CIRCUITS
23
Q23.1. Reason: No, this is not a complete circuit. A connection from the outer metal case of the bulb to the
negative terminal of the battery would complete the circuit and light the bulb.
Assess: A complete circuit can be made in a few different ways with just one wire, one bulb, and one battery.
In fact, if your bulb is rated for 9 V you can even do it without the wire since both terminals of a 9 V battery are
on the same end.
Q23.2. Reason: The current is the same at every point in the circuit. There is no place to get more charge
carriers and there is no place to store charge carriers. For every charge carrier that enters the battery, another one
leaves the battery, hence the rate of flow of charge carriers is the same everywhere.
Assess. The water analogy used in the previous chapter is useful here. If you have a water hose filled with
water and you add one drop at one end, one drop will exit the other end. The rate of flow of water at every point
in the hose is one drop.
Q23.3. Reason: (a) I out = I in because of conservation of charge. There are no branch points and the charge
carriers must go somewhere (they don’t build up).
(b) As we saw in the previous part, the current must be the same everywhere in the wires and resistors. Since
R = ΔV/I and the current is the same for all, then we look at the graph to see what the potential drop across each
resistor is: Δ V3 > Δ V1 > Δ V2 . Therefore R3 > R1 > R2 .
Assess: Greater resistances have a greater voltage drop across them for a given current.
Q23.4. Reason: The power dissipated by a resistor is determined by P = I 2R . Since the current is the same in
both resistors, the larger resistor, R1, will dissipate more power.
Assess: If the current is the same in two resistors, the one with the larger resistance will dissipate more power.
Q23.5. Reason: Since the resistors are connected in parallel across the battery they have the same potential
difference (voltage drop) across them. Hence, we use this version of the power equation: P =
( ΔV )2
R
.
The resistor with the lower resistance will dissipate more power (since R is in the denominator); that is R2 .
Assess:
You might wonder why we don’t apply P = I 2R for the power instead of P = (Δ V ) 2 / R ; that would
seem to give the opposite answer. It is because I is not the same for the two resistors, so P = I 2R doesn’t allow a
comparison based on R since we don’t know the currents. We used P = ( ΔV ) 2 / R because we know that ΔV is the
same for both resistors.
Think about this result in terms of light bulbs. The conclusion is that the resistance of a 100 W light bulb is lower
that the resistance of a 60 W light bulb (assuming 120 V in both cases). Measure the resistance of some light
bulbs with an ohmmeter.
Q23.6.
Reason: When the switch is closed, bulb B will go out. The charge carriers have been given an
essentially resistance-free alternative and they will take it. Bulb A will get brighter. To see why, let’s examine
the expression I = V / R . When the switch is closed, as a result of the negligible resistance of the switch and wire,
as far as charge carriers are concerned the resistance has decreased by a factor of two and the electric potential
supplied to the circuit has not changed, so the current will increase by a factor of two and the bulb will appear
much brighter.
Assess: Less resistance will result in more current when the electric potential supplied to the circuit remains the same.
23-1
23-2
Chapter 23
Q23.7. Reason: (a) Since both points a and c are at the same potential as the positive terminal, and both
points b and d are at the same potential as the negative terminal, then Δ Vab = Δ Vcd .
(b) I1 > I2 = I3
Assess: Because the three identical resistors in series have a total resistance three times that in the left circuit,
the current in the right circuit will be 1/3 the current in the left circuit, and so the voltage drop across each of the
three resistors will be 1/3 the voltage drop across the lone resistor, but the total potential difference will
be 3 × (1/3) and Δ Vab = ΔVcd , just as we found from our earlier reasoning.
Q23.8. Reason: (a) Since the batteries are identical we can state Δ Vab = ΔVcd = Δ Vef .
(b) Since the current is the same everywhere in a simple circuit like the one on the left, we can
write I1 = I 2 = V / R . The current I3 is determined by
I3 =
V
V
V
=
= 3 = 3 I1
Rtotal 1/3 R
R
Since the current I3 is split equally into three currents, we can write
1
1
I 4 = I 5 = I 3 = (3 I1 ) = I1
3
3
Putting this all together we have
I 3 > I1 = I 2 = I 4 = I 5
Assess: Ohm’s law, the junction rule, and a knowledge of how resistors add in parallel is needed in order to
solve this problem.
Q23.9. Reason: (a) The current coming from the left into the junction will split according to the ratio of the
resistances it sees on the branches of the junctions. The lower the resistance of a branch, the more current will go
there. Since the 3 Ω branch has one third the resistance as the other branch, three times as much current will go
down the 3 Ω branch. So the fraction of current going through the 3 Ω resistor will be three quarters of the total.
(b) If the 9 Ω resistor were replaced with a larger resistor, then even less current would go through it, so the
fraction going through the 3 Ω resistor would increase.
Assess: Take the question to the extreme: If the 9 Ω resistor were replaced with a much, much larger resistor,
then almost no current would go through it, and nearly all the current would go through the 3 Ω resistor. In fact, if
you want to you could consider a single resistor to be in parallel with a resistor of infinite resistance.
Q23.10. Reason: Since resistors in series add as RTotal = R1 + R2 + R3 and R2 = 50 Ω, there is no way the total
resistance can be less then or equal to 50 Ω. The total resistance of the three resistors is greater than 50 Ω.
Assess: The total resistance of three resistors in series will be greater than any of the resistors.
Reason: Charges have more pathways with the three resistors than they would with just the 200 Ω
resistor, so the total resistance must be less than 200 Ω . Adding resistors in parallel always decreases the total
resistance for this reason.
Assess: The result would be the same if only one of the unknown resistors were in parallel instead of two.
We can check the original result by using the formula for the equivalent resistance. The three resistors in parallel
are equivalent to a resistor of resistance Req .
Q23.11.
1
1
1
1 2
1
400 Ω + R
= +
+ = +
=
Req R 200 Ω R R 200 Ω (200 Ω) R
Req =
(200 Ω) R
200 Ω
=
400 Ω + R 1 + 400R Ω
(
)
Now take limits: as R → 0 Ω, Req → 0 Ω , and as R → ∞ Ω, Req → 200 Ω . So Req is between 0 Ω and 2 00 Ω .
Circuits 23-3
Q23.12.
Reason: I1 is the total current in the circuit and it has the largest value. Notice that I1 splits into I2
and I3, hence they are both smaller than I1. Since R3 is larger than R2, I3 will be smaller than I2. This allows us to
rank the currents as follows: I1 > I 2 > I 3 .
Assess: The current in a branch of the circuit depends on the resistance of that branch of the circuit.
Q23.13. Reason: The brightness of the bulbs is determined by how much power is dissipated in each bulb,
so we use the power equation. When the resistances of the bulbs we are comparing are the same we compare
currents through the bulbs: P = I 2R .
Bulb A is in parallel with the series combination of B and C. Since bulbs B and C are in series, they have the
same current through them, so they will be equally bright. The resistance of the B-C combination is greater than
the resistance of bulb A, so the current through bulb A is greater than the current through bulbs B and C.
Therefore I 2 R for bulb A is larger than for bulbs B and C: PA > PB = PC . This is the ranking of the brightness.
Assess: It often helps novices to re-draw the diagram to see more easily which bulbs are in parallel and which
are in series.
Q23.14. Reason: All of the current in the circuit goes through bulb D, so it is the brightest. The same
current goes through bulbs A and B, so they are equally bright. Since all the bulbs are identical, the parallel
branch with bulbs A and B has twice the resistance as the parallel branch with bulb C and hence half the current.
This allows us to rank the brightness of the bulb as follows: D > C > A = B.
Assess: The brightness of the bulb depends on the current and the current depends on the resistance.
Q23.15. Reason: The brightness of the bulbs is determined by how much power is dissipated in each bulb,
so we use the power equation. When the resistances of the bulbs we are comparing are the same (as is the case
here) we compare currents through the bulbs.
P = I 2R
Bulbs D and E have the same current through them so they are equally bright.
We need to examine the resistance of each major branch. They can be compared in one’s head to arrive
at RABC < RDE because of the parallel bulbs B and C. But we can also be more quantitative about it. Call the
resistance of each bulb R; then the resistance of the ABC branch is RABC = (3/2) R and the resistance of the DE
branch is RDE = 2 R.
Because ΔV is the same across each major branch we can use Ohm’s law over both branches
Δ V = I ABC RABC = I DE RDE .
I ABC
R
2R
4
= DE
=
I DE
RABC (3/2) R 3
The current in bulb A is IABC and so we see that bulb A is brighter than bulbs D and E.
I B = I C = (1/2) IA = (1/2) I ABC = (2/3) I DE , so bulbs B and C are not as bright as bulbs D and E.
PA > PD = PE > PB = PC . This is the ranking of the brightness.
Assess: When you have a thorough understanding and a bit of practice then this question can be answered
almost at a glance. All of the details we worked through make sense intuitively.
Q23.16.
Reason: (a) All the current in the circuit goes through bulb A, so it is the brightest. Since the bulbs
are identical, the current through A splits equally to go through bulbs B and C, so they are equally bright but less
bright than A. This reasoning allows us to rank the brightness of the bulb as follows: A > B = C.
(b) When a wire is placed between points 1 and 2, bulb C goes out. We say that the bulb has been shorted out.
This means that the charge carriers have found an easier (essentially no resistance) route between points 1 and 2
23-4
Chapter 23
than going through the bulb. Since no charge carriers travel through the bulb, it does not light up—it has gone
out. After the wire is placed between points 1 and 2, bulbs A and B are now in series, so they have the same
current and are equally bright
Assess: The brightness of the bulb depends on the current and the current depends on the resistance of the
bulbs. By now you should be feeling fairly confident analyzing simple circuits.
Q23.17.
Reason:
Applying Kirchhoff’s loop law around the outside edge of the circuit,
Σi ΔVi = E − IR − ΔV12 = 0 ⇒ ΔV12 = E − IR
That is, the potential difference ΔV12 between points 1 and 2 is potential supplied by the battery minus the potential lost
in resistor R . When bulb B is in place, a current I exists through the resistor and the bulb. In that case, Δ V12 is less
than E. But if bulb B is removed the current no longer exists in that branch ( I = 0 A). Thus Δ V12 (no bulb) = E. The
answer is that the potential difference ΔV12 increases.
Assess: Removing bulb B means that there is no resistor between points 1 and 2. “No resistor” is not the same
as “no resistance.” “No resistor” means an insulator with R = ∞ Ω, not R = 0 Ω.
Ohm’s laws applies to resistors, not to empty spaces. Although I = 0 A when bulb B is removed, you cannot
use IR to conclude that ΔV12 = 0 V because there is no longer a resistor between points 1 and 2 to which Ohm’s
law could be applied.
Q23.18. Reason: (a) The total current splits and some goes through the parallel branch containing the point a
(the left branch) and some goes through the parallel branch containing the point b (the right branch). Notice that
the resistance of each branch is the same (3R), as a result the current splits equally between the two branches.
Let’s call the current in the two branches I. This current in the two branches causes a potential drop of 2IR in the
resistor 2R in the left branch and a potential drop of IR in the resistor R in the right branch. As a result, the
electric potential at point b is greater (it is more positive) than the electric potential at point a and Δ Vab ≠ 0.
