Chapter 18
Direct-Current Circuits
Problem Solutions
18.1
18.2
From
V
I R r , the internal resistance is
r
V
I
9.00 V
0.117 A
R
72.0
4.92
(a) When the three resistors are in series, the equivalent resistance of the circuit is
Req R1 R2 R3 3 9.0
27
(b)
The term inal potential d ifference of the battery is applied across the series
com bination of the three 9.0 resistors, so the current supplied by the battery an d
the current through each resistor in the series com bination is
I
V
Req
(c) If the three 9.0
resistance is
1
Req
12 V
27
0.44 A
are now connected in parallel w ith each other, the equivalent
1
9.0
1
9.0
1
9.0
3
9.0
or
Req
9.0
3
3.0
When this parallel com bination is connected to the battery, the potential d ifference
across each resistor in the com bination is V 12 V , so the current through each of
the resistors is
I
V
R
12 V
9.0
1.3 A
142
Direct-Current Circuits
18.3
143
For the bulb in use as intend ed ,
V
Rbulb
2
2
120 V
P
192
75.0 W
N ow , presum ing the bulb resistance is unchanged ,
the current in the circuit show n is
I
V
Req
120 V
192
0.800
0.800
0.620 A
and the actual pow er d issipated in the bulb is
P I 2 Rbulb
18.4
0.620 A
2
192
73.8 W
(a) When the 8.00- resistor is connected across the 9.00-V term inal potential
d ifference of the battery, the current through both the resistor and the battery is
V
R
I
9.00 V
8.00
1.13 A
(b) The relation betw een the em f and the term inal potential d ifference of a battery
supplying current I is V
Ir , w here r is the internal resistance of the battery.
Thus, if the battery has r 0.15 and m aintains a term inal potential d ifference of
V 9.00 V w hile supplying the current found above, the em f of this battery m ust
be
V
18.5
Ir 9.00 V
1.13 A 0.15
9.00 0.17
(a) The equivalent resistance of the tw o parallel
resistors is
Rp
1
7.00
1
10.0
1
4.12
Thus,
Rab
R4
Rp
R9
4.00 4.12 9.00
17.1
9.17
144
CH APTER 18
V
(b) I ab
ab
Rab
Also,
18.6
34.0 V
1.99 A , so I 4
17.1
V
I ab Rp
p
V
Then,
I7
and
I10
p
R7
V
R10
p
1.99 A 4.12
1
6.0
1
12
8.18 V
8.18 V
7.00
1.17 A
8.18 V
10.0
0.818 A
(a) The parallel com bination of the 6.0
of
1
R p1
I9 1.99 A
2 1
12
or
and 12
R p1
resistors has an equivalent resistance
12
3
Sim ilarly, the equivalent resistance of the 4.0
1
Rp 2
1
4.0
1
8.0
2 1
8.0
or
Rp 2
4.0
and 8.0
parallel com bination is
8
3
The total resistance of the series com bination betw een points a and b is then
Rab
R p1 5.0
Rp 2
4.0
5.0
8.0
3
35
3
(b) If Vab 35 V , the total current from a to b is I ab
Vab Rab 35 V 35
and the potential d ifferences across the tw o parallel com binations are
Vp1
I ab Rp1
3.0 A 4.0
12 V and
Vp 2
I ab R p 2
3.0 A
8.0
3
3
3.0 A
8.0 V
so the ind ivid ual currents through the various resistors are:
I12
Vp1 12
1.0 A ;
I6
I8
Vp 2 8.0
1.0 A ;
and
Vp1 6.0
I4
2.0 A ;
Vp 2 4.0
I5
2.0 A
I ab
3.0 A ;
Direct-Current Circuits
18.7
The equivalent resistance of the parallel com bination of three id entical resistors is
1
Rp
1
R1
1
R2
1
R3
3
or
R
Rp
R
3
The total resistance of the series com bin ation betw een points a and b is then
Rab
18.8
(a)
R
Rp
R
R
3
2R
7
R
3
The equivalent resistance of this first parallel
com bination is
1
R p1
1
10.0
1
5.00
or
Rp1
3.33
(b) For this series com bination,
Rupper
Rp1 4.00
7.33
(c) For the second parallel com bination,
1
Rp 2
1
Rupper
1
3.00
1
7.33
1
3.00
or
Rp 2
2.13
(d ) For the second series com bination (and hence the entire resistor netw ork)
Rtotal
2.00
Rp 2
2.00
2.13
4.13
(e) The total current supplied by the battery is
(f)
The potential d rop across the 2.00
V2
R2 I total
2.00
1.94 A
I total
V
Rtotal
8.00 V
4.13
1.94 A
resistor is
3.88 V
(g) The potential d rop across the second parallel com bination m ust be
Vp 2
V
V2
8.00 V 3.88 V
(h) So the current through the 3.00
4.12 V
resistor is
I total
Vp 2
R3
4.12 V
3.00
1.37 A
145
146
18.9
CH APTER 18
(a) Turn the circuit given in Figure P18.9 90° counterclockw ise to observe that it is
equivalent to that show n in Figure 1 below . This red uces, in stages, as show n in the
follow ing figures.
From Figure 4,
I
V
R
25.0 V
1.93 A
12.9
(b) From Figure 3,
P
Wtotal
t
W1 W2
t
7.5 10 12 J 4.1 10
4.5 10 3 s
11
J
1.1 10
(a) From Figures 1 and 2, the current through the 20.0
I 20
18.10
(a)
V
Rbca
ba
5.68 V
25.0
8
W
11 nW
resistor is
0.227 A
By Ohm ’ s law , the current in A is I A
R . The equivalent resistance of the
series com bination of bulbs B and C is 2R . Thus, the current in each of these bulbs
is I B IC
2R .
(b) B and C have the same brightness because they carry the same current.
Direct-Current Circuits
(c)
18.11
147
A is brighter than B or C because it carries twice as much current.
The equivalent resistance is Req
R Rp , w here R p is the total resistance of the three
parallel branches;
Rp
1
120
Thus, 75
1
40
30
R
1
1
R 5.0
1
30
R2
R 5.0
65
R 35
w hich red uces to R2
10
1
1
R 5.0
30
R 5.0
R 35
2
R 150
R 35
R 2 475
2
0 or R 55
0.
