142
Chapter 18
PROBLEM SOLUTIONS
18.1
From ΔV = I ( R + r ) , the internal resistance is
r=
18.2
ΔV
9.00 V
−R=
− 72.0 Ω = 4.92 Ω
I
0.117 A
(a)
When the three resistors are in series, the equivalent resistance of the circuit is
Req = R1 + R2 + R3 = 3 ( 9.0 Ω ) = 27 Ω
(b)
The terminal potential difference of the battery is applied across the series combination of
the three 9.0 Ω resistors, so the current supplied by the battery and the current through
each resistor in the series combination is
I=
(c)
ΔV 12 V
=
= 0.44 A
Req 27 Ω
If the three 9.0 Ω are now connected in parallel with each other, the equivalent
resistance is
1
1
1
1
3
=
+
+
=
Req 9.0 Ω 9.0 Ω 9.0 Ω 9.0 Ω
or
Req =
9.0 Ω
= 3.0 Ω
3
When this parallel combination is connected to the battery, the potential difference across
each resistor in the combination is ΔV = 12 V, so the current through each of the
resistors is
I=
18.3
ΔV 12 V
=
= 1.3 A
R
9.0 Ω
For the bulb in use as intended,
Rbulb =
( ΔV )2
P
=
(120 V)2
75.0 W
= 192 Ω
Now, assuming the bulb resistance is unchanged,
the current in the circuit shown is
I=
120 V
ΔV
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
P = I 2 Rbulb = ( 0.620 A )2 (192 Ω ) = 73.8 W
18.4
(a)
When the 8.00-Ω resistor is connected across the 9.00-V terminal potential difference of
the battery, the current through both the resistor and the battery is
I=
(b)
ΔV 9.00 V
=
= 1.13 A
R
8.00 Ω
The relation between the emf and the terminal potential difference of a battery supplying
current I is ΔV = ε − Ir , where r is the internal resistance of the battery. Thus, if the battery
has r = 0.15 Ω and maintains a terminal potential difference of ΔV = 9.00 V while supplying the current found above, the emf of this battery must be
ε = ΔV + Ir = 9.00 V + (1.13 A ) ( 0.15 Ω ) = ( 9.00 + 0.17 ) Ω =
56157_18_ch18_p110-170.indd 142
9.17 Ω
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Direct-Current Circuits
18.5
(a)
143
The equivalent resistance of the two parallel
resistors is
1 ⎞
⎛ 1
Rp = ⎜
+
⎝ 7.00 Ω 10.0 Ω ⎟⎠
−1
= 4.12 Ω
Thus,
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
(b)
18.6
(a)
I ab =
( ΔV )ab
Rab
=
34.0 V
= 1.99 A, so I 4 = I 9 = 1.99 A
17.1 Ω
Also,
( ΔV ) p = I ab Rp = (1.99 A ) ( 4.12 Ω ) = 8.18 V
Then,
I7 =
and
I10 =
( ΔV ) p
R7
( ΔV ) p
R10
=
8.18 V
= 1.17 A
7.00 Ω
=
8.18 V
= 0.818 A
10.0 Ω
The parallel combination of the 6.0 Ω and 12 Ω resistors has an equivalent resistance of
1
1
1
2 +1
=
+
=
Rp1 6.0 Ω 12 Ω 12 Ω
or
Rp1 =
12 Ω
= 4.0 Ω
3
Similarly, the equivalent resistance of the 4.0 Ω and 8.0 Ω parallel combination is
1
1
1
2 +1
=
+
=
Rp 2 4.0 Ω 8.0 Ω 8.0 Ω
or
Rp2 =
8Ω
3
The total resistance of the series combination between points a and b is then
Rab = Rp1 + 5.0 Ω + Rp 2 = 4.0 Ω + 5.0 Ω +
(b)
8.0
35
Ω=
Ω
3
3
If ΔVab = 35 V, the total current from a to b is I ab = ΔVab Rab = 35 V ( 35 Ω 3) = 3.0 A
