to the right,
and that would be rolling down the hill, as we expect.
Chapter 22 Solutions: Conceptual/Multiple Choice #24, 26, 27; Problems #3, 26, 38, 40 Q22.24. Reason: The resistance of a wire may be determined by R = ρ L/A. In this case the resistivity is a constant. The
resistivity is a function of the material used to manufacture the wire and that will not change as
the wire is stretched. At this point the problem is complicated by the fact that as the length of the wire increases, the crosssectional area decreases. The problem can be simplified by realizing that the cross sectional area can be related to the
length. This can be done by realizing that the mass and volume of the wire remain constant as it is stretched. We aren’t
creating or destroying any wire, we are just changing its shape. We could use the mass as the quantity that is constant, but
that would result in our using the mass density and volume of the wire—so let’s just agree to use the volume. Recall that
the volume of the wire is determined by V = LA. We can use this expression for the volume to eliminate either the length L
or the area A; let’s agree to eliminate the area. In that case the expression for the resistance becomes
R = ρ L/A = ρ L/(V /L) = ρ L2 /V.
Since the resistivity and the volume are constant, we see that as the wire is stretched the resistance will increase.
Furthermore we are able to say how much it will increase. For example, if the length of the wire doubles, the resistance
will increase by a factor of four.
Based on this information, as the wire is stretched, the resistivity remains the same and the resistance increases. The
correct choice is C.
Assess: The question was simplified by reducing the expression for the resistance to just one variable. We used length as
the variable. You should repeat the solution using the area and see if you obtain the same result.
Q22.26. Reason: The power output for the speaker may be determined by P = ΔV 2/R. Since the resistance of the
speaker is a constant, the power output depends on the square of the electric potential difference. If you increase the
potential difference across the speaker by a factor of 2 you will double the power output of the speaker. This new
potential difference may be calculated by ΔVnew = (1+ 2 )ΔVold = 2.41(5.0 V) = 7.1 V.
The correct choice is A.
Assess: We have obtained this value by inspecting the basic relationship between power, resistance, and electric
potential difference.
Q22.27. Reason: Reading about energy and time in the question makes us think of power = energy/time. The other
information in the question gives us voltage and current, and we also know that P = I ΔV. So we tie these ideas together.
time =
energy energy
5000 J
=
=
= 2777 s ≈ 46 min
power
I ΔV
(1.2 A)(1.5 V)
The correct choice is D.
Assess: Rechargeable batteries are often rated in mA ⋅ h, which, assuming the standard voltage 1.2 V for most
rechargeables, allows us to calculate the energy stored. For example, for a 2000 mA ⋅ h battery (a typical value for a
NiMH AA battery)
energy = PΔt = I ΔV × Δt = (I Δt) × ΔV = (2000 mA ⋅ h)(1.2 V)(3600 s/h) = 8.6 kJ
This result is comparable to the one in the question, with the times (46 min versus 1 h), currents (1.2 A versus 2 A), and
voltages (1.5 V versus 1.2 V), a little different but in the same general range.
P22.3. Prepare: Current is defined in terms of the amount of charge ΔQ passing through a cross section of a wire in a
time interval Δt: I = ΔQ/Δt.
We will also need to know how many electrons are in a coulomb. In Chapter 20 we learned that the charge on one electron
is − e = −1.6 × 10−19 C, so there are 1/1.6 × 10−19 electrons in one coulomb.
Solve: Solve the equation for ΔQ.
$ 1 electron '
ΔQ = I Δt = (1.0 A)(30 s) = 30 C &
= 1.9 × 1020 electrons
% 1.6 × 10−19 C )(
Assess: It is hard to get our minds around such large numbers, but try we must. That many electrons really do go past a
point (cross section) of the wire in those conditions. The electrons generally aren’t traveling very fast, but a lot of them go
by to make up a 1.0 A current. And remember that when electrons are the charge carriers, they go in the opposite direction
from the direction of the current.