R 45
Only the positive solution is physically acceptable, so R
55
18.12 (a) The equivalent resistance of the parallel com bination
1
Rp
is
1
R
1
R
R
2
Rp
or
Rab
and the total resistance betw een a and b is
Vab
(b) From Pab
(c) When
I ab
Vab
max
Rab
2
I ab
R
Vab
, w e have
Vab
resistors is I p
then Ps
2
Rab
Vab
R Rp
max
Rab Pab
max
R
2
3R 2
3R
24 W
2
6 R volts
, the current in the series resistor on the left is
6 R
3R 2
4
and the current through each of the tw o parallel
R
I ab 2 2
4
max
R
R . The pow er d elivered to the series resistor on the left is
2
R
parallel resistors is P p
16 W , and the pow er d elivered to each of the tw o
R
2
p
I R
2
R
2
R
4W .
148
18.13
CH APTER 18
The resistors in the circuit can be com bined in the stages show n below to yield an
equivalent resistance of Rad
.
63 11
V
From Figure 5, I
ad
Rad
Then, from Figure 4,
V
18 V
63 11
bd
3.14 A
I Rbd
3.14 A 30 11
8.57 V
N ow , look at Figure 2 and observe that
V
I2
so
3.0
V
be
bd
2.0
I 2 Rbe
Finally, from Figure 1,
8.57 V
1.71 A
5.0
1.71 A 3.0
I12
5.14 V
V
R12
be
5.14 V
12
0.43 A
Direct-Current Circuits
18.14
149
(a) The resistor netw ork connected to the battery in
Figure P18.14 can be red uced to a single equivalent
resistance in the follow ing steps. The equivalent
resistance of the parallel com bination of
the P fd
V
2
1.8 V
fd
2
0.36 W and
R fd
9.0
6.00- resistors is
1
Rp
Pcd
1
3.00
V
1
6.00
2
3.0 V
cd
Rcd
3
6.00
or
2
1.5 W
6.0
This resistance is in series w ith the 4.00 and the other 2.00 resistor, giving a
total equivalent resistance of
Req 2.00
Rp 4.00
8.00
(b) The current in the Pdb
V
2
db
6.0 V
2
Rdb
6.0
supplied by the battery and is equal to
V 18.0 V
I total
2.25 A
Req 8.00
6.0 W resistor is the total current
(c) The pow er the battery d elivers to the circuit is
P=
18.15
V I total
(a) Connect two 50-
18.0 V 2.25 A
resistors in parallel to get 25
com bination in series with a 20(b) Connect two 50-
40.5 W
resistor for a total resistance of 45
resistors in parallel to get 25
Also, connect two 20-
. Then connect that parallel
.
resistors in parallel to get 10
.
Then, connect these two parallel combinations in series to obtain 35
.
.
150
CH APTER 18
18.16 (a)
The equivalent resistance of the parallel
com bination betw een points b and e is
1
Rbe
1
12
1
24
or
Rbe
8.0
The total resistance betw een points a and e is
then
Rae Rab Rbe 6.0
8.0
14
The total current supplied by the battery (and also the current in the 6.0
is
Vae 42 V
I total I 6
3.0 A
Rae
14
resistor)
The potential d ifference betw een points b and e is
Vbe
so P
V
Rbe I total
2
120 V
8.0
3.0 A
24 V
2
and
I 24
(b) Applying the junction rule at point b yield s
I6
Req
360
40.0 W
Using the loop rule on loop abdea gives
Vbe
Rbde
I12
I 24
42 6I6 24I24
and using the loop rule on loop bcedb gives
12I12
24 V
24
Substituting Equations [2] and [3] into [1] yield s 7I 24
7.0
Then, Equations [2] and [3] yield
and
I6
3.0 A
[1]
0
0 or I 6
24I 24
1.0 A
7.0 4I 24 [2]
0 or I12
or
I12
2I 24
[3]
I 24 1.0 A
2.0 A
Direct-Current Circuits
18.17
151
Going counterclockw ise around the upper loop,
applying Kirchhoff’ s loop rule, gives
15.0 V
or
7.00 I1
5.00 2.00 A
15.0 V 10.0 V
7.00
I1
0.714 A
From Kirchhoff’ s junction rule, I1
so
I2
2.00 A I1
0
I2 2.00 A 0
2.00 A 0.714 A
1.29 A
Going around the low er loop in a clockw ise d irection gives
2.00 I 2
or
18.18
2.00
5.00 2.00 A
1.29 A
0
5.00
2.00 A
12.6 V
Observe that the center branch of this circuit, that is the branch containing points a and
b, is not a continuous cond ucting path, so no current can flow in this branch. The only
current in the circuit flow s counterclockw ise around the perim eter of this circuit. Going
counterclockw ise around the this outer loop and applying Kirchhoff’ s loop rule gives
8.0 V
or
I
2.0
I
12 V 8.0 V
20
3.0
I +12 V
10
I
5.0
I
0
0.20 A
N ow , w e start at point b and go around the upper panel of the circuit to point a, keeping
track of changes in potential as they occur. This gives
Vab Va Vb
Since
18.19
Vab
4.0 V+ 6.0
0
3.0
0.20 A
12 V
10
0.20 A
0 , point a is 5.4 V higher in potential than point b
(a) Applying Kirchhoff’ s loop rule, as you go clockw ise around the loop, gives
20.0 V
or I
2 000 I 30.0 V
3.00 10 3 A
3.00 mA
2 500 I 25.0 V
500 I
0,
5.4 V
152
CH APTER 18
(b) Start at the ground ed point and m ove up the left sid e, record ing changes in
potential as you go, to obtain
VA
20.0 V
or VA
3.00 10
3
A
30.0 V
1000
3.00 10
19.0 V
2
(c) Popen
2 000
2
Reqopen
3R
(The upper end is at the high er potential.)
18.20
Follow ing the path of I1 from a to b, and record ing
changes in potential gives
Vb Va
24 V
6.0
3.0 A
6.0 V
N ow , follow ing the path of I2 from a to b, and
record ing changes in potential gives
Vb Va
3.0
6.0 V , or I 2
I2
2.0 A
Thus, I2 is d irected from from b toward a and has m agnitud e of 2.0 A.