and the potential differences across the two parallel combinations are
⎛ 8.0 ⎞
ΔVp1 = I ab Rp1 = ( 3.0 A ) ( 4.0 Ω ) = 12 V and ΔVp 2 = I ab Rp 2 = ( 3.0 A ) ⎜
Ω⎟ = 8.0 V
⎝ 3
⎠
so the individual currents through the various resistors are:
56157_18_ch18_p110-170.indd 143
I12 = ΔVp1 12 Ω = 1.0 A ;
I 6 = ΔVp1 6.0 Ω = 2.0 A ; I 5 = I ab = 3.0 A ;
I 8 = ΔVp 2 8.0 Ω = 1.0 A ;
and
I 4 = ΔVp 2 4.0 Ω = 2.0 A
3/19/08 1:35:36 AM
144
18.7
Chapter 18
The equivalent resistance of the parallel combination of three identical resistors is
1
1
1
1
3
=
+
+
=
Rp R1 R2 R3 R
or
Rp =
R
3
The total resistance of the series combination between points a and b is then
Rab = R + Rp + R = 2 R +
18.8
(a)
R 7
= R
3
3
The equivalent resistance of this first parallel
combination is
1
1
1
=
+
Rp1 10.0 Ω 5.00 Ω
or
Rp1 = 3.33 Ω
(b)
For this series combination,
Rupper = Rp1 + 4.00 Ω = 7.33 Ω
(c)
For the second parallel combination,
1
1
1
1
1
=
+
=
+
Rp 2 Rupper 3.00 Ω 7.33 Ω 3.00 Ω
(d)
Rp2 = 2.13 Ω
or
For the second series combination (and hence the entire resistor network),
Rtotal = 2.00 Ω + Rp 2 = 2.00 Ω + 2.13 Ω = 4.13 Ω
(e)
The total current supplied by the battery is
I total =
(f)
ΔV
8.00 V
=
= 1.94 A
Rtotal 4.13 Ω
The potential drop across the 2.00 Ω resistor is
ΔV2 = R2 I total = ( 2.00 Ω ) (1.94 A ) = 3.88 V
(g)
The potential drop across the second parallel combination must be
ΔVp2 = ΔV − ΔV2 = 8.00 V − 3.88 V = 4.12 V
(h)
56157_18_ch18_p110-170.indd 144
So the current through the 3.00 Ω resistor is I total =
ΔVp 2
R3
=
4.12 V
= 1.37 A .
3.00 Ω
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Direct-Current Circuits
18.9
(a)
145
Turn the circuit given in Figure P18.9 90° counterclockwise to observe that it is equivalent
to that shown in Figure 1 below. This reduces, in stages, as shown in the following figures.
Figure 1
Figure 2
Figure 3
Figure 4
From Figure 4,
I=
(b)
ΔV 25.0 V
=
= 1.93 A
R
12.9 Ω
From Figure 3,
( ΔV )ba = IRba = (1.93 A ) ( 2.94 Ω ) =
(a)
From Figures 1 and 2, the current through the 20.0 Ω resistor is
I 20 =
18.10
(a)
5.68 V
( ΔV )ba
Rbca
=
5.68 V
= 0.227 A
25.0 Ω
By Ohm’s law, the current in A is I A = ε R . The equivalent resistance of the series
combination of bulbs B and C is 2R. Thus, the current in each of these bulbs is
I B = IC = ε 2 R .
(b)
(c)
56157_18_ch18_p110-170.indd 145
B and C have the same brightness because they carry the same current .
A is brighter than B or C because it carries twice as much current .
3/19/08 1:35:38 AM
146
18.11
Chapter 18
The equivalent resistance is Req = R + Rp, where Rp is the total resistance of the three parallel
branches;
1
1
⎛ 1
⎞
Rp = ⎜
+
+
⎝ 120 Ω 40 Ω R + 5.0 Ω ⎟⎠
75 Ω = R +
Thus,
( 30 Ω ) ( R + 5.0 Ω )
R + 35 Ω
−1
=
1
⎛ 1
⎞
=⎜
+
⎝ 30 Ω R + 5.0 Ω ⎟⎠
−1
=
( 30 Ω ) ( R + 5.0 Ω )
R + 35 Ω
R 2 + ( 65 Ω ) R + 150 Ω 2
R + 35 Ω
which reduces to R 2 − (10 Ω ) R − 2 475 Ω 2 = 0 or ( R − 55 Ω ) ( R + 45 Ω ) = 0 .