P22.26. Prepare: The current I in a wire when a potential difference is applied to the ends of the wire can be obtained
from Equation 22.6, I = ΔV /R, and Equation 22.8, R = ρ L/A. The resistivity of nichrome from Table 22.1 is 1.5 × 10−6 Ω ⋅ m.
Solve:
# A & (3.0 V)π (0.40 × 10−3 m)2
I = ΔV %
=
= 2.0 A
$ ρ L (' (1.5 × 10−6 Ω ⋅ m)(0.50 m)
Assess: The resistivity of nichrome is small, so a current of 2.0 A is reasonable.
P22.38. Prepare: Electric power, potential, and resistance are related by P = V 2 / R. Electric potential, current, and
resistance are related by Ohm’s Law R = V / I.
Solve: (a) The resistance of the wire in the blanket is R = V 2 / P = 4.6 Ω (b) The current in the electric blanket is
I = V / R = 3.9A
P22.40. Prepare: The relationships needed for this problem are P = IV , P = ΔU / Δt and Q = I Δt.
Solve: (a) The power is P = IV = 360 W.
(b) The total energy is ΔU = PΔt = 0.36 J.
(c) The total charge that flows is Q = I Δt = 8.0 × 10−4 C.
Assess: Given the electric potential, current, and time pulse, these are reasonable values for the power, total energy, and
total charge.
Chapter 23 Solutions: Conceptual/Multiple Choice #31, 32, 34; Problems #26, 39, 40, 44, 48, 50 Q23.31. Reason: The power dissipated is determined by P = I 2 R. Since the resistors are in series, they each have the
same current. As a result, the larger resistor will dissipate the most power. The correct choice is B.
Assess: If the current in two resistors is the same, the power dissipated will depend only on the resistance of
the resistors.
Q23.32. Reason: The wattage ratings on the bulbs are only meaningful when the voltage context is taken into account.
Let’s first calculate the resistance of each bulb by assuming they are in the normal household situation (parallel across 120
V).
For the bulb labeled 40 W:
R=
(ΔV)2 (120 V)2
=
= 360 Ω
P
40 W
R=
(ΔV)2 (120 V)2
=
= 240 Ω
P
60 W
For the bulb labeled 60 W:
The resistances will not change when we put them in the new series circuit. What is different is that the current is the same
through both bulbs, rather than having the potential difference across them be the same. So we use for the power (i.e., the
brightness of the bulbs) P = I 2 R. When the currents are the same, the one with the greater resistance will dissipate the most
power and glow brighter; this is the bulb labeled as 40 W.
The correct choice is B.
Assess: It sounds backward for the 40 W bulb to glow brighter than the 60 W bulb, but that is the case when they are put
in series and have the same current through them. The 60 W bulb only dissipates 60 W when it is connected across a 120
V source.
Q23.34. Reason: First use Ohm’s law to find Req .
Req =
ΔV 8.0 V
=
= 4.0 Ω
I
2.0 A
Now we use the formula for resistors in parallel:
1
1
1
1
=
+
+
Req R1 R2 R3
Solve for R3 where Req = 4.0 Ω, R1 = 10 Ω, and R2 = 15 Ω.
−1
−1
" 1
" 1
1
1%
1
1 %
R3 = $
−
− ' =$
−
−
= 12 Ω
# 4.0 Ω 10 Ω 15 Ω '&
# Req R1 R2 &
The correct choice is B.
Assess: We know the equivalent resistance is less than any of the individual resistors in parallel, and our answer fits that
criterion.
P23.26. Prepare: Assume ideal batteries and wires and ohmic resistors. Label the resistors from left to right 1, 2, 3, and
4 (so R2 = 10 Ω ). Compute the total resistance by first adding R3 + R4 = 10 Ω . Then that combination in parallel with
R2 gives 5.0 Ω . Lastly, Rtot = R1 + 5.0 Ω = 10 Ω .
Solve: The current through the battery (and R1 ) is then I = ΔV / Rtot = (10 V) / (10 Ω) = 1.0 A. From the junction law
and symmetry that current splits up evenly in the other two branches, so I 2 = I 3 = I 4 = 0.50A.