(b) Since point b is at a higher potential than point a, if they are connected there will be an electric current from
point b to point a.
Assess: The solution of the question requires a good understanding of electric potential difference.
Reason: (a) When the bulb is removed the resistance of that branch becomes ∞. The equivalent
resistance of the parallel combination increases and is the resistance of the single − R branch. Since the resistance
of the parallel combination has increased, the current decreases.
(b) When the bulb is removed there is no potential difference across the resistor that is in series with the bulb
because there is no current through the resistor (ΔV = IR). So the potential difference between 1 and 2 is now the
potential difference of the parallel group. That is, removing the bulb increases the potential difference between 1
and 2.
Assess: The second part is like Question 22.17, and we arrived at the same answer here.
Q23.19.
Q23.20. Reason:
(a) A good voltmeter has very high resistance which causes the current in the incorrect circuit to be nearly zero.
(b) Because there is no current there is no potential difference (voltage drop) across the resistor, so the voltmeter
measures the emf of the battery, or 9 V.
(c) Put the voltmeter in parallel with the resistor rather than in series with it.
Assess: Voltmeters should be in parallel with the circuit element we want to know the potential difference across.
Q23.21. Reason:
(a) Because a good ammeter has very low resistance, the current would nearly all go through the ammeter, so the
current in the 5. 0 Ω resistor is nearly zero.
(b) Put the ammeter in series with the resistor rather than in parallel with it. The current will be the same
anywhere in the one-loop circuit, so it doesn’t matter exactly where the ammeter is placed in series.
Assess: Ammeters should be placed in series in the circuit where we want to know the current.
Q23.22.
Reason:
The equivalent capacitance of each grouping is determined as follows:
Circuits 23-5
Group 1: In this group we have three capacitors in series and the equivalent capacitance is 1/ Ceq = 1/ C1 + 1/ C2 +
1/ C3 or Ceq = C/3.
Group 2: In this group we have three capacitors in parallel and the equivalent capacitance is
Ceq = C1 + C2 + C3 = 3C.
Group 3: In this group we have two capacitors in parallel and this combination is in series with the third capacitor. First
determine the equivalent capacitance of the parallel combination and then consider this equivalent capacitance to be in
series with the third capacitor. The equivalent parallel capacitance is CParallel = C + C = 2C. The equivalent capacitance
for the group is determined by 1/ Ceq = 1/ CParallel + 1/ C = 1/(2C ) + 1/ C = 3/(2C ) or Ceq = 2C /3 .
Group 4: In this group we have two capacitors in series and this combination is in parallel with the third capacitor. First
determine the equivalent capacitance of the series combination and then consider this equivalent capacitance to be in
parallel with the third capacitor. The equivalent series capacitance is 1/ Cseries = 1/ C + 1/ C = 2/ C or CSeries = C /2. The
equivalent capacitance for the group is determined by Ceq = Cseries + C = C /2 + C = 3 C /2.
This allows us to rank the capacitance of the groups as follows: C2 = 3C > C4 = 3C /2 > C3 = 2 C /3 > C1 = C /3.
Assess: The solution of this question requires a good knowledge of how capacitors add in parallel and in series.
Q23.23.
Reason: (a) Both bulbs are glowing because current is in both wires as the capacitor is being charged.
(b)They are equally bright because the currents are the same. The same amount of positive charge is flowing off
the bottom plate and onto the positive plate in the same time.
(c) The brightness of the bulbs decreases with time. As the charge on the plates increases, the potential differences along
the wire decrease and so does the current in the wire. Less current means the bulbs aren’t as bright.
Assess: The bulbs eventually go completely out when the current nears zero as the capacitor becomes fully
charged and the potential difference across the plates becomes the same as the emf of the battery.
Q23.24.
Reason:
The electric potential differenced across the capacitor at any time is
Δ VC = ( ΔVC )o e −t / RC = (Δ VC )o e− t /τ
where τ = RC.
After a time equal to one time constant (i.e., t = τ = RC ), the electric potential difference across the capacitor is
Δ VC = (Δ VC )o e −t / RC = (Δ VC )o e −t /τ = (ΔVC )o e−τ /τ = (Δ VC )o e −1 = 0.37(Δ VC )o
This says that after one time constant the electric potential difference across the capacitor has fallen to 37% of its
initial value. Given this, the strategy is to mark off the t = 0 value for each case, then find the time for the electric
potential difference across the capacitor to drop down to 37% of that value for each case. This time is a time
constant. The case with the greatest time has the greatest resistance since the capacitor is the same for all three
cases. The figure shows this procedure on the graph of Δ VC vs. t.
Checking the figure we see that the time constants are related as follows:
τ 2 > τ 3 > τ1
Assess:
or
R2C > R3C > R1C
or
R2 > R3 > R 1
The greatest resistance will result in the greatest time constant and hence the longest time.
23-6
Chapter 23
Q23.25. Reason: Given that the resistors are all identical, the pair in parallel (circuit A) has a lower
resistance than the pair in series (circuit B). The current will be greater, therefore, in circuit A where the
resistance is less. Since the current is greater in A, it will discharge the capacitor in the shortest amount of time.
Assess: Don’t let the fact that the capacitor is drawn between the two resistors in circuit B fool you. The three
elements are all still in series, and it would be the same if the resistors were drawn in series next to each other.
Reason: The rate of flashing is controlled by the time constant (τ = RC ). The flashing may be
slowed down by increasing the time constant. This may be done by increasing either (or both) the resistor and the
capacitor in the circuit.
Assess: The time constant depends on both the resistance and the capacitance, if either one (or both) are varied,
the time constant will vary.
Q23.26.
Q23.27. Reason: Increasing the dielectric constant increases the capacitance. Since τ = RC , increasing the
capacitance increases the charging and discharging time.
Assess: One can therefore determine skin moisture by measuring the charging and discharging time.
Q23.28.
Reason: (a) The new resistor has half the length of the old resistor, so the new resistance is half the old.
The capacitance depends on A the area, and halving the distance between Ranvier nodes would halve A and
therefore also halve C.
(b) The new time constant is τ ′ = R′C ′ = ( 12 R )( 12 C ) = 14 τ .
(c) Also note that Δ xnode = 12 Δ xnode:
v′ =
Δ x′node 12 Δ xnode
= 1
= 2v
τ′
4τ
So the speed of propagation is twice what it was.
Assess: If the distance between the nodes of Ranvier is halved, the propagation speed is doubled. Autoimmune
diseases such as multiple sclerosis and ALS attack and destroy one’s myelin so that the nerve impulses travel
much slower.
Q23.29.
Reason: The resistance of an axon between one node and the next is fixed and fairly constant. The
myelin insulation increases the separation between the inner conducting fluid and the outer conducting fluid.
Since the capacitance of a capacitor depends inversely on the electrode spacing, the myelin reduces the
capacitance of the membrane. Reducing the capacitance reduces the time constant. Since the time constant is a
measure of how much time it takes for a signal to jump from one node to the next, an axon sheathed in myelin
will cause faster signal propagation.
Assess: The true understanding of most biological systems requires knowledge of physics.
Q23.30.
Reason:
The resistances of the resistors in series add, so that Req = 4. 0 Ω + 6. 0 Ω = 10 Ω.
The potential difference across the set of resistors is 10 V.
The current is I = Δ V / R = 10 V/(10 Ω) = 1 A.
The correct choice is A.
Assess: The current would be greater if either resistor were replaced by an ideal wire.
Reason: The power dissipated is determined by P = I 2R . 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.
Q23.31.
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=
(Δ V) 2 (120 V) 2
=
= 360 Ω
P
40 W
Circuits 23-7
For the bulb labeled 60 W:
R=
(Δ V) 2 (120 V) 2
=
= 240 Ω
P
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 2R. 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.
Q23.33.
Reason: Since the original wire has a resistance of R, when it is cut in half each segment has a resistance
R/2. When these two segments of wire are connected together, we have two resistors (of resistance R/2) in parallel and
the equivalent resistance is 1/ Req = 1/( R/2) + 1/( R/2) = 4/ R or Req = R /4. The correct choice is A.
Assess: When resistors are connected in parallel, the equivalent resistance is smaller than the smallest resistor
in the parallel combination.
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
⎜
⎟
⎝ Req R1 R2 ⎠ ⎝ 4 .0 Ω 10 Ω 15 Ω ⎠
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.
Q23.35.
Reason: The capacitance of the parallel combination is CParallel = C + C = 2C.
The capacitance of the series combination is 1/ CSeries = 1/ C + 1/ C or CSeries = C/2. This calculation shows that the
combination has four times the capacitance as the series combination. The correct choice is B.
Assess: This question requires a good knowledge of how capacitors add in parallel and in series.
Q23.36. Reason: Since the resistance depends on the length (in this case the cell wall thickness), the
resistance will increase if the membrane of the cell doubles in thickness. Since the capacitance depends inversely
on the cell wall thickness, if the thickness is doubled the capacitance will decrease. Since the resistance increases
and the capacitance decreases, the correct answer is B.
Assess: The analysis of this biological system requires an understanding of some of the basic principles of
electricity.
Q23.37. Reason: Assume a spherical cell as in Examples 23.13 and 23.14. Reducing the diameter will
decrease the surface area, which decreases the cross sectional area of the resistor (increasing the resistance) and
decreases the area of the capacitor (decreasing the capacitance). Therefore the correct answer is B.
Assess: In τ = RC = ρ ( L / A)(ε 0 A / d ) the area cancels so τ remains unchanged.
Problems
23-8
Chapter 23
Prepare: From the circuit in Figure P23.1, we see that 50 Ω and 100 Ω resistors are connected in
series across the battery. Another resistor of 75 Ω is also connected across the battery.
Solve:
P23.1.
Assess: The series resistors have the same current in them, but the parallel resistors have the same potential
difference across them.
P23.2. Prepare: In Figure P23.2, the positive terminal of the battery is connected to a resistor. The other end
of that resistor is connected to resistor and a capacitor in parallel.
Solve:
Assess: The series elements have the same current in them, but the parallel elements have the same potential
difference across them.
Prepare: Circuits are generally drawn with straight lines and 90 ° angles for the wires. We also
usually orient the battery with the positive terminal up so the higher potential part of the circuit is toward the top
of the paper.
Solve:
P23.3.
Assess: The elements that are in series with each other have the same current in them; the elements that are in
parallel with each other have the same potential difference across them.
P23.4. Prepare: Please refer to Figure P23.4. We will use Ohm’s law for each resistor and assume that the
connecting wires are ideal. The direction of the current is from higher potential to the lower potential.
Solve: The current in the 2 Ω resistor is I1 = 6 V/(2 Ω) = 3 A to the left. The current in the 5 Ω resistor is I 2 =
10 V/(5 Ω) = 2 A downward. Using Kirchhoff’s junction law, Equation 23.1, we see that I = I1 + I2 = 3 A + 2 A = 5
A. This current flows toward the junction, that is, downward.
P23.5. Prepare: Please refer to Figure P23.5. We will use Ohm’s law (ΔV = IR) for the resistor and the
bulb, use Tactics Box 23.1 for Kirchoff’s loop law, and assume that the connecting wires are ideal.