Only the positive solution is physically acceptable, so R = 55 Ω .
18.12
(a)
The equivalent resistance of the parallel combination
is
1
1 1
= +
Rp R R
or
Rp =
R
2
and the total resistance between a and b is
R
Rab = R + Rp = R + = 3R 2
2
(b)
From
Pab =
( ΔVab )max =
(c)
( ΔVab )2 , we have
Rab
Rab (Pab )max =
3R
( 24 W) = 6 R volts
2
When ΔVab = ( ΔVab )max , the current in the series resistor on the left is
I ab =
( ΔVab )max
Rab
=
6 R
4
and the current through each of the two parallel
=
3R 2
R
resistors is I p = I ab 2 = 2
R . The power delivered to the series resistor on the left
2
is then
⎛ 4 ⎞
R = 16 W , and the power delivered to each of the two
⎝ R ⎟⎠
Ps = I ab2 R = ⎜
2
⎛ 2 ⎞
parallel resistors is P p = I p2 R = ⎜
R= 4 W .
⎝ R ⎟⎠
56157_18_ch18_p110-170.indd 146
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Direct-Current Circuits
18.13
147
The resistors in the circuit can be combined in the stages shown below to yield an equivalent
resistance of Rad = ( 63 11) Ω.
Figure 1
Figure 2
Figure 3
Figure 4
( ΔV )ad
From Figure 5,
I=
Then, from Figure 4,
( ΔV )bd
Rad
=
18 V
( 63 11) Ω
Figure 5
= 3.14 A
= I Rbd = ( 3.14 A ) ( 30 11 Ω ) = 8.57 V
Now, look at Figure 2 and observe that
I2 =
so
( ΔV )bd
8.57 V
=
= 1.71 A
3.0 Ω + 2 .0 Ω 5.0 Ω
( ΔV )be = I 2 Rbe = (1.71 A ) ( 3.0 Ω ) = 5.14 V
Finally, from Figure 1,
56157_18_ch18_p110-170.indd 147
I12 =
( ΔV )be
R12
=
5.14 V
= 0.43 A
12 Ω
3/19/08 1:35:39 AM
148
18.14
Chapter 18
(a)
The resistor network connected to the battery in
Figure P18.14 can be reduced to a single equivalent
resistance in the following steps. The equivalent
resistance of the parallel combination of the 3.00 Ω
and 6.00 Ω resistors is
1
1
1
3
=
+
=
Rp 3.00 Ω 6.00 Ω 6.00 Ω
2.00 Ω
+
18.0 V
−
4.00 Ω
6.00 Ω
3.00 Ω
Rp = 2.00 Ω
or
This resistance is in series with the 4.00 Ω and the other 2.00 Ω resistor, giving a total
equivalent resistance of Req = 2.00 Ω + Rp + 4.00 Ω = 8.00 Ω .
(b)
The current in the 2.00 Ω resistor is the total current supplied by the battery and is equal to
I total =
(c)
ΔV 18.0 V
=
= 2.25 A
Req 8.00 Ω
The power the battery delivers to the circuit is
P = ( ΔV ) I total = (18.0 V ) ( 2.25 A ) = 40.5 W
18.15
(a)
Connect two 50-Ω resistors in parallel to get 25 Ω . Then connect that parallel combination
in series with a 20-Ω resistor for a total resistance of 45 Ω.
(b)
Connect two 50-Ω resistors in parallel to get 25 Ω.
Also, connect two 20-Ω resistors in parallel to get 10 Ω.
Then, connect these two parallel combinations in series to obtain 35 Ω.