Applying Kirchhoff’ s junction rule at point a gives
I3
18.21
(a)
I1
I2
3.0 A
2.0 A
1.0 A
Applying Kirchhoff’ s junction rule at point a gives
I3
[1]
I1 I 2
Using the loop rule on the low er loop yield s
12 12I2 16I3
0 or
I2 1
4I3
3
[2]
3
A
Direct-Current Circuits
153
Applying the loop rule to loop form ing the outer perim eter of the circuit gives
24 28I1 16I3
or
0
I1
24 16 I 3
28
[3]
Substituting Equations [2] and [3] into [1] yield s I 3
m ultiplying by 84 to elim inate fractions: 84I3
to
and gives
24 16 I 3
28
1
4I3
, and
3
72 48I3 84 112I3 w hich red uces
0.64 A . 244I3 156
I3
Then, Equation [2] gives I 2
0.15 A and Equation [3] yield s I1
0.49 A .
(b) The pow er d elivered to each of the resistors in this circuit is:
P28 I12 R28
and
18.22
0.49 A
Rupper
2
80.0
28
6.7 W ; P12
P16 I32 R16
0.64 A
I 22 R12
2
16
0.15 A
2
12
0.27 W
6.6 W
(a) The 30.0 and 50.0 resistors in the upper branch are in series, and ad d to give a
total resistance of for this path. This 80.0 resistance is in parallel w ith the
80.0 resistance of the m id d le branch, and the rule for com bining resistors in
parallel yield s a total resistance of Rab 40.0 betw een points a and b. This
resistance is in series w ith the 20.0 resistor, so the total equivalent resistance of
the circuit is
Req
20.0
Rab
20.0
40.0
60.0
(b) The current supplied to this circuit by the battery is I total
(c) The pow er d elivered by the battery is
2
Ptotal Req I total
V
Req
60.0
12 V
60.0
0.20 A
0.20 A
2
2.4 W
154
CH APTER 18
(d ) The potential d ifference betw een points a and b is
Vab
Rab I total
40.0
0.20 A
8.0 V
so the pow er d elivered to the 50.0
2
P50 R50 I upper
18.23
50.0
0.10 A
Vab
I upper
and the current in the upper branch is
Rupper
8.0 V
80.0
0.10 A
resistor is
2
0.50 W
(a) We nam e the currents I1, I2 , and I3 as show n.
Applying Kirchhoff’ s loop rule to loop abcfa ,
gives
R2 I2 R1I1 0
1
2
or
3I2
and
I1 5.00 mA 1.50I2
2I1 10.0 mA
[1]
Applying the loop rule to loop edcfe yield s
R3 I3
3
and
I3
2
R2 I2
0 or 3I 2
4I 3
20.0 mA
[2]
5.00 mA 0.750I2
Finally, applying Kirchhoff’ s junction rule at junction c gives
I2
[3]
I1 I3
Substituting Equations [1] and [2] into [3] yield s
I2
5.00 mA 1.50I2 5.00 mA 0.750I2
and I 2
3.08 mA . Then [1] gives I1
or
3.25I2 10.0 mA
0.380 mA , and from [2] I3
2.69 mA .
(b) Start at point c and go to point f, record ing changes in potential to obtain
Vf
Vc
or
V
2
cf
R2 I 2
60.0 V
3.00 103
3.08 10 3 A
69.2 V and point c is at the higher potential
69.2 V
Direct-Current Circuits
18.24
155
(a) Applying Kirchhoff’ s loop rule to the circuit gives
3.00 V
or R
0.255
3.00 V
0.600 A
0.153
0.255
R 0.600 A
0.153
0
4.59
(b) The total pow er input to the circuit is
Pinput
1
I
2
Ploss I 2 r1 r2
1.50 V 1.50 V 0.600 A
0.600 A
2
0.255
1.80 W
0.153
0.147 W
Thus, the fraction of the pow er input that is d issipated intern ally is
Ploss
Pinput
18.25
(a)
(b)
0.147 W
1.80 W
0.081 6 or 8.16%
No. Som e sim plification could be m ad e by recognizing that the 2.0 and 4.0
resistors are in series, ad d ing to give a total of 6.0 ; and the 5.0 and 1.0
resistors form a series com bination w ith a total resistance of 6.0 . The circuit
cannot be sim plified any further, and Kirchhoff’ s rules m ust be used to analyze the
circuit.
Applying Kirchhoff’ s junction rule at junction a
gives
[1]
I1 I 2 I3
Using Kirchhoff’ s loop rule on the upper loop
yield s
24 V 2.0 4.0 I1 3.0 I3 0
or
[2]
I3 8.0 A 2 I1
and for the low er loop,
red uces to
I2
12 V
3.0 I3
12 V 3.0 8.0 A 2 I1
6.0
1.0 5.0 I 2
or
Substituting Equations [2] and [3] into [1] gives
Then, Equation [3] gives I 2
2.5 A ,
I2
0 . Using Equation [2], this
6.0 A I1
I1
3.5 A
and [2] yield s I3 1.0 A
[3]
156
18.26
CH APTER 18
Using Kirchhoff’ s loop rule on the outer
perim eter of the circuit gives
12 V
0.01 I1
0.06 I3
0
I1 1.2 103 A 6 I3
or
[1]
For the rightm ost loop, the loop rule gives
10 V
or
I2
1.00 I 2
0.06 I3
0
[2]
0.06 I3 10 A
Applying Kirchhoff’ s junction rule at either junction gives
I1
I2
[3]
I3
Substituting Equations [1] and [2] into [3] yield s 7.06I3 1210 A and
I3 171 A in starter . Then Equation [2] gives I 2
18.27
(a)
0.26 A in dead battery .
No. This m ulti-loop circuit d oes not contain any resistors in series (i.e., connected
so all the current in one m ust pass through the other) nor in parallel (connected so
the voltage d rop across one is alw ays the sam e as that across the other). Thus, this
circuit cannot be sim plified any further, and Kirchhoff’ s rules m ust be used to
analyze it.
(b) Assum e currents I1, I2 , and I3 in the d irections show n.