Solve: Let us assign clock direction to the current in the circuit.
(a): The Kirchoff’s loop law is:
Σ(Δ V )i = Δ Vbat + Δ Vresistor + ΔVbulb = E + (V2 − V1 ) + (V3 − V2 ) = 0
3.0 V − I (2.0 Ω) − I (1.0 Ω) = 0 ⇒ I = 1.0 A.
Circuits 23-9
So,
Δ V12 = V2 − V1 = − I (2.0 Ω) = −(1.0 A)(2.0 Ω) = −2.0 V
Δ V23 = V3 − V2 = − I (1.0 Ω) = −(1.0 A)(1.0 Ω) = − 1.0 V
ΔV34 = V4 − V3 = 0 V
(b) When the bulb is removed from the socket, no current flows in the circuit. Thus, Δ V12 = V2 − V1 = I (2.0 Ω) =
(0 A)(2.0 Ω) = 0 V. This means that points 1 and 2 are at the same potential as the positive terminal of the
battery. For the same reason, Δ V34 = V4 − V3 = 0, implying that points 3 and 4 are at the same potential as the
negative terminal of the battery. Finally, Δ V23 = V3 − V2 = V4 − V1 = −3.0 V.
Assess: The potential at point 1 is higher than at point 4. This is what you would have expected.
P23.6. Prepare: Please refer to Figure P23.6. The batteries and the connecting wires are ideal. Choose the
current I to be in the clockwise direction. If I ends up being a positive number, then the current really does flow
in this direction. If I is negative, the current really flows counterclockwise. There are no junctions, so I is the
same for all elements in the circuit.
Solve: (a) With the 9 V battery being labeled 1 and the 6 V battery being labeled 2, Kirchhoff’s loop law is
∑ ΔV = ΔV
i
bat 1
+ ΔVR + ΔVbat 2 = +E1 − IR − E2 = 0 ⇒ I =
E1 − E2 9 V − 6 V
=
= 0 .10 A
30 Ω
R
Note the signs: Potential is gained in battery 1, but potential is lost both in the resistor and in battery 2. Because I
is positive, we can say that I = 0.10 A flows from left to right through the resistor.
(b)
Assess: The graph shows 9 V gained in battery 1, ΔVR = −IR = 3 V lost in the resistor, and another 6 V
lost in battery 2. The final potential is the same as the initial potential, as required.
P23.7. Prepare: Assume ideal connecting wires and an ideal battery for which ΔVbat = E. We assume a
clockwise direction for I . Note that the choice of the current’s direction is in general arbitrary; if we guess
wrong the sign of the current will let us know. (In this simple case it is easy to see that the 6 . 0 V battery will
triumph over the 3. 0 V battery and the current will really go counter-clockwise.) Label the 3.0 V battery as 1
and the 6 . 0 V battery as 2.
Solve:
(a) Kirchhoff’s loop law, going clockwise from the negative terminal of the 3. 0 V battery is
ΣΔVi = ΔVbat1 + ΔVR + Δ Vbat 2 = 0 = 3 . 0 V − (18 Ω) I − 6 . 0 V
I=
3.0 V
1
= − A ≈ − 0 . 17 A
−18 Ω
6
The negative sign tells us the direction of the current is really to the left, contrary to our assumption.
(b)
23-10 Chapter 23
Assess: The final potential is the same as the initial potential, as required.
P23.8. Prepare: Please refer to Figure P23.8. Define the current I as a clockwise flow. There are no junctions,
so conservation of current tells us that the same current flows through each circuit element. As we go
around the circuit in the direction of the current, potential is gained in the battery (ΔVbat = Ebat = +15 V) and
potential is lost in the resistors (ΔVresistor = −IR).
Solve:
(a) From Kirchhoff’s loop law,
ΣΔVi = ΔVbat + ΔV10 + ΔV20 = 0
0 = Ebat – IR1 – IR2 = Ebat – I(R1 + R2) ⇒ I =
Ebat
15 V
=
= 0.50 A
R1 + R2 30 Ω
Now that we know the current, we can find the potential difference across each resistor:
ΔV10 = IR1 = (0.50 A)(10 Ω) = 5 V
ΔV20 = IR2 = (0.50 A)(20 Ω) = 10 V
(b)
Assess: The graph shows a 15 V gain in the battery, a 5 V loss in the 10 Ω resistor, and a loss of 10 V in
the 20 Ω resistor. The final potential is the same as the initial potential, as required.
P23.9. Prepare: Please refer to Figure P23.9. The graph shows a 2 V loss, a 6 V gain, and then a 4 V loss.
The final potential is the same as the initial potential, as required. Define the current I as a clockwise flow. As we
go around the circuit in the direction of the current, potential will be gained in the battery (ΔVbat = Ebat = +6 V)
and potential will be lost in the resistors (ΔVresistor = −IR). Because the current I in the circuit is 2.0 A, the two
resistors are 2 V/(2.0 A) = 1 Ω and 4 V/(2.0 A) = 2 Ω.
Solve: The circuit diagram is shown.
Circuits 23-11
Assess:
A 6 V gain and voltage loss of 2 V and 4 V indicate that we are dealing with a series circuit.
P23.10.
Prepare: Please refer to Figure P23.10. The three resistances in both (a) and (b) are series resistors.
We will thus use Equation 23.8 to find the equivalent resistance.
Solve: (a) The equivalent resistance is Req = 2 Ω + 3 Ω + 6 Ω = 11 Ω.
(b) The equivalent resistance is Req = 3 Ω + 3 Ω + 3 Ω = 9 Ω.
Assess: We must learn how to combine series and parallel resistors.
P23.11. Prepare: Please refer to Figure P23.11. The three resistances in (a), (b), and (c) are parallel
resistors. We will thus use Equation 23.12 to find the equivalent resistance.
Solve: (a) The equivalent resistance is
−1
⎛ 1
1
1 ⎞
Req = ⎜
+
+
⎟ = 1.0 Ω
⎝ 2.0 Ω 3.0 Ω 6.0 Ω ⎠
(b) The equivalent resistance is
−1
⎛ 1
1
1 ⎞
Req = ⎜
+
+
⎟ = 1.0 Ω
3.0
3.0
3.0
Ω
Ω
Ω⎠
⎝
(c) The equivalent resistance is
−1
Assess:
⎛ 1
1
1 ⎞
Req = ⎜
+
+
⎟ = 0.5 Ω
2.0
1.0
2.0
Ω
Ω
Ω⎠
⎝
We must learn how to combine series and parallel resistors.
P23.12. Prepare: The copper wire and the iron wire are connected in series. The composite resistance is
simply the equivalent resistance of RCu and RFe. The resistivities of copper and iron from Table 22.1 are
1.7 × 10−8 Ω ⋅ m and 9.7 × 10−8 Ω ⋅ m, respectively.
Solve:
ρ L
ρ L
R = RCu + RFe = Cu Cu + Fe Fe
ACu
AFe
Using the data for lengths and areas of cross section,
(1.7 × 10 −8 Ω ⋅ m)(0.20 m) (9.7 × 10 −8 Ω ⋅ m)(0.60 m)
R=
+
= 4.33 × 10−3 Ω + 74.1 × 10−3 Ω = 78 mΩ
π (0.5 × 10−3 m) 2
π (0.5 × 10−3 m) 2
Assess:
We didn’t expect a large resistance for the two highly conducting wires of copper and iron.
P23.13. Prepare:
For resistors in parallel,
⎛ 1
⎞
1
Req = ⎜ +
+ ⋅ ⋅ ⋅⎟
⎝ R1 R2
⎠
−1
Adding four of the resistors in parallel gives
−1
−1
⎛ 1
1
1
1 ⎞ ⎛ 4 ⎞
Req = ⎜
+
+
+
⎟ =⎜
⎟ = 0 . 25 k Ω
1
0
k
1
0
k
1
0
k
1
0
kΩ ⎠ ⎝ 1.0 kΩ ⎠
.
Ω
.
Ω
.
Ω
.
⎝
Solve: We can put four of the resistors in parallel. The total resistance is now 0 . 25 k Ω.
Assess:
There are other ways to arrive at the same Req using more from our collection of 1 . 0 k Ω resistors.
23-12 Chapter 23
P23.14. Prepare: When resistors are combined in parallel, the combination is always less than the smallest
resistor. As a result, the way to obtain the smallest resistance is to connect the resistors in parallel.
Solve: The equivalent resistance when the six resistors are connected in parallel is obtained by
1
1 1 1 1 1 1 6
= + + + + + =
or RTotal = R /6 = 1000 Ω /6 = 167 Ω
RTotal R R R R R R R
Assess: Notice that the total resistance is less than the smallest resistor. The smallest resistor is 1000 Ω and
the total resistance when the six resistors are connected in parallel is 167 Ω.
P23.15. Prepare: The resistance of the three 6.0 Ω resistors in parallel is
−1
−1
⎛ 1
1
1 ⎞ ⎛ 3 ⎞
Req = ⎜
+
+
⎟ =⎜
⎟ = 2 .0 Ω
⎝ 6.0 Ω 6.0 Ω 6.0 Ω ⎠ ⎝ 6.0 Ω ⎠
Solve: That parallel combination in series with the 3. 0 Ω resistor adds to an equivalent resistance of 5. 0 Ω.
So that’s the answer: the three 6.0 Ω resistors in parallel with each other, and then that combination in series with
the 3 . 0 Ω resistor.
Assess: There are typical resistances that one can buy; generally one can’t (cheaply) buy resistors with every
possible value of resistance. So it is common to combine the resistors you have in combinations of series and
parallel to arrive at a different value of equivalent resistance called for in the circuit.
P23.16. Prepare: When resistors are connected in parallel the combination has less resistance than any of the
individual resistors. Two 1 . 0 k Ω resistors in parallel have an equivalent resistance of 0. 50kΩ.
Solve: We can connect three such parallel pairs in series so that the total resistance is
Rtot = 0 . 50 k Ω + 0 . 50 k Ω + 0 . 50 k Ω = 1 . 5 k Ω
Assess: It is often necessary to combine standard value resistors in creative ways to arrive at a non-standard
required resistance.
P23.17. Prepare: The connecting wires are ideal with zero resistance. We have to reduce the circuit to a
single equivalent resistor by continuing to identify resistors that are in series or parallel combinations.
Solve:
For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance is
1
1
1
=
+
⇒ Req 1 = 18 Ω
Req 1 30 Ω 45 Ω
For the second step, resistors 42 Ω and Req 1 = 18 Ω are in series. Therefore,
Req 2 = Req 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω
For the third step, the resistors 40 Ω and Req 2 = 60 Ω are in parallel. So,
1
1
1
=
+
⇒ Req 3 = 24 Ω
Req 3 60 Ω 40 Ω
The equivalent resistance of the circuit is 24 Ω.
Assess: Have a good understanding of how series and parallel resistors combine to obtain equivalent resistors.
P23.18. Prepare: The connecting wires are ideal with zero resistance. For the first step, the 10 Ω and 30 Ω
resistors are in series; for the second step, the 60 Ω and 40 Ω resistors are in parallel; and for the third step, the
Circuits 23-13
24 Ω and 10 Ω resistors are in series.