18.16
(a)
The equivalent resistance of the parallel
combination between points b and e is
1
1
1
=
+
Rbe 12 Ω 24 Ω
or
Rbe = 8.0 Ω
The total resistance between points a and e is then
Rae = Rab + Rbe = 6.0 Ω + 8.0 Ω = 14 Ω
The total current supplied by the battery (and also the current in the 6.0 Ω resistor) is
ΔVae 42 V
=
= 3.0 A
Rae
14 Ω
The potential difference between points b and e is
I total = I 6 =
ΔVbe = Rbe I total = ( 8.0 Ω ) ( 3.0 A ) = 24 V
so I12 =
ΔVbe 24 V
=
= 2.0 A
Rbce 12 Ω
and
I 24 =
ΔVbe 24 V
=
= 1.0 A
Rbde 24 Ω
continued on next page
56157_18_ch18_p110-170.indd 148
3/19/08 3:30:36 AM
Direct-Current Circuits
(b)
Applying the junction rule at point b yields I 6 − I12 − I 24 = 0
[1]
Using the loop rule on loop abdea gives +42 − 6 I 6 − 24 I 24 = 0
or
I 6 = 7.0 − 4 I 24
[2]
and using the loop rule on loop bcedb gives −12 I12 + 24 I 24 = 0
or
I12 = 2 I 24
[3]
Substituting Equations [2] and [3] into [1] yields 7 I 24 = 7.0
Then, Equations [2] and [3] yield I 6 = 3.0 A
18.17
149
and
or
I 24 = 1.0 A
I12 = 2.0 A
Going counterclockwise around the upper loop,
applying Kirchhoff’s loop rule, gives
+15.0 V − ( 7.00 ) I1 − ( 5.00 ) ( 2 .00 A ) = 0
or
I1 =
15.0 V − 10.0 V
= 0.714 A
7.00 Ω
ε
From Kirchhoff’s junction rule, I1 + I 2 − 2 .00 A = 0
so
I 2 = 2 .00 A − I1 = 2 .00 A − 0.714 A = 1.29 A
Going around the lower loop in a clockwise direction gives
+ ε − ( 2 .00 ) I 2 − ( 5.00 ) ( 2 .00 A ) = 0
or
18.18
ε = ( 2 .00 Ω ) (1.29 A ) + ( 5.00 Ω ) ( 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 conducting path, so no current can flow in this branch. The only current in the circuit
flows counterclockwise around the perimeter of this circuit. Going counterclockwise around the
this outer loop and applying Kirchhoff’s loop rule gives
−8.0 V − ( 2.0 Ω ) I − ( 3.0 Ω ) I +12 V − (10 Ω ) I − ( 5.0 Ω ) I = 0
or
I=
12 V − 8.0 V
= 0.20 A
20 Ω
Now, we 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
ΔVab = Va − Vb = − 4.0 V + ( 6.0 Ω ) ( 0 ) − ( 3.0 Ω ) ( 0.20 A ) + 12 V − (10 Ω ) ( 0.20 A ) = +5.4 V
Since ΔVab > 0, point a is 5.4 V higher in potential than point b .
18.19
(a)
Applying Kirchhoff’s loop rule, as you go clockwise around the loop, gives
+ 20.0 V − ( 2 000 ) I − 30.0 V − ( 2 500 ) I + 25.0 V − ( 500 ) I = 0,
or
I = 3.00 × 10 −3 A = 3.00 mA
continued on next page
56157_18_ch18_p110-170.indd 149
3/19/08 3:30:37 AM
150
Chapter 18
(b)
Start at the grounded point and move up the left side, recording changes in potential as you
go, to obtain
)
VA = + 20.0 V − ( 2 000 Ω ) ( 3.00 × 10 −3 A − 30.0 V − (1 000 Ω ) ( 3.00 × 10 −3 A
or
(c)
)
VA = − 19.0 V
( ΔV )1 500 = (1 500 Ω ) ( 3.00 × 10 −3 A ) =
4.50 V
(The upper end is at the higher potential.)
18.20
Following the path of I1 from a to b, and recording changes
in potential gives
Vb − Va = + 24 V − ( 6.0 Ω ) ( 3.0 A ) = + 6.0 V
Now, following the path of I 2 from a to b, and
recording changes in potential gives
Vb − Va = − ( 3.0 Ω ) I 2 = + 6.0 V , or I 2 = − 2 .0 A
Thus, I 2 is directed from from b toward a and has magnitude of 2.0 A.