Then, using Kirchhoff’ s junction rule at junction a gives
I3
[1]
I1 I 2
Applying Kirchhoff’ s loop rule on the low er loop,
10.0 V
or
I2
5.00 I 2
20.0 I3
2.00 A 4 I3
0
[2]
and for the loop around the perim eter of the circuit,
or
I1 0.667 A 0.667I3
[3]
20.0 V 30.0I1 20.0I3
0
Direct-Current Circuits
Substituting Equations [2] and [3] into [1]: I3
0.667 A 0.667I3 2.00 A 4 I3
w hich red uces to
and gives
5.67I3
Then, Equation [2] gives I 2
2.67 A
0.116 A
and from [3]
I1
I3
157
0.471 A
0.353 A
All currents are in the d irections ind icated in the circuit d iagram given above.
18.28
(a) Going counterclockw ise around the upper loop,
Kirchhoff’ s loop rule gives
11I12 12 7I12 5I18 18 8I18
or
18I12 13I18
0
[1]
30
(b) Going counterclockw ise around the low er loop:
5I36 36 7I12 12 11I12
or
5I36 18I12
0
[2]
24
(c) Applying the junction rule at the nod e in the left end of the circuit gives
I18
I12
[3]
I 36
(d ) Solving Equation [3] for I 36 yield s
I 36
I18
(e) Substituting Equation [4] into [2] gives 5 I18
or
(f)
5I18
23I12
I12
162 , and I12
30 18I12 13 w hich yield s
2.88 A
24
24 23I12 5 . Substituting this into
Equation [1] and sim plifying gives 389I12
(g) Equation [4] gives I36
18I12
[5]
24
Solving Equation [5] for I18 yield s I18
Equation [2], I18
[4]
I12
I18
2.88 A .
0.416 A , or I 36
3.30 A .
0.416 A . Then, from
158
CH APTER 18
(h) The negative sign in the answ er for I12 m eans that this current flow s in the opposite
d irection to that show n in the circuit d iagram and assum ed d uring this solution.
That is, the actual current in the m id d le branch of the circuit flow s from right to left
and has a m agnitud e of 0.416 A.
18.29
Applying Kirchhoff’ s junction rule at
junction a gives
I3
[1]
I1 I 2
Using Kirchhoff’ s loop rule on the leftm ost
loop yield s
3.00 V
so
I1
4.00 I3
5.00 I1 12.0 V 0
9.00 A 4 I3 5.00
or
I1 1.80 A 0.800 I3
[2]
and for the rightm ost loop,
3.00 V
and
I2
4.00 I3
3.00 2.00 I 2 18.0 V 0
15.0 A 4 I3 5.00
or
I2
3.00 A 0.800 I3
[3]
Substituting Equations [2] and [3] into [1] and sim plifying gives 2.60I3 4.80 and
I3 1.846 A . Then Equations [2] and [3] yield I1 0.323 A and I 2 1.523 A .
Therefore, the potential d ifferences across the resistors are
18.30
V2
I 2 2.00
3.05 V , V3
V4
I3 4.00
7.38 V , and V5
The tim e constant is
RC
I 2 3.00
I1 5.00
1.62 V
RC . Consid ering units, w e find
Ohms Farads
Volts
Amperes
Coulombs
Volts
Coulombs
Coulombs Second
or
4.57 V
RC has units of tim e.
Coulombs
Amperes
Second
Direct-Current Circuits
18.31
(a) The tim e constant is:
(b) At t
18.32
(a)
, q 0.632Qmax
RC
(b) Qmax
(c) Ps
18.33
Qmax
100
C
I Req
nR
6
6
2
F
1.88 s
F 12.0 V
1.90 10
2.00 ms
4
1.80 10
C
180 C
nR
1.0 106
RC
6
6
2
V
5.0 10 6 F 30 V
C
0.632 25.0 10
F 9.00 V
nR
25.0 10
2.00 10 3 s
F
2
V
2
s
0.632 C
20.0 10
20.0 10
75.0 103
RC
1.5 10 4 C , and
5.0 10
6
F
5.0 s
Thus, at t 10 s 2
Q Qmax 1 e
18.34
t
1.5 10
4
2
C 1 e
1.3 10
The charge on the capacitor at tim e t is Q Qmax 1 e
Q C
We are given that,
Therefore,
V
V and P1
e
1.0 s
R
10
12
C
, w here
2
. Thus,
12 V , and at P p
1
t
4
12 10
12
V
nP1
t
1 e
n
1
or e
6.0
V
R
t
or e
t
1
2
,
V 10 V
0.75 e 1.0 s
6.0
V
4
C
159
160
CH APTER 18
Taking the natural logarithm of each sid e of the equation gives
1.0 s
Since the tim e constant is
C
18.35
R
0.56 s
RC , w e have
0.56 s
12 103
4.7 10
5
F
47 F
(a) The charge rem aining on the capacitor after tim e t is q Qe
Thus, if q 0.75Q , then and ,
or
(b)
18.36
1.0 s
ln 6.0
or
ln 6.0
(a)
t
ln 0.75
C
RC , so
I max
R
R
(b) Qmax
1.5 s
250 103
R
.
0.43 s
6.0 10
6
F
6.0 F
, so the resistance is
48.0 V
0.500 10-3 A
I max
The tim e constant is
C
1.5 s ln 0.75
t
R
0.960 s
9.60 104
C
9.60 104
RC , so the capacitance is found to be
1.00 10
10.0 F 48.0 V
5
F
10.0 F
480 C , so the charge stored in the capacitor at
t 1.92 s is
Q Qmax 1 e
18.37
t
480 C 1 e
1.92 s
0.960 s
480 C 1 e
2
415 C
The current d raw n by a single 75-W bulb connected to a 120-V source is . Thus, the
num ber of such bulbs that can be connected in parallel w ith this source before the total
current d raw n w ill equal 30.0 A is
n
30.0 A
I1
30.0 A
120 V
75 W
48
Direct-Current Circuits
18.38
(a) The equivalent resistance of the parallel com bination is
1
R1
Req
1
R2
1
1
R3
1
150
1
25
1
50
1
15
so the total current supplied to the circuit is
V
R
I total
120 V
15
8.0 A
(b) Since the appliances are connected in parallel,
the voltage across each one is
(c)
V
I lamp
Rlamp
V
(d ) Pheater
18.39
120 V
150
2
From P
V
2
V
R
120 V .
0.80 A
120 V
Rheater
V
2
5.8 102 W
25
R , the resistance of the elem ent is
2
240 V
P
2
3 000 W
19.2
When the elem ent is connected to a 120-V source, w e find that
(a)
I
(b) P
V
R
120 V
19.2
V I
6.25 A , and
120 V 6.25 A
750 W
161
162
18.40
CH APTER 18
(a) The current d raw n by each appliance operating separately is
Coffee Maker:
Toaster:
Waffle Maker:
I
I
P
P
V
1100 W
120 V
V
I
P
V
1 200 W
120 V
10 A
9.2 A
1 400 W
120 V
12 A
(b) If the three appliances are operated sim ultaneously, they w ill d raw a total current
of I total 10 9.2 12 A 31 A .