Solve:
For the first step, the 10 Ω and 30 Ω resistors are in series and the equivalent resistance is 40 Ω. For the second
step, the 60 Ω and 40 Ω resistors are in parallel and the equivalent resistance is
−1
⎡ 1
1 ⎤
+
⎢
⎥ = 24 Ω
40
60
Ω
Ω⎦
⎣
For the third step, the 24 Ω and 10 Ω resistors are in series and the equivalent resistance is 34 Ω.
Assess: Have a good understanding of how series and parallel resistances are combined to get an
equivalent resistance.
P23.19. Prepare: From Kirchhoff’s junction law we can easily determine that the current through the
unknown resistor must be 2 .5 A − 1 . 5 A = 1 .0 A down. Also, the potential difference across the unknown resistor
is the same as across the 300 Ω resistor. Assume the resistors obey Ohm’s law.
Solve: The potential difference across the 3 00 Ω resistor is Δ V = IR = (1 .5 A)(300 Ω) = 300 V. The unknown
resistor must have a resistance of
R=
ΔV 300 V
=
= 300 Ω
I
1.0 A
Assess: We used a combination of Kirchhoff’s junction law and Ohm’s law to solve this one.
P23.20. Prepare: Assume ideal batteries and wires. Use Kirchhoff’s loop law (going clockwise from the
lower left corner) and solve for I . Due to the orientation of both batteries, the current must be going clockwise,
so the potential differences shown for the two resistors need to be subtracted.
Solve:
ΣΔVi = 4 . 5 V + 3. 0 V − 3 . 5 V − I (150 Ω) − 2 . 0V = 0V ⇒
I=
2V
= 13 mA
150 Ω
Assess: This is a small but reasonable current.
P23.21. Prepare: The two resistors are in parallel and so have the same voltage drop ΔV = IR across
them: Δ Vleft = Δ Vright.
Solve:
(a) We want to solve for I r ight.
Δ Vleft = ΔVr ight
I left Rleft = I r ight Rr ight
I r ight =
I left Rleft (3.0 A)(2.0 Ω)
=
= 2.0 A
3.0 Ω
R r ight
(b) Now use Kirchhoff’s junction law at the top junction.
Σ I in = Σ I out
I = 3A + 2 A
23-14 Chapter 23
So I = 5.0 A.
Assess: We see that the resistor with the greater resistance has less current, as we would expect in this parallel
situation.
When the two currents rejoin at the bottom junction the sum is again 5. 0 A.
P23.22. Prepare: Assume ideal connecting wires and an ideal battery. The three resistors are in series and we
can use Equation 23.8 to find the equivalent resistance.
Solve: As shown in Figure P23.22, a potential difference of 5.0 V causes a current of 100 mA through the three
series resistors. The situation is the same if we replace the three resistors with an equivalent resistor Req. That is,
a potential difference of 5.0 V across Req causes a current of 100 mA through it. From Ohm’s law,
5.0 V
Δ VR
Req =
⇒ R + 15 Ω + 10 Ω =
⇒ R + 25 Ω = 50 Ω ⇒ R = 25 Ω
I
100 mA
Assess:
Learn how to combine resistors and apply Ohm’s law.
P23.23. Prepare: Please refer to Figure P23.23. Assume that the connecting wire and the battery are ideal. The
middle and right branches are in parallel, so the potential difference across these two branches must be the same.
Solve: The currents in the middle and right branches are known, so the potential differences across the two
branches, using Ohm’s law, are ΔVmiddle = (3.0 A)R = ΔVright = (2.0 A)(R + 10 Ω).
This is easily solved to give R = 20 Ω. The middle resistor R is connected directly across the battery, thus (for an
ideal battery, with no internal resistance) the potential difference ΔVmiddle equals the emf of the battery. That is
E = ΔVmiddle = (3.0 A)(20 Ω) = 60 V.
Assess: Note that the potential difference across parallel resistors is the same.
P23.24.
Prepare: Please refer to Figure P23.24. The batteries are ideal, the connecting wires are ideal, and
the ammeter has a negligibly small resistance.
Solve: Kirchhoff’s junction law tells us that the current flowing through the 2.0 Ω resistance in the middle
branch is I1 + I2 = 3.0 A. We can therefore determine I1 by applying Kirchhoff’s loop law to the left loop. Starting
clockwise from the lower left corner,
+ 9.0 V – I1(3.0 Ω) – (3.0 A)(2.0 Ω) = 0 V ⇒ I1 = 1.0 A ⇒ I2 = (3.0 A – I1) = (3.0 A – 1.0 A) = 2.0 A
Finally, to determine the emf ε , we apply Kirchhoff’s loop law to the right loop and start counterclockwise from the
lower right corner of the loop:
E − I2(4.5 Ω) − (3.0 A)(2.0 Ω) = 0 V ⇒ E – (2.0 A)(4.5 Ω) – 6.0 V = 0 V ⇒ E = 15.0 V
Assess:
The currents and the emf look reasonable.
P23.25. Prepare: Assume ideal batteries and wires and ohmic resistors. Label the top resistor as 1 and the
other three from left to right 2, 3, and 4. The total resistance is
Rtot = 5 . 0 Ω +
5.0 Ω
= 6 . 667 Ω
3
Solve: The current through the battery (and R1 ) is then I = Δ V / Rtot = (10 V) / (6 . 667 Ω) = 1 . 5 A. From the
junction law and symmetry that current splits up evenly in the other three resistors, so I 2 = I 3 = I 4 = 0 . 50 A.
Now Δ V1 = I1R1 = (1 . 5 A)(5 .0 Ω) = 7. 5 V. The loop law using the battery, R1 ,
Δ V2 = 10 V − 7 . 5 V = 2 . 5 V and similarly for Δ V3 and Δ V4 .
R I (A) Δ V (V)
R1 1.5 7.5
R2 0.50 2.5
R3 0.50 2.5
R4 0.50 2.5
and
R2
shows that
Circuits 23-15
Assess: From symmetry we expect the results to be the same for R2 , R3 , and R4 .
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 = Δ V / 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 . 5 0A.
Now
Δ V1 = I1R1 = (1 . 0 A)(5 . 0 Ω) = 5 . 0 V.
The
loop
law
using
the
battery,
R1 and R2 ,
shows
that
Δ V2 = 10 V − 5 . 0 V = 5 . 0 V. The potential difference of 5.0 V is split between R 3 and R4 so Δ V3 = Δ V4 = 2 . 5 V.
R I (A) Δ V (V)
R1 1.5 5.0
R2 0.50 5.0
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 .
P23.27. Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit
in Figure P23.27 using the laws of series and parallel resistances. We have labeled the resistors as R1 = 6.0 Ω, R2
= 15 Ω, R3 = 6.0 Ω, and R4 = 4.0 Ω. Having reduced the circuit to a single equivalent resistance Req, we will
reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and
potential difference of each resistor.
Solve:
R3 and R4 are combined to get R34 = 10 Ω, and then R34 and R2 are combined to obtain R234:
1
1
1
1
1
⇒ R234 = 6 Ω
=
+
=
+
R234 R2 R34 15 Ω 10 Ω
Next, R234 and R1 are combined to obtain
Req = R234 + R1 = 6.0 Ω + 6.0 Ω = 12 Ω
From the final circuit,
I=
E
24 V
=
= 2.0 A
Req 12 Ω
Thus, the current through the battery and R1 is IR1 = 2.0 A and the potential difference across R1 is I(R1) = (2.0 A)
(6.0 Ω) = 12 V.
23-16 Chapter 23
As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors
must have the same potential difference ΔV.
In Step 1 of the previous figure, Req = 12 Ω is returned to R1 = 6.0 Ω and R234 = 6.0 Ω in series. Both resistors
must have the same 2.0 A current as Req. We then use Ohm’s law to find
ΔVR1 = (2 .0A)(6.0 Ω) = 12 V
ΔVR234 = (2.0 A)(6.0 Ω) = 12 V
As a check, 12 V + 12 V = 24 V, which was ΔV of the Req resistor. In Step 2, the resistance R234 is returned to R2
and R34 in parallel. Both resistors must have the same ΔV = 12 V as the resistor R234. Then from Ohm’s law,
I R2 =
12 V
= 0.8 A
15 Ω
I R34 =
12 V
= 1.2 A
10 Ω
As a check, IR2 + IR34 = 2.0 A, which was the current I of the R234 resistor. In Step 3, R34 is returned to R3 and R4
in series. Both resistors must have the same 1.2 A as the R34 resistor. We then use Ohm’s law to find
(ΔV)R3 = (1.2 A)(6.0 Ω) = 7.2 V
(ΔV)R4 = (1.2 A)(4.0 Ω) = 4.8 V
As a check, 7.2 V + 4.8 V = 12 V, which was ΔV of the resistor R34.
Resistor
R1
R2
R3
R4
Potential difference (V)
12
12
7.2
4.8
Current (A)
2.0
0.8
1.2
1.2
The three steps as we rebuild our circuit are shown.
Assess: This problem requires a good understanding of how to first reduce a circuit to a single equivalent
resistance and then to build up a circuit.
P23.28. Prepare: The circuit reduction process shows that Req = 10 Ω . So the current through the battery (and
the first resistor) is I = ΔV / Req = (10 V) / (10 Ω) = 1. 0 A.
Solve: The potential difference across the first resistor is Δ V = IR = (1 .0 A)(5 . 0 Ω) = 5 . 0 V. Use the loop law
around the left-most loop to deduce that Δ V14 = 5 . 0V.
Moving on, the equivalent resistance of the four right-most resistors (those not in the left-most loop) is
Req = 10 Ω , so the junction law tells us the current splits evenly at point 1, giving 0. 50 A through the middle
horizontal resistor. Then the potential difference across that middle horizontal resistor is
Δ V = IR = (0 . 50 A)(5 . 0 Ω) = 2 . 5 V. Use the loop law around the loop containing the battery and point 2 (but not
point 3) to deduce that Δ V24 = 2 . 5 V.
Because both downstream branches from point 3 have the same resistance ( 10 Ω ) the current splits evenly there
too, giving the current through the right-most horizontal resistor of I 3 = 0 . 2 5 A. The potential difference across
that resistor is then Δ V = IR = (0 .25 A)(5 .0 Ω) = 1 .25 V. The loop law around the outer loop then gives
Δ V34 = 1 . 25 V ≈ 1 . 3 V .
Assess: Checking with other calculations (such as different applications of the loop law) give the same results.
P23.29.
Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the
circuit in Figure P23.29 using the laws of series and parallel resistances. Having reduced the circuit to a single
Circuits 23-17
equivalent resistance, we will reverse the procedure and “build up” the circuit using the loop law and the junction
law to find the current and potential difference of each resistor.
Solve:
From the last circuit in the diagram,
E
12 V
=
= 2.0 A
6.0 Ω 6.0 Ω
Thus, the current through the battery is 2.0 A. As we rebuild the circuit, we note that series resistors must have
the same current I and that parallel resistors must have the same potential difference ΔV.