Applying Kirchhoff’s junction rule at point a gives
I 3 = I1 + I 2 = 3.0 A + ( −2 .0 A ) = 1.0 A
18.21
(a)
Applying Kirchhoff’s junction rule at point a gives
I 3 = I1 + I 2
[1]
Using the loop rule on the lower loop yields
+12 − 12 I 2 − 16 I 3 = 0
or
I2 = 1 −
4 I3
3
[2]
Applying the loop rule to loop forming the outer
perimeter of the circuit gives
+24 − 28 I1 − 16 I 3 = 0
or
I1 =
24 − 16 I 3
28
Substituting Equations [2] and [3] into [1] yields I 3 =
[3]
24 − 16 I 3
4I
+ 1 − 3 , and
28
3
multiplying by 84 to eliminate fractions: 84 I 3 = 72 − 48 I 3 + 84 − 112 I 3 which reduces
to 244 I 3 = 156 and gives I 3 = 0.64 A .
Then, Equation [2] gives I 2 = 0.15 A and Equation [3] yields I1 = 0.49 A .
(b)
The power delivered to each of the resistors in this circuit is:
P28 = I12 R28 = ( 0.49 A )2 ( 28 Ω ) = 6.7 W ; P12 = I 22 R12 = ( 0.15 A )2 (12 Ω ) = 0.27 W
and
56157_18_ch18_p110-170.indd 150
P16 = I 32 R16 = ( 0.64 A )2 (16 Ω ) = 6.6 W
3/19/08 1:35:42 AM
Direct-Current Circuits
18.22
(a)
151
The 30.0 Ω and 50.0 Ω resistors in the upper
branch are in series, and add to give a total resistance
of Rupper = 80.0 Ω for this path. This 80.0 Ω
resistance is in parallel with the 80.0 Ω resistance
of the middle branch, and the rule for combining
resistors in parallel yields a total resistance of
Rab = 40.0 Ω between points a and b. This resistance
is in series with the 20.0 Ω resistor, so the total
equivalent resistance of the circuit is
Req = 20.0 Ω + Rab = 20.0 Ω + 40.0 Ω = 60.0 Ω
(b)
The current supplied to this circuit by the battery is I total =
(c)
The power delivered by the battery is
(d)
The potential difference between points a and b is
ΔV
12 V
=
= 0.20 A .
Req 60.0 Ω
2
2
Ptotal = Req I total
= ( 60.0 Ω ) ( 0.20 A ) = 2.4 W .
ΔVab = Rab I total = ( 40.0 Ω ) ( 0.20 A ) = 8.0 V
ΔVab
8.0 V
=
= 0.10 A
Rupper 80.0 Ω
so the power delivered to the 50.0 Ω resistor is
and the current in the upper branch is I upper =
2
2
P50 = R50 I upper
= ( 50.0 Ω ) ( 0.10 A ) = 0.50 W
18.23
(a)
We name the currents I1 , I 2 , and I 3 as shown.
Applying Kirchhoff’s loop rule to loop abcfa,
gives + ε1 − ε 2 − R2 I 2 − R1 I1 = 0
or
and
ε
ε
ε
3I 2 + 2 I1 = 10.0 mA
I1 = 5.00 mA − 1.50 I 2
[1]
Applying the loop rule to loop edcfe yields
+ ε 3 − R3 I 3 − ε 2 − R2 I 2 = 0
or
and
3I 2 + 4 I 3 = 20.0 mA
I 3 = 5.00 mA − 0.750 I 2
[2]
Finally, applying Kirchhoff’s junction rule at junction c gives
I 2 = I1 + I 3
[3]
Substituting Equations [1] and [2] into [3] yields
I 2 = 5.00 mA − 1.50 I 2 + 5.00 mA − 0.750 I 2
or
3.25 I 2 = 10.0 mA
and I 2 = 3.08 mA . Then [1] gives I1 = 0.380 mA , and from [2] I 3 = 2.69 mA .