(c)
No. The total current required exceed s the lim it of the circuit breaker, so they
cannot be operated sim ultaneously. In fact, w ith a 15 A lim it, no tw o of these
appliances could be operated at the sam e tim e w ithout trip ping the breaker.
Direct-Current Circuits
18.41
163
(a) The area of each surface of this axon m em brane is
A L 2 r
0.10 m 2
10 10
6
m
2
10
6
m2
and the capacitance is
C
0
A
d
3.0 8.85 10
12
2 10 6 m 2
1.0 10-8 m
C2 N m 2
1.67 10
8
F
In the resting state, the charge on the outer surface of the m em brane is
Qi
C
V
1.67 10
i
8
F 70 10
3
V
1.17 10
9
C
1.2 10
9
C
The num ber of potassium ions required to prod uce this charge is
1.17 10 9 C
1.6 10-19 C
Qi
e
NK
7.3 109 K + ions
and the charge per unit area on this surface is
Qi
A
1.17 10 9 C
1e
2 10-6 m2 1.6 10-19 C
10 20 m2
1 Å2
1e
8.6 104 Å 2
1e
290 Å
2
This correspond s to a low charge d ensity of one electronic charge per s quare of sid e
290 Å, com pared to a norm al atom ic spacing of one atom per several Å 2 .
(b) In the resting state, the net charge on the inner surface of the m em brane is
Qi
1.17 10 9 C , and the net positive charge on this surface in the excited state
is
Qf
C
V
f
1.67 10
8
F
30 10
3
V
5.0 10
10
C
The total positive charge w hich m ust pass through the m em brane to prod uce the
excited state is therefore
164
CH APTER 18
Q Qf
Qi
5.0 10
10
C
1.17 10
9
C
1.67 10
9
C
1.7 10
9
C
correspond ing to
1.67 10 9 C
1.6 10-19 C Na + ion
Q
e
N Na +
1.0 1010 Na + ions
(c) If the sod ium ions enter the axon in a tim e of
Q
t
I
1.67 10
2.0 10
9
3
C
s
8.3 10
7
A
t 2.0 ms , the average current is
0.83 A
(d ) When the m em brane becom es perm eable to sod ium ions, the initial influx of
sod ium ions neutralizes the capacitor w ith no requ ired energy input. The energy
input required to charge the now neutral capacitor to the potential d ifference of the
excited state is
W
18.42
1
C
2
V
2
f
1
1.67 10
2
8
F 30 10
3
V
2
7.5 10
12
J
The capacitance of the 10 cm length of axon w as found to be C 1.67 10 8 F in the
solution of Problem 18.41.
(a) When the m em brane becom es perm eable to potassium ions, these ions flow out of
the axon w ith no energy input required until the capacitor is neutralized . To
m aintain this outflow of potassium ions and charge the now n eutral capacitor to the
resting action potential requires an energy input of
W
1
C
2
V
2
1
1.67 10
2
8
F 70 10
3
V
2
4.1 10
11
J .
(b) As found in the solution of Problem 18.41, the charge on the inner surface of the
m em brane in the resting state is 1.17 10 9 C and the charge on this surface in the
excited state is 5.0 10 10 C . Thus, the positive charge w hich m ust flow out of the
axon as it goes from the excited state to the resting state is
Q 5.0 10
10
C 1.17 10 9 C 1.67 10 9 C ,
and the average current d uring the 3.0 m s required to return to the resting state is
PL 3.60 W
Direct-Current Circuits
18.43
From Figure 18.28, the d uration of an action potential pulse is 4.5 m s. From the solution
Problem 18.41, the energy input required to reach the excited state is W1 7.5 10 12 J .
The energy input required d uring the return to the resting state is found in Problem
18.42 to be W2 4.1 10 11 J . Therefore, the average pow er input required d uring an
action potential pulse is
P
18.44
165
Wtotal
t
7.5 10 12 J 4.1 10
4.5 10 3 s
W1 W2
t
Using a single resistor
11
3 d istinct values: R1
J
1.1 10
, R2
2.0
8
W
4.0
11 nW
, R3
6.0
2 resistors in Series
2 ad d itional d istinct values: R4 2.0
6.0
8.0 , and
R5 4.0
6.0
10 . N ote: 2.0 and 4.0 in series duplicates R3 above .
2 resistors in Parallel
R6
R7
R8
2.0
2.0
4.0
and 4.0
and 6.0
and 6.0
3 resistors in Series
R9
2.0
3 ad d itional d istinct values:
1 ad d itional d istinct value:
4.0
3 resistors in Parallel
R10
in parallel 1.3
in parallel 1.5
in parallel 2.4
2.0 , 4.0
6.0
12
1 ad d itional d istinct value:
and 6.0
in parallel 1.1
1 resistor in Parallel w ith Series com bination of the other 2:
ad d itional values:
R11 Rp 2.0 , 4.0 and 6.0 in series 1.7
R12
Rp
4.0
, 2.0
and 6.0
in series
2.7
R12
Rp
6.0
, 2.0
and 4.0
in series
3.0
1 resistor in Series w ith Parallel com bination of the other 2:
P1 I12 R
R15
Rs
4.0 , 2.0
and 6.0
in parallel
5.5
R16
Rs
6.0 , 2.0
and 4.0
in Parallel
7.3
3
3 ad d itional values:
Thus, 16 distinct values of resistance are possible using these three resistors.
166
CH APTER 18
18.45
The resistive netw ork betw een a an b red uces, in the stages show n below , to an
equivalent resistance of Req 7.5
.