I=
In Step 1, the 6.0 Ω resistor is returned to a 3.0 Ω and 3.0 Ω resistor in series. Both resistors must have the same
2.0 A current as the 6.0 Ω resistance. We then use Ohm’s law to find
ΔV3 = (2.0 A)(3.0 Ω) = 6.0 V
As a check, 6.0 V + 6.0 V = 12 V, which was ΔV of the 6.0 Ω resistor. In Step 2, one of the two 3.0 Ω resistances
is returned to the 4.0 Ω, 48 Ω, and 16 Ω resistors in parallel. The three resistors must have the same ΔV = 6.0 V.
From Ohm’s law,
6V
6V
6V
I4 =
= 1.5 A
I 48 =
= 0.125 A I16 =
= 0.375 A
4Ω
48 Ω
16 Ω
Resistor
Potential difference (V)
Current (A)
3.0 Ω
6.0
4.0 Ω
6.0
1.5
48 Ω
6.0
0.13
16 Ω
6.0
0.38
2.0
Assess: The larger the resistance, the smaller the current it carries. This is what we would have expected.
P23.30. Prepare: The total resistance is
Req = 1. 0 kΩ + Rph
and the current in the circuit is
I = Δ V / Req = (9 .0 V) / (1 . 0 k Ω + Rph ) .
Solve:
(a)
Δ V1.0 kΩ = IR1.0 kΩ =
9.0 V
(1 . 0 k Ω)
1 . 0 k Ω + Rph
For Rph = 0. 56 Ω
Δ V1.0 kΩ =
9.0 V
(1 . 0 k Ω) = 9 .0 V
1 .0 k Ω + 0 . 56 Ω
Δ V1.0 kΩ =
9.0 V
(1 . 0 k Ω) = 1 . 8 V
1 .0 k Ω + 4 .0 kΩ
For Rph = 4. 0 kΩ
23-18 Chapter 23
For Rph = 20 k Ω
Δ V1.0 kΩ =
9.0 V
(1 .0 k Ω) = 0 .43 V
1 .0 k Ω + 20 k Ω
(b) As the light increases the voltmeter reading increases.
Assess: This is a simple circuit for a light meter.
P23.31. Prepare: We know the potential difference across the photoresistor must be less than 9.0 V, but not
tiny either.
Solve:
(a)
(b) The ammeter will read
I=
ΔV
9.0 V
=
= 2 . 571 mA ≈ 2 . 6 mA
Req 1 . 0 k Ω + 2 .5 k Ω
The voltmeter will read
Δ Vph = IRph = (2 . 571 mA)(2 . 5 k Ω) = 6 . 4 V
Assess: The answer agrees with our prediction in the prepare step.
P23.32. Prepare: Capacitors in parallel follow Equation 23.17.
Solve:
The equivalent capacitance is
Ceq = C1 + C2 + C3 = 6.0 μF + 10 μF + 16 μF = 32 μF
Assess:
P23.33.
Solve:
This value is the sum of all capacitances, so the result is reasonable.
Prepare: Capacitors in series follow Equation 23.19.
The equivalent capacitance is
−1
−1
⎛ 1
⎛ 1
1
1 ⎞
1
1 ⎞
−6
+
+
+
Ceq = ⎜ +
⎟ =⎜
⎟ = 3.04 × 10 F = 3.0 μ F
C
C
C
μ
μ
μ
6.0
F
10
F
16
F
2
3 ⎠
⎝
⎠
⎝ 1
Assess:
This value is the smallest of all capacitances, so the result is reasonable.
P23.34.
Prepare: Two capacitors in parallel combine to give greater capacitance according to Equation
23.17.
Solve: Since we want a capacitance of 50 μF and we have a 30 μF capacitor, we must connect the second
capacitor in parallel with the 30 μF capacitor. That is,
C + 30 μF = 50 μF ⇒ C = 50 μF − 30 μF = 20 μF
Assess:
We must learn how the series and parallel circumstances combine.
P23.35. Prepare: Two capacitors in series combine to give less capacitance according to Equation 23.19.
Circuits 23-19
Solve: Since we have a 75 μF capacitor and we want a 50 μF capacitance, we must connect the second
capacitor in series with the 75 μF capacitor. The capacitance of the second capacitor is calculated as follows:
1
1
1
+ =
⇒ C = 150 μF
75 μ F C 50 μ F
Assess:
We must learn how to combine series and parallel capacitances.
P23.36.
Prepare: Please refer to figure P23.36. The pictorial representation shows how to find the
equivalent capacitance of the three capacitors shown in the figure.
Solve:
Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is
1
Ceq 12
=
1
1
1
1
1
+
=
+
=
⇒ Ceq 12 = 12 μF
C1 C2 20 μ F 30 μ F 12 μ F
Then, Ceq 12 and C3 are in parallel. So,
Ceq = Ceq 12 + C3 = 12 μF + 25 μF = 37 μF
Assess:
We must understand well how to combine series and parallel capacitance.
P23.37. Prepare: Please refer to Figure P23.37. The following visual overview shows how to find the
equivalent capacitance of the three capacitors shown in the figure.
Solve:
Because C1 and C2 are in parallel, their equivalent capacitance Ceq 12 is
Ceq 12 = C1 + C2 = 20 μF + 60 μF = 80 μF
Then, Ceq 12 and C3 are in series. So,
1
1
1
1
1
9
80
=
+
=
+
= ( μ F) −1 ⇒ Ceq =
μ F = 8.9 μ F
C eq Ceq 12 C3 80 μ F 10 μ F 80
9
Assess:
Learn well how to combine series and parallel capacitances.
P23.38. Prepare: Assume ideal battery, wires, and capacitors.
Solve:
(a) The two on the right are in parallel, so we add them to get 3 . 5 μ F . That in series with 4 . 0 μ F gives
−1
⎛ 1
1 ⎞
Ceq = ⎜
+
⎟ = 1 . 867 μ F ≈ 1 .9 μ F
F
μ
μF ⎠
3
5
4
0
.
.
⎝
(b) The amount of charge that flows while charging is given by
Q = CE = (1 . 867 μ F)(12 V) = 22 μ C
Assess: The values are typical.
23-20 Chapter 23
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 = C ΔV = (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 k Ω + 1.0 k Ω = 2.0 k Ω . Thus, the time constant for the
discharge of the capacitors is
τ = RCeq = (2.0 k Ω )(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.
P23.41.
Prepare: The circuit in Figure P23.41 has an equivalent circuit with resistance Req and capacitance
C = 4.0 μ F. Assume ideal wires as the capacitor discharges through the two 1 kΩ resistors.
Solve: The equivalent resistance is
1
1
1
⇒ Req = 500 k Ω
=
+
Req 1.0 k Ω 1.0 k Ω
Thus, the time constant for the discharge of the capacitors is
τ = ReqC = (500 Ω)(4.0 μ F) = 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.
P23.42. Prepare: We use Ohm’s law to determine the initial leakage resistance although the leakage current
decreases as the capacitor discharges.
Solve:
(a) The leakage resistance is
R=
2.5 V
ΔV
=
= 10 k Ω
0 .25 mA
I
(b) First compute the time constant for the circuit. τ = RC = (10 k Ω)(0 . 47 F) = 4700 s. Then solve for t , the
time for the voltage to drop to 1.0 V.
Δ V = Δ V0 e −t /τ
⇒ t = −τ ln
ΔV
1.0 V
= −(4700 s)ln
= 4306 s ≈ 72 min
2.5 V
ΔV 0
Assess: This seems reasonable; the capacitor leaks significantly in about an hour.
P23.43. Prepare: 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/τ.
Solve: The time constant is τ = RC = (1.0 × 10−3 Ω)(10 × 10−6 F) = 0.010 s.
The initial charge on the capacitor is Q0 = 20 μC and it decays to 10 μC in time t. That is,
Circuits 23-21
⎛ 10 μ C ⎞
t
10 μC = (20 μC)e−t/0.010 s ⇒ ln ⎜ 2 μ C ⎟ = − 0.010 s ⇒ t = (0.010 s) ln 2 = 6.9 ms
⎝
⎠
Assess:
A time constant of a few milliseconds is reasonable.
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/τ.
Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a
charge Q0 = C ΔV = Cε = (2 μ F)(9 V) = 18 μ C.
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
Δ VR
9V
=
= 0.18 A = 180 mA
I0 =
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 /τ, where τ = RC = (50 Ω)(2 μ F) = 100 μs.
Thus, the charge is Q = (18 μ C)e −50 μ s/(100μs) = (18 μ C)e−0.5 = 10.9 μ C = 11 μC.
The resistor current is I = I 0e−t /τ = (180 mA)e −50 μs/(100μ s) = 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.
P23.45. Prepare: Before the action potential the membrane potential is approximately − 70 mV and at the peak
of depolarization it is + 40 mV. We’ll use positive values since we only need the strength of the electric field.
Solve: Before the action potential,
E=
ΔV 70 mV
=
= 7 .8 MV/m
d
9 . 0 nm
E=
ΔV 40 mV
=
= 4 . 4 MV/m
d
9 . 0 nm
At the peak of depolarization,
Assess: The high values are due to the thinness of the membrane.
P23.46. Prepare: For the resistance follow Example 23.13; for the capacitance follow Example 23.14.
Solve:
Rmembrane =
Cmembrane =
κε 0 A
d
=
ρL
A
=
(3 . 6 × 107 Ω ⋅ m)(9 . 0 × 10−9 m)
= 6 . 446 × 107 Ω
4π (2 . 0 × 10−5 m) 2
9 . 0(8 . 85 × 10 −12 C 2 / (N ⋅ m 2 ))4π (2 . 0 × 10−5 m) 2
= 4 . 448 × 10−11 F
9 .0 × 10−9 m
τ = RC = (6. 446 × 107 Ω)(4. 448 × 10−11 F) = 2. 9 ms
Assess: Our result is similar to the one obtained in the chapter.
P23.47. Prepare: We want to combine the equation in the text for the speed of nerve impulse propagation,
v = Δ xnode / τ , with the time constant equation, τ = Raxon Cmembrane .
We are given Δ xnode = 1 . 0 mm = 1 . 0 × 10−3 m, v = 55 m/s, and R = 25 M Ω = 25 × 106 Ω .
Solve:
23-22 Chapter 23
v=
Δ xnode
τ
=
Δ xnode
RC
Solve this for C , which is the quantity sought.
C=
Δ xnode
1 . 0 × 10−3 m
=
= 7 . 272 × 10−13 F ≈ 0 . 73 pF
Rv
(25 × 106 Ω)(55 m/s)
Assess: The answer is smaller than, but in the same ballpark as, the value of the capacitance given in the
text, C ≈ 1 . 6 pF. This is as we would expect, given that the speed of the nerve impulse in this problem is
twice the speed calculated in the text.
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
τ
=
0. 80 × 10−3 m
= 33 m/s
(20 M Ω)(1. 2 pF)
Assess: This result is similar to the one in the chapter. The units do work out since Ω ⋅ F = s.
P23.49. Prepare: Speed may be determined by v = Δx /Δt. The speed at which a nerve impulse travels down
the axon is v = Δxnode / τ , where τ = RC is the time constant.
Solve: (a) The speed of propagation of the nerve impulse may be determined by
v = Δx /Δt = 0.24 m/(4 × 10−3 s) = 60 m/s.