(b)
Start at point c and go to point f, recording changes in potential to obtain
)
)
V f − Vc = − ε 2 − R2 I 2 = − 60.0 V − ( 3.00 × 10 3 Ω ( 3.08 × 10 −3 A = − 69.2 V
or
56157_18_ch18_p110-170.indd 151
ΔV
cf
= 69.2 V and point c is at the higher pootential
3/19/08 1:35:43 AM
152
18.24
Chapter 18
(a)
Applying Kirchhoff’s loop rule to the circuit gives
+ 3.00 V − ( 0.255 Ω + 0.153 Ω + R ) ( 0.600 A ) = 0
or R =
(b)
3.00 V
− ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω
0.600 A
The total power input to the circuit is
Pinput = (ε1 + ε 2 ) I = (1.50 V + 1.50 V) ( 0.600 A ) = 1.880 W
Ploss = I 2 ( r1 + r2 ) = ( 0.600 A )2 ( 0.255 Ω + 0.153 Ω ) = 0.147 W
Thus, the fraction of the power input that is dissipated internally is
Ploss 0.147 W
=
= 0.081 6 or 8.16%
Pinput 1.80 W
18.25
(a)
No. Some simplification could be made by recognizing that the 2.0 Ω and 4.0 Ω resistors are in series, adding to give a total of 6.0 Ω; and the 5.0 Ω and 1.0 Ω resistors form
a series combination with a total resistance of 6.0 Ω. The circuit cannot be simplified any
further, and Kirchhoff’s rules must be used to analyze the circuit.
(b)
Applying Kirchhoff’s junction rule at junction a gives
I1 = I 2 + I 3
[1]
Using Kirchhoff’s loop rule on the upper loop yields
+ 24 V − ( 2 .0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0
or
I 3 = 8.0 A − 2 I1
[2]
and for the lower loop,
+ 12 V + ( 3.0 ) I 3 − (1.0 + 5.0 ) I 2 = 0
Using Equation [2], this reduces to
I2 =
12 V + 3.0 ( 8.0 A − 2 I1 )
6.0
or
I 2 = 6.0 A − I1
[3]
Substituting Equations [2] and [3] into [1] gives I1 = 3.5 A .
Then, Equation [3] gives I 2 = 2 .5 A , and [2] yields I 3 = 1.0 A .
56157_18_ch18_p110-170.indd 152
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Direct-Current Circuits
18.26
153
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
+ 12 V − ( 0.01) I1 − ( 0.06 ) I 3 = 0
or
I1 = 1.2 × 10 3 A − 6 I 3
[1]
For the rightmost loop, the loop rule gives
+10 V + (1.00 ) I 2 − ( 0.06 ) I 3 = 0
or
I 2 = 0.06 I 3 − 10 A
[2]
Applying Kirchhoff’s junction rule at either junction gives
I1 = I 2 + I 3
[3]
Substituting Equations [1] and [2] into [3] yields 7.06 I 3 = 1 210 A and I 3 = 171 A ( in starter ) .
Then Equation [2] gives I 2 = 0.26 A ( in dead battery ) .
18.27
(a)
No. This multi-loop circuit does not contain any resistors in series (i.e., connected so all
the current in one must pass through the other) nor in parallel (connected so the voltage
drop across one is always the same as that across the other). Thus, this circuit cannot be
simplified any further, and Kirchhoff’s rules must be used to analyze it.
(b)
Assume currents I1 , I 2 , and I 3 in the directions shown.
Then, using Kirchhoff’s junction rule at junction a gives
I 3 = I1 + I 2
[1]
Applying Kirchhoff’s loop rule on the lower loop,
+ 10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
or
I 2 = 2 .00 A − 4 I 3
[2]
and for the loop around the perimeter of the circuit, 20.0 V − 30.0 I1 − 20.0 I 3 = 0
or
I1 = 0.667 A − 0.667 I 3
[3]
Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667 I 3 + 2 .00 A − 4 I 3
which reduces to 5.67 I 3 = 2.67 A and gives I 3 = 0.471 A .
Then, Equation [2] gives I 2 = 0.116 A and from [3] I1 = 0.353 A .
All currents are in the directions indicated in the circuit diagram given above.