18.46
(a)
R
V
I
6.0 V
3.0 10 3 A
2.0 103
2.0 k
(b) The resistance in the circuit consists of a series
com bination w ith an equivalent resistance of P2
The em f of the battery is then
IReq
(c)
V3
IR3
3.0 10
3
A 5.0 103
3.0 10 3 A 3.0 103
I 22 R .
15 V
9.0 V
(d ) In this solution, we have assumed that we have ideal devices in the circuit. In particular,
w e have assum ed that the battery has negligible internal resistance, the voltm eter
has an extrem ely large resistance and d raw s negligible current, and the am m eter
has an extrem ely low resistance and a negligible voltage d rop across it.
Direct-Current Circuits
18.47
(a) The resistors com bine to an equivalent resistance of Req
Vab
Req
(b) From Figure 5, I1
15 V
15
15
167
as show n.
1.0 A
Then, from Figure 4,
Vac
Vdb
I2
From Figure 3,
From Figure 2,
Ved
Then, from Figure 1,
I5
and
(c) From Figure 2,
6.0 V and
I1 6.0
V fd
9.0
Vce
I3
Vcd
6.0
3.0 V
6.0
I3 3.6
I4
Ved
9.0
I3 2.4
Vcd
I1 3.0
3.0 V
0.50 A
1.8 V
Ved
6.0
1.8 V
9.0
1.8 V
6.0
0.30 A
0.20 A
1.2 V . All the other need ed potential d ifferences
w ere calculated above in part (b). The results w ere
Vac
Vdb
6.0 V ; Vcd
3.0 V ; and
V fd
Ved
1.8 V
168
CH APTER 18
(d ) The pow er d issipated in each resistor is found from P
follow ing results:
Pac
6.0 V
ac
Rac
V
Ped
V
2
2
1.8 V
ed
0.54 W
2
3.0 V
Rcd
6.0
V
P fd
V
Pdb
1.2 V
2
fd
1.8 V
2
db
Rdb
6.0 V
6.0
2
R
2
120 V
P
2
240
60.0 W
As connected , the parallel com bination of
R2 and R3 is in series w ith R1 . Thus, the
equivalent resistance of the circuit is
Req
1
R2
R1
1
R3
1
240
1
240
The total pow er d elivered to the circuit is
P
V
Req
2
120 V
360
2
40.0 W
1
240
1
360
R w ith the
2
0.60 W
2
0.36 W
9.0
(a) From P
V R , the resistance of each of the
three bulbs is given by
V
2
2.4
R fd
2
1.5 W
2
ce
Rce
2
6.0
cd
V
Pce
6.0 W
6.0
Red
Pcd
18.48
2
V
V
2
6.0 W
Direct-Current Circuits
V
Req
(b) The current supplied by the source is I
120 V
360
169
1
A . Thus, the potential
3
d ifference across R1 is
V
1
1
A
3
I R1
240
80.0 V
The potential d ifference across the parallel com bination of R2 and R3 is then
V
18.49
(a) From
I
V
2
V
3
source
V
1
120 V 80.0 V
40.0 V
I r Rload , the current supplied w hen the head lights are the entire load is
r
12.6 V
0.080+5.00
Rload
2.48 A
The potential d ifference across the head lights is then
V
I Rload
2.48 A 5.00
12.4 V
(b) The starter m otor connects in parallel w ith the head lights. If I hl is the current
supplied to the head lights, the total current d elivered by t he battery is
I Ihl 35.0 A
The term inal potential d ifference of the battery is
V
V r w hile the current to the head lights is Ihl
35.0 A becom es
I
I
Ihl
V
r
V
5.00
I r , so the total current is
V 5.00 . Thus,
35.0 A
w hich yield s
V
35.0 A r
1 r 5.00
12.6 V
35.0 A 0.080
1+ 0.080
5.00
9.65 V
170
18.50
CH APTER 18
(a) After stead y-state cond itions have been
reached , there is no current in the branch
containing the capacitor.
Thus, for R3 :
I R3
0 steady-state
For the other tw o resistors, the stead y-state
current is sim ply d eterm ined by the 9.00-V em f
across the 12.0-k and 15.0-k resistors in series:
For R1 and R2 :
I
R1 R2
R1
R2
9.00 V
(12.0 k
15.0 k )
333 A steady-state
(b) When the stead y-state has been reached , the potential d ifference across C is the
sam e as the potential d ifference across R2 because there is no change in potential
across R3 . Therefore, the charge on the capacitor is
Q C
V
C I R2
18.51
R2
10.0 F 333 10
6
A 15.0 103
50.0 C
(a) When sw itch S is open, all three bulbs are in
series and the equivalent resistance is
Reqopen R R R 3R .
When the sw itch is closed , bulb C is shorted
across and no current w ill flow through that
bulb. This leaves bulbs A and B in series w ith an equivalent resistance of
Reqclosed R R 2R .
2
(b) With the sw itch open, the pow er d elivered by the battery is Popen
w ith the sw itch closed , Pclosed
2
Reqclosed
2
Reqopen
2
3R
, and
2R .
(c) When the sw itch is open, the three bulbs have equal brightness. When S is closed ,
bulb C goes out, while A and B remain equal at a greater brightness than they had w hen
the sw itch w as open.
Direct-Current Circuits
18.52
171
With the sw itch open, the circuit m ay be red uced as follow s:
With the sw itch closed , the circuit red uces as show n below :
Since the equivalent resistance w ith the sw itch closed is one-half that w hen the sw itch is
open, w e have
1
R 50
2
R 18
18.53
When a generator w ith em f
voltage is V
I r.