(b) The reason the second proposed method is inaccurate is that it neglects to account for the time it takes
between the stimulus and the peak of the action potential. The first method is better because it measures the peak
of the action potential at the elbow and the same thing at the wrist.
Assess: We must be thoughtful and careful in designing experiments to make sure we measure what we want.
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.
P23.51.
Prepare: Please refer to Figure P23.51. Assume ideal connecting wires and an ideal battery. The
power dissipated by each resistor can be calculated from Equation 22.14, PR = I2R, provided we can find the
current through the resistors.
Solve: Let us choose a clockwise direction for the current and solve for the value of I by using Kirchhoff’s
loop law. Going clockwise from the negative terminal of the battery,
∑ (ΔV ) = ΔV
i
bat
+ ΔVRI + ΔVR2 = 0 ⇒ + 9V − IR1 − IR2 = 0 ⇒ I =
i
9.0 V
9.0 V
1
=
= A
R1 + R2 12 Ω + 15 Ω 3
The power dissipated by resistors R1 and R2 is:
2
⎛1 ⎞
PR1 = I 2 R1 = ⎜ A ⎟ (12 Ω) = 1.3 W
⎝3 ⎠
Assess:
P23.52.
2
⎛1 ⎞
PR2 = I 2 R2 = ⎜ A ⎟ (15 Ω) = 1.7 W
⎝3 ⎠
For a relatively small current of ( 13 )A, power dissipated by 12 Ω and 15 Ω resistors is reasonable.
Prepare: We will assume ideal connecting wires and an ideal power supply. The two light bulbs
are basically two resistors in series. Our strategy is to find the current that flows through the two bulbs and then
use Equation 22.14 to find the power dissipated by each bulb.
Circuits 23-23
Solve:
A 75 W (120 V) light bulb has a resistance of
R=
Δ V 2 (120 V) 2
=
= 192 Ω
P
75 W
The combined resistance of the two bulbs is Req = R1 + R2 = 192 Ω + 192 Ω = 384 Ω.
The current I flowing through Req is
I=
ΔV 120 V
=
= 0.3125 A
Req 384 Ω
Because Req is a series combination of R1 and R2, the current 0.3125 A flows through R1 and R2. Thus, PR1 = I2R1
= (0.3125 A)2(192 Ω) = 19 W = PR2.
Assess: Please note that a light bulb rated at 75 W (120 V) delivers 75 W power only when it is connected
across a source of 120 V. As we see in this problem, the power delivered by two series bulbs across 120 V is
considerably reduced compared to their rated power.
P23.53.
Prepare:
Assume ideal connecting wires and an ideal power supply.
Solve: We have two resistors in series such that Req = Rbulb + Rcontacts. Rbulb can be found from the fact that we
have a 100 W (120 V) bulb:
Rbulb =
Δ V 2 (120 V) 2
=
= 144 Ω
P
100 W
We have a total resistance of Req = 144 Ω + 5 Ω = 149 Ω. The current flowing through Req is
I=
ΔV 120 V
=
= 0.8054 A
Req 149 Ω
Because Req is a series combination of Rbulb and Rcontacts, this current flows through both the bulb and the contacts.
Thus,
Pbulb = I2Rbulb = (0.8054 A)2(144 Ω) = 93 W
Assess: The corroded leads change the circuit’s total resistance and reduce the current below that at which the
bulb was rated. So, it makes sense for it to operate at less than full power.
Prepare: The 10 Ω internal resistance of the battery is in series with the external 2.0 Ω resistor.
So the total resistance of the circuit is 1.0 Ω + 2.0 Ω = 3.0 Ω.
As a preliminary calculation, use Ohm’s law to find the current in the circuit.
P23.54.
I=
Solve:
ΔV 1 . 5 V
=
= 0 . 50 A
R
3.0 Ω
(a) The potential difference between the terminals of the battery is
23-24 Chapter 23
Δ Vbatt = E − IRinternal = 1 .5 V − (0 . 50 A)(1.0 Ω) = 1 .0 V
(b) The total power dissipated is P = I 2 R = (0 . 50 A) 2 (3.0 Ω) = 0 . 75 W. The power dissipated internally in the
battery is P = I 2 R = (0. 50 A)2 (1.0 Ω) = 0. 25 W, or 1/3 of the total.
Assess: While ideal batteries are sources of constant emf, real batteries with internal resistance have a potential
difference between the terminals that depends on the external resistance (the load).
P23.55.
Prepare:
The current in the circuit may be obtained by I = E/( r + R ) and the power dissipated by
the external resistor is P = I 2R.
Solve: When the external resistor is R = 0.25 Ω the current in the circuit is I = E/( r + R) = 1.5 V/(1.25 Ω ) = 1.2 A.
The power dissipated by the 0.25 Ω external resistor is P0.25Ω = I 2 R = (1.2 A) 2 (0.25 Ω) = 0.36 W.
If we continue this process for all of the resistors specified, we obtain the following values:
P0.25Ω = 0.36 W
P0.50 Ω = 0.50 W
P1.0 Ω = 0.56 W
P2.0 Ω = 0.50 W
P4.0 Ω = 0.36 W
Notice that we get the most power delivered to the external resistor by the battery when the external resistor is
equal to the internal resistance of the battery.
Assess: This is a common phenomenon with electrical devices. They deliver the most power to the external
circuit when the resistance of the external circuit is equal to the internal resistance of the device.
P23.56. Prepare: The power supply needs a larger emf than the 1.5 V of the battery. 10 min = 600 s.
Solve:
(a) The power supply needs an emf that is greater than the battery’s by Δ V = IR for the desired current.
E = 1 . 5 V + (0 . 75 A)(1 . 0 Ω) = 2 . 25 V ≈ 2 .3 V
(b) The energy going into the battery is
E = Pt = IVt = (0 . 75 A)(2 . 25 V)(600 s) = 1. 0 kJ
The energy dissipated by the internal resistance of the battery is
Pt = I 2 Rt = (0 . 75 A) 2 (1 . 0 Ω)(600 s) = 340 J
Assess: Of the energy put into the battery to recharge it about one third is dissipated as thermal energy.
P23.57.
Prepare: First note that the current in the battery in the parallel case is (due to Kirchhoff’s junction
law) I par = 3 .2 A + 1 .8 A = 5 . 0 A.
Call the unknown (but constant) emf of the battery ε . Call the resistor that carries the 3.2 A current R1, and call
the resistor that carries the 1.8 A current R2.
Using Ohm’s law in the parallel case, we see that E = (3. 2 A) R1 = (1 . 8 A) R2 . This gives R1 = E/(3 . 2 A) and
R2 = E/(1 .8 A).
From the parallel case we also know that E = I par Req par .
As a reminder,
Req ser = R1 + R2
Solve:
=
⎛ 1
1 ⎞
Req par = ⎜ +
⎟
R
R
2 ⎠
⎝ 1
−1
Now we turn our attention to the series case.
−1
I ser =
and
E
Req ser
=
I par Req par
Req ser
−1
−1
⎛
⎞
⎛ 3. 2A 1. 8A ⎞
⎛ 5 .0A ⎞
⎛
⎞
(5 . 0A)⎜ 1 + 1 ⎟
+
(5. 0 A)⎜
(5. 0 A)⎜
(5 . 0 A) ⎜ E ⎟
⎟
⎟
R
R
.
E
E
E
5
0A
2 ⎠
⎝ 1
⎝
⎠ =
⎝
⎠ =
⎝
⎠
=
=
E + E
(5. 0A)E
(5. 0A) E
R1 + R2
3 .2A 1. 8A
(3. 2A)(1. 8A)
(3 . 2A)(1. 8A)
(3 .2 A)(1. 8 A)
= 1.2 A
5.0 A
Circuits 23-25
Assess: We knew that putting the two resistors in series would produce a bigger equivalent resistance than
when they were in parallel, so we expected the current in the series case to be less than the current in the parallel
case, and our answer conforms to this expectation.
P23.58. Prepare: Please refer to Fig. P23.58. The connecting wires are ideal, but the battery is not. We will
designate the current in the 5.0 Ω resistor I5 and the voltage drop ΔV5. Similar designations will be used for the
other resistors.
Solve: Since the 10 Ω resistor is dissipating 40 W,
P10 = I102 R10 = 40 W ⇒ I10 =
P10
=
R10
40 W
= 2.0 A ⇒ ΔV10 = I10R10 = (2.0 A)(10 Ω) = 20.0 V
10 Ω
The 20 Ω resistor is in parallel with the 10 Ω resistor, so they have the same potential difference: ΔV20 = ΔV10 = 20.0
V. From Ohm’s law,
I 20 =
Δ V20 20.0 V
=
= 1.0 A
R20
20 Ω
The combined current through the 10 Ω and 20 Ω resistors first passes through the 5.0 Ω resistor. Applying
Kirchhoff’s junction law at the junction between the three resistors,
I5 = I10 + I20 = 1.0 A + 2.0 A = 3.0 A ⇒ ΔV5 = I5R5 = (3.0 A)(5.0 Ω) = 15 V
Knowing the currents and potential differences, we can now find the power dissipated:
P5 = I5ΔV5 = (3.0 A)(15.0 V) = 45
P23.59.
Prepare:
P20 = I20ΔV20 = (1.0 A)(20.0 V) = 20 W
Use Kirchhoff’s loop law Σ( ΔV )i = 0, where the voltage drop across the resistor is IR and
Δ VC = Q / C.
Solve:
Q
=0
C
E − Q /C 20 V − 200 μ C/(20 μ F) 20 V − 10 V
=
=
= 500 Ω
R=
20 mA
20 mA
I
Σ(ΔV )i = E − IR −
Assess:
500 Ω is a reasonable value for resistance.
P23.60.
Prepare: The connecting wires are ideal with zero resistance. A visual overview of how to reduce
the circuit to an equivalent resistance is shown.
Solve: In the first step, the resistors 100 Ω, 100 Ω, and 100 Ω in the top branch are in series. Their combined
resistance is 300 Ω. In the middle branch, the two resistors, each 100 Ω, are in series. So, their equivalent
resistance is 200 Ω. In the second step, the three resistors are in parallel. Their equivalent resistance is
1
1
1
1
⇒ Req = 54.5 Ω
=
+
+
Req 300 Ω 200 Ω 100 Ω
23-26 Chapter 23
The equivalent resistance of the circuit is 55 Ω.
P23.61.
Solve:
Prepare: A visual overview of how to reduce the circuit to an equivalence resistance is shown below.
(a) The three resistors in parallel have an equivalent resistance of
1
1
1
1
=
+
+
⇒ Req = 4.0 Ω
Req 12 Ω 12 Ω 12 Ω
(b) One resistor in parallel with two series resistors has an equivalent resistance of
1
1
1
1
1
1
=
+
=
+
=
⇒ Req = 8.0 Ω.
Req 12 Ω + 12 Ω 12 Ω 24 Ω 12 Ω 8 Ω
(c) One resistor in series with two parallel resistors has an equivalent resistance of
−1
⎛ 1
1
1 ⎞
= 12 Ω + ⎜
+
⎟ = 12 Ω + 6 Ω = 18.0 Ω
Ω
Ω⎠
Req
12
12
⎝
(d) The three resistors in series have an equivalent resistance of
12 Ω + 12 Ω + 12 Ω = 36 Ω
Assess:
Learn how to combine series and parallel resistors.