Now
ΔV1 = I1 R1 = (1.0 A)(5.0 Ω) = 5.0 V.
The
loop
law
using
the
battery,
R1 and R2
,
shows
ΔV2 = 10 V − 5.0 V = 5.0 V. The potential difference of 5.0 V is split between R 3 and R4 so ΔV3 = ΔV4 = 2.5 V.
R I (A) ΔV (V)
R1 1.5
5.0
R2 0.50 5.0
that
R3 0.50 2.5
R4 0.50 2.5
Assess: From symmetry we expect the results to be the same for R3 and R4 .
P23.39. Prepare: For capacitors in series we know the equivalent capacitance is less than any of the individual
capacitances. The charge on capacitors in series is the same.
Solve:
(a) All three are in series, so
−1
" 1
1
1 %
Ceq = $
+
+
= 0.5714 µ F ≈ 0.57 µ F
# 4.0 µ F 2.0 µ F 1.0 µ F '&
(b) The charge on each capacitor is the same as we would calculate on an equivalent capacitor.
Q = CΔV = (0.5741 µ F)(12 V) = 6.9 µC
Assess: Indeed, the equivalent capacitance is less than the smallest capacitor.
P23.40.
Prepare: The circuit in Figure P23.40 has an equivalent circuit with resistance Req and capacitance C = 1.0
µ F. Assume ideal wires as the capacitor discharges through the two 1 k resistors.
Solve: The equivalent resistance is Req = 1.0 kΩ + 1.0 kΩ = 2.0 kΩ . Thus, the time constant for the discharge of the
capacitors is
τ = RCeq = (2.0 kΩ )(1.0 µF) = 2.0 × 10 3 s = 2.0 ms
−
Assess: The capacitor will be almost entirely discharged 5τ = 5 × 2.0 ms = 10 ms after the switch is closed.
P23.44.
Prepare: The capacitor discharges through a resistor. The switch in the circuit in Figure P23.44 is in position
a. When the switch is in position b the circuit consists of a capacitor and a resistor. Current and voltage during a capacitor
discharge are given by Equations 23.22. Because the charge on a capacitor is Q = CΔV, the decay of the capacitor charge
is given by Q = Q0 e t/ .
Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a charge
Q0 = CΔV = Cε = (2 µ F)(9 V) = 18 µC.
−
τ
Immediately after the switch is moved to the b position, the charge on the capacitor is Q0 = 18 µC. The current through the
resistor is
I0 =
ΔVR
9V
=
= 0.18 A = 180 mA
R
50 Ω
Note that as soon as the switch is closed, the potential difference across the capacitor ΔVC appears across the 50 Ω resistor.
(b) The charge Q0 decays as Q = Q0 e t / , where τ = RC = (50 Ω)(2 µ F) = 100 µs.
−
Thus, the charge is Q = (18 µC)e
τ
−50 µs/(100µs)
= (18 µC)e−0.5 = 10.9 µC = 11 µC.
The resistor current is I = I 0 e−t /τ = (180 mA)e−50 µs/(100µs) = 110 mA.
(c) Likewise, the charge is Q = 2.4 µC and the current is I = 24 mA.
Assess: All of these values seem reasonable.
P23.48. Prepare: Follow the text and assume the conduction speed is approximately the distance between the nodes
divided by the RC time constant.
Solve:
v=
Lnode
0.80 × 10−3 m
=
= 33 m/s
τ
(20 MΩ)(1.2 pF)
Assess: This result is similar to the one in the chapter. The units do work out since Ω ⋅ F = s.
P23.50. Prepare: Halving the thickness of the myelin would double the capacitance of the membrane.
Solve: The resistance between axons is not changed, so τ ! = R(2C) = 2τ .
v! =
Lnode 1
1
= v = (40 m/s) = 20 m/s
2τ
2
2
Assess: Thicker myelin helps conduction speed.