If
, w hich yield s R
and internal resistance r supplies current I, its term inal
V 110 V w hen I 10.0 A , then
Given that
V 106 V w hen I 30.0 A , yield s
Subtracting Equation [2] from [1] gives 4.0 V
Equation [1] yield s
18.54
14
110 V
10.0 A r
[1]
106 V
30.0 A r
[2]
20.0 A r , or
112 V .
At tim e t, the charge on the capacitor w ill be Q Qmax 1 e
RC
2.0 106
3.0 10
6
F
When Q 0.90 Qmax , this gives 0.90 1 e
or
e
giving t
t
0.10
r 0.20
Thus,
6.0 s ln 0.10
t
14 s
ln 0.10
6.0 s
t
t
w here
. Then,
172
18.55
CH APTER 18
(a) For the first m easurem ent, the equivalent circuit is
as show n in Figure 1. From this,
Rab
R1
Ry
Ry
2 Ry
1
R1
2
so Ry
[1]
For the second m easurem ent, the equivalent circuit
is show n in Figure 2. This gives
Rac
R2
1
Ry
2
[2]
Rx
Substitute [1] into [2] to obtain
1 1
R1
2 2
R2
(b) If R1 13
Rx , or Rx
and R2
R2
1
R1
4
, then Rx
6.0
Since this exceed s the lim it of 2.0
18.56
2.8
, the antenna is inadequately grounded .
Assum e a set of currents as show n in the circuit
d iagram at the right. Applying Kirchhoff’ s loop
rule to the leftm ost loop gives
75
or
5.0 I
30 I
I1
0
[1]
7 I 6 I1 15
For the rightm ost loop, the loop rule gives
40 R I1
30 I
I1
0 , or I
7
3
R
I1
30
[2]
Substituting Equation [1] into [2] and sim plifying gives
310 I1 7 I1R
[3]
450
Also, it is know n that PR
I12 R 20 W , so I1 R
20 W
I1
[4]
Direct-Current Circuits
173
Substitution of Equation [4] into [3] yield s
140
I1
310 I1
310 I12 450 I1 140 0
or
450
450
Using the quad ratic form ula: I1
yield ing I1 1.0 A and I1
2
4 310 140
2 310
0.452 A . Then, from R =
for the resistance R. These are:
18.57
450
R 20
20 W
, w e find tw o possible values
I12
or R 98
When connected in series, the equivalent resistance is Req
the current is I s
V
,
R1 R2
Rn
n R . Thus,
V n R , and the pow er consum ed by the series
Req
configuration is
V
Ps I Req
2
s
nR
2
2
V
nR
2
nR
For the parallel connection, the pow er consum ed by each ind ivid ual resistor is
P1
V
R
2
, and the total pow er consum ption is
P p nP1
P
Therefore, s
Pp
n
V
2
R
V
nR
2
R
n
V
2
1
or Ps
n2
1
Pp
n2
174
18.58
CH APTER 18
Consid er a battery of em f
connected betw een
points a and b as show n. Applying Kirchhoff’ s
loop rule to loop acbea gives
1.0 I1
or
I3
1.0 I1 I3
0
[1]
2 I1
Applying the loop rule to loop adbea gives
3.0 I 2
or
5.0 I 2
I3
0
[2]
8 I 2 5 I3
For loop adca , the loop rule yield s
3.0 I 2
1.0 I3
1.0 I1
I1
0 or I 2
Substituting Equation [1] into [3] gives
I3
[3]
3
I2
I1
[4]
3
N ow , substitute Equations [1] and [4] into [2] to obtain 18I1
I1
13
27
. Then, Equation [4] gives I 2
13
27
9
27
4
27
18.59
I1
I2
13
27
4
27
17
27
. Therefore,
(a) and (b) - With R the value of the load resistor, the
current in a series circuit com posed of a 12.0 V battery,
an internal resistance of 10.0 , and a load resistor is
I
12.0 V
R 10.0
and the pow er d elivered to the load resistor is
PL I R
2
144 V 2 R
R 10.0
2
, w hich red uces to
, and [1] yield s I 3
Then, applying Kirchhoff’ s junction rule at junction a gives
I
26
3
1
27
.
Direct-Current Circuits
175
Som e typical d ata values for the graph
are
R( )
1.00
5.00
10.0
15.0
20.0
25.0
30.0
P L (W)
1.19
3.20
3.60
3.46
3.20
2.94
2.70
The curve peaks at PL
18.60
3.60 W at a load resistance of R 10.0
.
The total resistance in the circuit is
1
R1
R
1
1
R2
1
2.0 k
and the total capacitance is C C1 C2
Thus, Qmax
and
C
RC
5.0 F 120 V
1.2 103
1
1
3.0 k
1.2 k
2.0 F+3.0 F=5.0 F
600 C
6
5.0 10
F
6.0 10
3
s
6.0 s
1 000
The total stored charge at any tim e t is then
Q Q1 Q2
Qmax 1 e
t
or
Q1 Q2
600 C 1 e
1 000 t 6.0 s
[1]
Since the capacitors are in parallel w ith each other, the sam e potential d ifference exists
across both at any tim e.
Therefore,
V
C
Q1
C1
Q2
,
C2
or
Q2
C2
Q1 1.5 Q1
C1
[2]
Substituting Equation [2] into [1] gives
2.5Q1
600 C 1 e
Then, Equation [2] yield s
1 000 t 6.0 s
and
Q2 1.5 240 C 1 e
Q1
240 C 1 e
1 000 t 6.0 s
1 000 t 6.0 s
360 C 1 e
1 000 t 6.0 s
176
18.61
CH APTER 18
(a) Using the rules for com bining resistors in series and parallel, the circuit red uces as
show n below :
From the figure of Step 3, observe that
I
25.0 V
1.93 A
12.94 A
and
Vab
I 2.94
(b) From the figure of Step 1, observe that I1
18.62
1.93 A 2.94
Vab
25.0
5.68 V
25.0
5.68 V
0.227 A
(a) When the pow er supply is connected to points A and B, the circuit red uces as
show n below to an equivalent resistance of Req 0.099 9 .
From the center figure above, observe that I R1
and
I R2
I R3
I100
5.00 V
111
0.045 0 A
I1
5.00 V
0.100
45.0 mA
50.0 A
Direct-Current Circuits
(b) When the pow er supply is connected to points A and C, the circu it red uces as
show n below to an equivalent resistance of Req 1.09
.
From the center figure above, observe that I R1
I R3
and
I100
5.00 V
110
0.045 5 A
I R2
I1
5.00 V
1.10
4.55 A
45.5 mA
(c) When the pow er supply is connected to points A and D, the circuit red uces as
show n below to an equivalent resistance of Req 9.99
.
From the center figure above, observe that I R1
I100
and
18.63
5.00 V
100
0.050 0 A
I R2
50.0 mA
In the circuit d iagram at the right, note that all points
labeled a are at the sam e potential and equivalent
to each other. Also, all points labeled c are equivalent.