P23.62. Prepare: The current will be the same through all three pieces of wire. To find this current we need
the equivalent resistance. The cross-sectional area of all the wires is A = π (0 . 13mm) 2 = 5. 309 × 10−8 m 2 . First
find the resistance of each piece.
RCu =
ρCu LCu
RFe =
RW =
A
ρ Fe LFe
A
ρW LW
A
=
(1 . 7 × 10−8 Ω ⋅ m)(0 .10 m)
= 0 . 03202 Ω
5 . 309 × 10−8 m 2
=
(9 . 7 × 10−8 Ω ⋅ m)(0 . 12 m)
= 0 . 2193 Ω
5 . 309 × 10−8 m 2
=
(5 . 6 × 10−8 Ω ⋅ m)(0 . 18 m)
= 0 . 1899 Ω
5 . 309 × 10−8 m 2
Solve:
Req = RCu + RFe + RW = 0 .03202 Ω + 0 . 2193 Ω + 0 . 1899 Ω = 0 . 4411 Ω
The current is
Circuits 23-27
I=
ΔV
9.0 V
=
= 20 . 4 A
Req 0. 4411 Ω
We can now find the potential difference across each piece of wire.
Δ VCu = IRCu = (20. 4 A)(0. 03202 Ω) = 0. 62 V
Δ VFe = IRFe = (20 . 4 A)(0 . 2193 Ω) = 4 . 5 V
Δ VW = IRW = (20. 4 A)(0 . 1899 Ω) = 3 . 9 V
Assess: The sum of the three potential differences adds up to the emf of the battery.
P23.63. Prepare: Whatever the value of the resistors, if they are all the same we can put three of them in
series and divide the potential difference into thirds. Use the loop law and symmetry to see that the potential
difference across each of the three resistors must be 3.0 V.
Solve:
Assess: If we had enough resistors we could influence the current by putting in six resistors in series and then
we’d have 3.0 V across each adjacent pair, but less overall current.
P23.64.
Prepare: Please refer to Figure P23.64. Assume the batteries and the connecting wires are ideal.
Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The direction of the current is
counterclockwise because the 12 V battery “wins.”
(b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoff’s loop
law in the counterclockwise direction and starting at the lower right corner, ∑ΔVi = 12 V − I(12 Ω) − I(6.0 Ω) −
6.0 V – IR = 0.
Note that the IR terms are all negative because we’re applying the loop law in the direction of current flow, and
the potential decreases as current flows through a resistor. We can easily solve to find the unknown resistance R:
6.0 V − (18 Ω) I 6.0 V − (18 Ω)(0.25 A)
=
= 6.0 Ω
6.0 V − I(18 Ω) – IR = 0 ⇒ R =
I
0.25 A
(c) The power is P = I2R = (0.25 A)2(6 Ω) = 0.38 W.
(d)
23-28 Chapter 23
The potential difference across a resistor is ΔV = IR, giving ΔV6 = 1.5 V, and ΔV12 = 3.0 V. Starting from the
lower left corner, the graph goes around the circuit clockwise, opposite from the direction in which we applied
the loop law. In this direction, we speak of potential as lost in the batteries and gained in the resistors.
Prepare: Please refer to Figure P23.65. The ammeter has a resistance of 50.0 Ω. Because the
ammeter we have shows a full-scale deflection with a current of 500 μA = 0.500 mA, we must not allow a
current more than 0.500 mA to pass through the ammeter branch. Since we wish to measure a maximum current
of 50.0 mA, we must split the current in such a way that 0.500 mA flows through the ammeter branch and 49.500
mA flows through the resistor R.
Solve: (a) The potential difference across the ammeter and the resistor R is the same. Thus,
VR = Vammeter ⇒ (49.500 × 10−3 A)R = (0.500 × 10−3 A)(50.0 Ω) ⇒ R = 0.505 Ω
P23.65.
(b) Equivalent resistance is
Assess:
1
1
1
=
+
⇒ R = 0.500 Ω
Req 0.505 Ω 50.0 Ω
For a pair of parallel resistors, more current flows through the smaller resistor compared to the larger.
P23.66. Prepare: Please refer to Figure P23.66. We will assume ideal connecting wires. Because the
ammeter we have shows a full-scale deflection with a current of 500 μA, we must not pass a current greater than
this through the ammeter.
Solve: The maximum potential difference is 5 V and the maximum current is 500 μA. Using Ohm’s law, ΔV =
IAR ⇒ 5.0 V = (500 × 10−6 A)R ⇒ R = 10 kΩ.
P23.67. Prepare: The figure shows how to simplify the circuit in Figure P23.67 using the laws of series and
parallel resistances. We will reverse the procedure and “build up” the circuit using the loop law and junction law
to find the current and potential difference of each resistor. The battery and the connecting wires are ideal.
Circuits 23-29
Solve: Having found Req = 12 Ω, the current from the battery is I = (24 V)/(12 Ω) = 2.0 A. As we rebuild the
circuit, we note that series resistors must have the same current I and that parallel resistors must have the same
potential difference ΔV.
In Step 1 of the previous figure, the 12 Ω resistor is returned to 4.0 Ω and 8.0 Ω resistors in series. Both resistors must
have the same 2.0 A as the 12 Ω resistor. We use Ohm’s law to find ΔV4 = 8.0 V and ΔV8 = 16 V. As a check, 8.0 V +
16 V = 24 V, which was ΔV of the 12 Ω resistor. In Step 2, the 8 Ω resistor is returned to the 12 Ω and 24 Ω resistors
in parallel. Both resistors must have the same ΔV = 16 V as the 8.0 Ω resistor. From Ohm’s law, I12
= (16 V)/(12 Ω) = 4/3 A and I 24 = 2/3 A. As a check, I12 + I24 = 2.0 A, which was the current I of the 8.0 Ω resistor.
In Step 3, the 12 Ω resistor is returned to the two 6.0 Ω resistors in series. Both resistors must have the same 4/3 A as
the 12 Ω resistor. We use Ohm’s law to find ΔV6 = 8.0 V and ΔV6 = 8.0 V. As a check, 8.0 V + 8.0 V = 16 V, which
was ΔV of the 12 Ω resistor. Finally, in Step 4, the 6 Ω resistor is returned to the 8.0 Ω and 24 Ω resistors in parallel.
Both resistors must have the same ΔV = 8.0 V as the 6.0 Ω resistor. From Ohm’s law, I8 = (8.0 V)/(8.0 Ω) = 1.0 A and
I 24 = 1/3 A. As a check, I 8 + I 24 = 4/3 A, which was the current I of the 6.0 Ω resistor.
23-30 Chapter 23
Potential difference (ΔV)
8.0
8.0
8.0
8.0
16
Resistor
4.0 Ω
6.0 Ω
8.0 Ω
Bottom 24 Ω
Right 24 Ω
Current (A)
2.0
1.3
1.0
0.33
0.67
P23.68. Prepare: Capacitors in parallel add to a greater capacitance compared to individual capacitances.
On the other hand, capacitors in series add to a smaller capacitance compared to individual capacitances.
(a) Three capacitors in series:
Solve:
1
1
1
1
3
=
+
+
= ( μ F) −1 ⇒ Ceq = 4.0 μF
Ceq 12 μ F 12 μ F 12 μ F 12
(b) Two capacitors in parallel and the third in series with this parallel combination:
Ceq 12 = 12 μF + 12 μF = 24 μF ⇒
1
1
1
1
1
1
=
+
=
+
=
⇒ Ceq = 8.0 μF
Ceq Ceq 12 12 μ F 24 μ F 12 μ F 8.0 μ F
(c) Two capacitors in series and the third in parallel with this series combination:
1
Ceq 12
=
1
1
1
(μ F) −1 ⇒ Ceq 12 = 6.0 μF ⇒ Ceq = Ceq 12 + 12 μF = 6.0 μF + 12 μF = 18 μF
+
=
12 μ F 12 μ F 6.0
(d) Three capacitors in parallel:
Ceq = 12 μF + 12 μF +12 μF = 36 μF
Assess:
Learn how to combine series and parallel capacitances.
P23.69. Prepare: Please refer to Figure P23.69. While the switch is in position A, the capacitors C2 and C3
are uncharged. When the switch is placed in position B, the charged capacitor C1 is connected to C2 and C3. C2
and C3 are connected in series to form an equivalent capacitor Ceq 23. We will also assume that the battery is ideal.
Solve: While the switch is in position A, a potential difference of V1 = 100 V across C1 charges it to
Q1 = C1V1 = (15 × 10−6 F)(100 V) = 1500 μ C.
When the switch is moved to position B, this initial charge Q1 is redistributed. The charge Q1′ goes on C1 and the
charge Qeq 23 goes on Ceq 23. The voltage across C1 and Ceq 23 is the same and Q1′ + Qeq 23 = Q1 = 1 500 μ C. Combining
these two conditions, we get
Since Ceq 23 =
(
1
30 μ F
+ 301μ F
)
1500 μ C − Qeq 23
15 μ F
−1
1500 μ C − Qeq 23 Qeq 23
Q1′ Qeq 23
=
⇒
=
C1 Ceq 23
C1
Ceq 23
= 15 μ F, we can rewrite this equation as
=
Qeq 23
15 μ F
⇒ Qeq 23 = 750 μC ⇒ Q1′ = Q1 − Qeq 23 = 1500 μ C − 750 μ C = 750 μ C
Having found the charge Qeq 23, it is easy to see that Q2 = Q3 = 750 μC because Ceq 23 is a series combination of C2
and C3. Thus,
750 μ C
Q
Q 750 μ C
Q′ 750 μ C
ΔV2 = 2 =
= 25 V
Δ V3 = 3 =
= 25 V
ΔV1 = 1 =
= 50 V
30 μ F
30 μ F
15 μ F
C2
C3
C1
Circuits 23-31
Assess: the potential differences across and charges on the three capacitors are consistent with the ideas of a
closed circuit.
P23.70. Prepare: The capacitor discharges through the resistor and the wires are ideal. 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 = Qo e – t/τ .
(a) The capacitor discharges through the resistor R as Q = Qo e – t/τ For Q = Q0 / 2,
Solve:
t
Q0
⎛1⎞
= Q0e− t /15 ms ⇒ ln ⎜ ⎟ = −
⇒ t = −(0.015 s)ln(0.5) = 10.4 ms ≈ 10 ms
2
2
0.015
s
⎝ ⎠
(b) If the initial capacitor energy is U0, we want the time when the capacitor’s energy will be U = U 0 /2. Noting
that U0 = Q02 /(2C ), this means Q = Q0 / 2. Applying the equation for the discharging capacitor,
Q0
⎛ 1 ⎞
t
⎛ 1 ⎞
= Q0e−t /15 ms ⇒ ln ⎜
⇒ t = −(0.015 s)ln ⎜
⎟=−
⎟ = 5.2 ms
0.015 s
2
⎝ 2⎠
⎝ 2⎠
Assess: A time of 5.2 ms for the energy to become half of its original energy is typical of discharging
capacitors. Notice the answer to part (b) is half the answer to part (a).
P23.71.
Prepare: A capacitor discharges through a resistor and the wires are ideal. 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 Q = Qo e – t/τ .
Solve: A capacitor initially charged to Q0 decays as Q = Qo e – t/τ . We wish to find R so that a 1.0 μF capacitor
will discharge to 10% of its initial value in 2.0 ms. That is,
(0.10)Q0 = Q0e −2 ms/(R(1 μ F)) ⇒ ln(0.10) = −
2 × 10−3 s
−2.0 × 10−3 s
⇒ R=
= 870 Ω
−6
R (1.0 × 10 F)
(1.0 × 10−6 F)ln(0.10)
A time constant of τ = RC = (870 Ω)(1.0 × 10−6 F) = 0.87 ms is reasonable.
P23.72. Prepare: Equation 23.25
gives
the
potential
difference
across
capacitor: ΔVC = E(1 − e−t /τ )
The time constant is τ = RC .
Solve: We want the quantity in the parentheses to be 87% or 0.87, and then solve for t.
Assess:
a
charging
1 − e −t /τ = 0 . 87
1 − 0 . 87 = e− t /τ
0 .13 = e− t /τ
ln(0 . 13) = −t / τ
τ ln(0 . 13) = −t
t = −τ ln(0 . 13) = 2τ
This says we want 8. 0 s = 2τ = 2 RC .
R=
8.0 s
8.0 s
=
= 40,000 Ω = 40 k Ω
2C
2(100 μ F)
Assess: This answer is, indeed, the most resistance that can be in series and still get the flash charged in 8.0 s
or less; if the resistance were greater, then τ would be greater and it would take longer to charge to 87% of 25 V.
By the way, the specific value of E (which, in this case, is the same as the capacitor’s voltage rating) is not
important; the answer would apply to any emf as long as C = 100 μ F and the question is to charge the capacitor
to 87% of its maximum value.
P23.73.
Prepare: A capacitor discharges through a resistor and we will assume that the wires are ideal. The
discharge current or the resistor current follows Equation 23.22: I = I 0e−t / ( RC ).
23-32 Chapter 23
Solve: We wish to find the capacitance C so that the resistor current will decrease to 25% of its initial value in
2.5 ms. That is,
0.25 I 0 = I 0e−2.5 ms/((100 Ω )C) ⇒ ln(0.25) = −
Assess:
2.5 × 10−3 s
⇒ C = 18 μF
(100 Ω)C
A capacitance of 18 μF is typical of capacitors.
P23.74. Prepare: Please refer to Figure P23.74. In an RC circuit, the capacitor voltage discharge is given by
Equation 23.22, i.e., ΔV = ΔV0e−t/τ.
Solve: From Figure P23.74, we note that ΔV0 = 30 V and ΔV = 10 V at t = 2 ms. So,
10 V = (30 V)e −2 ms/(R (50×10
−6
F))
⎛ 10 V ⎞
2 × 10−3 s
2 × 10−3 s
R
⇒ ln ⎜
⇒
=
−
= 36 Ω
⎟=−
(50 × 10−6 F)ln ( 13 )
R (50 × 10−6 F)
⎝ 30 V ⎠
P23.75. Prepare: Please refer to Figure P23.75. We will assume the battery and the connecting wires to be ideal.
Solve: (a) If the switch has been closed for a long time, the capacitor is fully charged and there is no current
flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 Ω
resistor and a voltage of 40 V appears across the 40 Ω resistor. That is, maximum voltage across the capacitor is
40 V. Thus, the charge on the capacitor is
Q0 = ε C = (40 V)(2.0 × 10−6 F) = 80 μC
(b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two
resistances (10 Ω and 40 Ω) in series and discharges according to Q = Q0e−t/(RC). For Q = 0.10 Q0,
t
⎛ 0.10 ⎞
0.10 Q0 = Q0e−t/((50 Ω)(2 μF)) ⇒ ln ⎜⎝ 1 ⎟⎠ = − (50 Ω)(2 μ F) ⇒ t = −(50 Ω)(2 μ F)ln(0.10) = 0.23 ms
Assess:
A time constant of 0.23 ms is reasonable for the given circuit.
P23.76. Prepare: Vb = Va when the ratio of the potential drops in the middle branch is the same as the ratio of
the potential drops in the right branch; that is,
ΔV
C = 47000 Ω
E
47100 Ω
Solve:
(a)
Δ VC = E (1 − e −t / ( RC ) ) ⇒
R=
−t
− 12 s
=
C ln(1 − Δ V / E ) (100 μ F )ln(1 −
47 000 Ω
47100 Ω
)
= 19 k Ω
(b) We see that R and t are proportional, so to increase the time the resistance should be increased.
Assess: This is a reasonable value for a variable resistor.
P23.77. Prepare: The chapter says that when the sodium channels open, the potential inside the cell changes
from − 70 mV to + 40 mV relative to outside the cell. This Δ V = 110 mV will be accompanied by a movement of
charge Δ Q = C Δ V as the ions flow into the cell.
The number of channels is the total charge that moves divided by the amount of charge per channel (which is
given as 10,000 Na + ions per channel).
Circuits 23-33
Solve:
⎛ channel ⎞
⎛ charge ⎞
⎛ charge ⎞
# channels = total charge ÷ ⎜
= total charge × ⎜
= ΔQ × ⎜
⎟
⎟
⎟
⎝ channel ⎠
⎝ channel ⎠
⎝ 10,000 ions ⎠
⎛ channel ⎞
1e
⎛
⎞ ⎛ channel ⎞
= C ΔV × ⎜
⎟ = (89pF)(110mV) ⎜
⎟
⎟⎜
−19
⎝ 1.6 × 10 C ⎠ ⎝ 10,000 e ⎠
⎝ 10,000 e ⎠
1e
⎛
⎞ ⎛ channel ⎞
= (9. 79 × 10−12 C) ⎜
⎟ ⎜ 10,000 e ⎟ = 6100 channels
−19
1.6
1
0
C
×
⎝
⎠⎝
⎠
Assess:
Example 23.15 says there are “a great many channels” in a cell, and our answer bears this out.
P23.78. Prepare: The area of the cylindrical axon is determined from the diameter and length by A = π dL.
Knowing the surface area of the axon and the capacitance per square centimeter of the membrane area, we can
determine the capacitance of the axon. Knowing the capacitance of the axon and the resting potential, we can
determine the amount of electric energy stored in the axon by U = CV 2/2.
Solve: The surface area of the axon is A = π dL = π (5 × 10−4 m)(0.10 m) = 1.57 × 10−4 m 2.
The capacitance of the axon is
C = (1.0 × 10−6 F/cm 2 ) A = (1.0 × 10−6 F/cm 2 )(1.57 × 10−4 m 2 )(104 cm 2 /m 2 ) = 1.57 × 10−6 F.
The electric energy stored in the axon is U = CV 2 /2 = (1.57 × 10−6 F)(7.0 × 10−2 V) 2 /2 = 3.8 nJ.
Assess:
This is a very small amount of electric energy but the squid is not known for its electrical properties.
P23.79. Prepare: The capacitance of a dielectric-filled capacitor is given in Equation 21.21: C = K E0 A/ d and
then Equation 21.17 gives the charge: Q = C ΔVC .
In Example 23.14 we are told the dielectric is κ = 9.0. We are also given A = 6 . 0 × 10−9 m 2, d = 7 .0 × 10−9 m,
and ΔVC = 70 × 10−3 V.
The sodium ions have a + 1 charge.
Solve: (a) We can combine both equations.
Q = C Δ VC =
K E0 A
(9)(8 . 85 × 10−12 C 2 /N ⋅ m 2 )(6 . 0 × 10−9 m 2 )
ΔVC =
(70 × 10−3 V) = 4 .78 × 10−12 C ≈ 4.8 × 10−12 C
d
7 .0 × 10−9 m
(b)
⎛
⎞
1e
7
4 . 78 × 10−12 C ⎜
⎟ = 3.0 × 10 e
−19
1
6
10
C
.
×
⎝
⎠
or 3.0 × 107 sodium ions.
Assess: This tiny fraction of a Coulomb corresponds to 30 million sodium ions. There is a lot going on in each
living cell!
P23.80.
Prepare: When the defibrillator is attached to a person, a capacitor is discharged and the potential
difference across the capacitor and current through the resistor (the heart) should look like that for a capacitor
discharging in an RC circuit. For this case, the potential difference across a capacitor is given by Δ VC =
(Δ VC )o e −t / RC and the current through the resistor (the heart) is given by I = I oe −t / RC.
Solve: The set of graphs labeled A match the above two functions.
Assess: Knowing the general shape of a function is a valuable skill.
23-34 Chapter 23
P23.81.
Prepare:
In an RC circuit the time constant is τ = RC .
Solve:
τ = RC = (32 μ F)(100 Ω) = 3 . 2 ms
The correct choice is C.
Assess: We were told that the capacitor discharges quickly, and our answer agrees with that 3.2 ms is quick, but
reasonable.
Prepare: The initial current may be determined by I o = (Δ VC )o / R and the time constant may be
determined by τ = RC.
Solve: Examining the above expressions for the initial current and time constant, we see that the initial current
will decrease and the time constant will increase if the resistance increases. The correct choice is B.
Assess: The ability to look at an expression and answer a “what if ” question is a valuable skill.
P23.82.
Prepare: The time constant does not depend on the voltage; it is still τ = RC. So lowering the
voltage leaves the time constant unchanged.
Solve: The correct choice is B.
Assess: The time constant gives us an idea of how fast the capacitor will discharge, i.e., we know the potential
difference will decay to 37% of its initial value after one time constant, and that is true no matter what the value
of the initial voltage is.
P23.83.
P23.84.
Prepare: The number of electrocytes needed may be determined by knowing the total voltage and
the voltage per electrocyte: VTotal = NVelect .
Solve: The number of electrocytes needed is N = VTotal / Velect = 450 V/150 × 10−3 V = 3 × 103. The correct choice is D.
Assess: Since the voltage per electrocyte is so small, we expect that a large number will be needed in order to
obtain the desired total voltage.
P23.85.
Prepare:
The power delivered is P = I ΔV .
Solve:
P = I ΔV = (0 . 80 A)(500 V) = 400 W
The correct choice is A.
Assess: That’s a pretty respectable power output. As the question says, the eel can only do it for a short time,
so the total energy output won’t be too large.
The units work out: A × V = W.
P23.86. Prepare: Salt water is a better conductor and hence has a lower resistivity than fresh water. Stacks of
electrocytes in parallel for the ray will result in a lower voltage and more current than the series arrangement of
the electrocytes in the electric eel.
Solve: Knowing that salt water has a lower resistivity eliminates C and D. Knowing that the parallel
arrangement will result in a lower voltage and larger current eliminates B. This leaves A as the answer.
Assess: If we understand how individual cells (batteries) add in parallel and in series, we can transfer this
knowledge to this situation.
P23.87.
Prepare: The voltage of emfs add when they are in series, so for the fish to produce a larger voltage
there must be more of the electrocyte cells in series, i.e., added to each stack.
Solve: The correct choice is B.
Assess: If we were told that the fish develops more current in the pulses, then that would take more stacks in
parallel. But more voltage is achieved by adding more cells in series, just as with batteries.