To d eterm ine the voltm eter read ing, go from point e
to point d along the path ecd, keeping track of all
changes in potential to find :
Ved
Vd Ve
4.50 V 6.00 V
1.50 V
I R3
I1
5.00 V
11.1
0.450 A
177
178
CH APTER 18
Apply Kirchhoff’ s loop rule around loop abcfa to find
6.00
I
6.00
I3
or
0
I3
I
[1]
0.600 A 0.600I
[2]
Apply Kirchhoff’ s loop rule around loop abcda to find
6.00
I 6.00 V
10.0
I2
0
or
I2
Apply Kirchhoff’ s loop rule around loop abcea to find
6.00
I
4.50 V
5.00
I1
0
or
[3]
I1 0.900 A 1.20I
Finally, apply Kirchhoff’ s junction rule at either point a or point c to obtain
I
I3
[4]
I1 I 2
Substitute Equations [1], [2], and [3] into Equation [4] to obtain the current through the
am m eter. This gives
I
or
18.64
I
0.900 A 1.20I 0.600 A 0.600I
3.80I 1.50 A and
I 1.50 A 3.80
0.395 A
In the figure given below , note that all bulbs have the sam e resistance, R .
(a) In the series situation, Case 1, the sam e current I1 flow s through both bulbs. Thus,
the sam e pow er, P1 I12 R , is supplied to each bulb. Since the brightness of a bulb is
proportional to the pow er supplied to it, they w ill have the sam e brightness. We
conclud e that the bulbs have the same current, power supplied, and brightness .
(b) In the parallel case, Case 2, the sam e potential d ifference V is m aintained across
each of the bulbs. Thus, the same current I2
V R w ill flow in each branch of
this parallel circuit. This m eans that, again, the same power P2
each bulb, and the tw o bulbs w ill have equal brightness .
I 22 R is supplied to
Direct-Current Circuits
179
(c) The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the
current in this case is I1
V 2R . N ote that this is one half of the current I2 that
flow s through each bulb in the parallel circuit (Case 2). Since the pow er supplied is
proportional to the square of the current, the pow er supplied to each bulb in Case 2
is four tim es that supplied to each bulb in Case 1. Thus, the bulbs in Case 2 are
m uch brighter than those in Case 1.
(d ) If either bulb goes out in Case 1, the only cond ucting path of the circuit is broken
and all current ceases. Thus, in the series case, the other bulb must also go out . If one
bulb goes out in Case 2, there is still a continuous cond uctin g path through the
other bulb. N eglecting any internal resistance of the battery, the battery continues to
m aintain the sam e potential d ifference V across this bulb as w as present w hen
both bulbs w ere lit. Thus, in the parallel case, the second bulb remains lit w ith
unchanged current and brightness w hen one bulb fails.
18.65
(a) The equivalent capacitance of this parallel com bination is
Ceq
C1 C2
3.00 F+2.00 F=5.00 F
When fully charged by a 12.0-V battery, the total stored charge
before the sw itch is closed is
Q0
Ceq
V
5.00 F 12.0 V
60.0 C
Once the sw itch is closed , the tim e constant of the resulting RC circuit is
5.00 102
RCeq
5.00 F
2.50 10
3
s 2.50 ms
Thus, at t 1.00 ms after closing the sw itch, the rem aining total stored charge is
q Q0e
t
60.0 C e 1.00 ms 2.50 ms
60.0 C e
0.400
40.2 C
The potential d ifference across the parallel com bination of capacitors is then
V
q
Ceq
40.2 C
5.00 F
8.04 V
and the charge rem aining on the 3.00 F capacitor w ill be
q3
C3
V
3.00 F 8.04 V
24.1 C
180
CH APTER 18
(b) The charge rem aining on the 2.00 F at this tim e is
q2
q q3
40.2 C 24.1 C
or, alternately,
q2
C2
V
16.1 C
2.00 F 8.04 V
16.1 C
(c) Since the resistor is in parallel w ith the capacitors, it has the sam e potential
d ifference across it as d o the capacitors at all tim es. Thus, Ohm ’ s law gives
I
18.66
V
R
8.04 V
5.00 102
1.61 10
2
A
16.1 mA
The resistor netw ork show n at the right d oes not contain
any obvious com binations of resistors in series or
parallel. Thus, the m ethod of replacing such
com binations by single resistors to sim plify the netw ork
and find the equivalent resistance is not useful here. Still,
the problem is easily solved if one takes note of, and
utilizes, the sym m etry present in this netw ork.
Im agine a battery of em f
connected to points a and d,
supplying current I w hich enters the netw ork at a and
exits at d. When the current reaches a, it has tw o id entical paths that it could follow .
Since the netw ork is totally sym m etric about a horizontal line d raw n from a to d, the
current w ill split equally w ith I 2 flow ing from a to b and I 2 flow ing from a to c. This
m eans Vab
Vac I 2 R , so the potential d ifference betw een points b and c w ill be
zero. Therefore, no current w ill flow through the vertical resistor betw een these tw o
points, and currents of I 2 follow each of the paths b to d and c to d. There, they m erge
form ing current I again, w hich returns to the battery.
The potential d ifference betw een points a and d is
Vac
that IReq
Vcd
I
R
2
I
R
2
IReq . But, it m ay also be w ritten as
IR . Since both are equal to the em f of the battery, w e see
IR , and conclud e that the equivalent resistance of this netw ork is Req
R.
Direct-Current Circuits
18.67
181
(a) With 4.0 103 cells, each w ith an em f of 150 m V, connected in series, the total
term inal potential d ifference is
4.0 103 150 10 3 V
V
6.0 102 V
When d elivering a current of I 1.0 A , the pow er output is
P I
1.0 A 6.0 102 V
V
6.0 102 W
(b) The energy released in one shock is
E1 P
t
1
6.0 102 W 2.0 10 3 s
1.2 J
(c) The energy released in 300 such shocks is Etotal 300E1 300 1.2 J 3.6 102 J . For a
1.0-kg object to be given a gravitational potential energy of this m agnitud e, the
height the object m ust be lifted above the reference level is
h
PEg
3.6 102 J
mg
1.0 kg 9.80 m s2
37 m