18.28
(a)
Going counterclockwise around the upper loop,
Kirchhoff’s loop rule gives
−11I12 + 12 − 7 I12 − 5 I18 + 18 − 8 I18 = 0
or
18 I12 + 13I18 = 30
[1]
continued on next page
56157_18_ch18_p110-170.indd 153
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154
Chapter 18
(b)
Going counterclockwise around the lower loop:
−5 I 36 + 36 + 7 I12 − 12 + 11I12 = 0
or
(c)
5 I 36 − 18 I12 = 24
[2]
Applying the junction rule at the node in the left end of the circuit gives
I18 = I12 + I 36
[3]
I 36 = I18 − I12
(d)
Solving Equation [3] for I 36 yields
(e)
Substituting Equation [4] into [2] gives
or
(f)
[4]
5 ( I18 − I12 ) − 18 I12 = 24
5 I18 − 23I12 = 24
[5]
Solving Equation [5] for I18 yields I18 = ( 24 + 23I12 ) 5 . Substituting this into Equation [1]
and simplifying gives 389 I12 = −162, and I12 = − 0.416 A . Then, from Equation [2],
I18 = ( 30 − 18 I12 ) 13, which yields I18 = 2.88 A .
(g)
(h)
18.29
Equation [4] gives I 36 = 2.88 A − ( − 0.416 A ) , or I 36 = 3.30 A .
The negative sign in the answer for I12 means that this current flows in the opposite
direction to that shown in the circuit diagram and assumed during this solution. That is,
the actual current in the middle branch of the circuit flows from right to left and has a
magnitude of 0.416 A.
Applying Kirchhoff’s junction rule at
junction a gives
I 3 = I1 + I 2
[1]
Using Kirchhoff’s loop rule on the leftmost
loop yields
−3.00 V − ( 4.00 ) I 3 − ( 5.00 ) I1 + 12.0 V = 0
so
I1 = ( 9.00 A − 4 I 3 ) 5.00 or I1 = 1.80 A − 0.800 I 3
[2]
and for the rightmost loop,
−3.00 V − ( 4.00 ) I 3 − ( 3.00 + 2 .00 ) I 2 + 18.0 V = 0
and
I 2 = (15.0 A − 4 I 3 ) 5 .00 or I 2 = 3.00 A − 0.800 I 3
[3]
Substituting Equations [2] and [3] into [1] and simplifying gives 2.60 I 3 = 4.80 and
I 3 = 1.846 A. Then Equations [2] and [3] yield I1 = 0.323 A and I 2 = 1.523 A.
Therefore, the potential differences across the resistors are
ΔV2 = I 2 ( 2 .00 Ω ) = 3.05 V , ΔV3 = I 2 ( 3 .00 Ω ) = 4.57 V
ΔV4 = I 3 ( 4 .00 Ω ) = 7.38 V , and ΔV5 = I1 ( 5 .00 Ω ) = 1.62 V
56157_18_ch18_p110-170.indd 154
3/19/08 1:35:47 AM
Direct-Current Circuits
18.30
155
The time constant is τ = R C. Considering units, we find
⎛ Volts ⎞ ⎛ Coulombss ⎞ ⎛ Coulombs ⎞
R C → ( Ohms ) ( Farads ) = ⎜
⎜
⎟=
⎝ Amperes ⎟⎠ ⎝ Volts ⎠ ⎜⎝ Amperes ⎟⎠
⎛
⎞
Coulombs
=⎜
= Second
⎝ Coulombs Second ⎟⎠
or
18.31
18.32
18.33
τ = R C has units of time.
)
)
(a)
The time constant is: τ = RC = ( 75.0 × 10 3 Ω ( 25.0 × 10 −6 F = 1.88 s
(b)
At t = τ , q = 0.632Qmax = 0.632 ( Cε ) = 0.632 ( 25.0 × 10 −6 F (12.0 V) = 1.90 × 10 −4 C
(a)
τ = R C = (100 Ω ) ( 20.0 × 10 −6 F ) = 2 .00 × 10 −3 s = 2 .00 ms
(b)
Qmax = Cε = ( 20.0 × 10 −6 F ( 9.00 V) = 1.80 × 10 −4 C = 180 μC
(c)
⎛ 1⎞
Q = Qmax (1 − e−t τ = Qmax (1 − e−τ τ = Qmax ⎜ 1 − ⎟ = 114 μC
⎝ e⎠
)
)
)
)
)
Qmax = Cε = ( 5.0 × 10 −6 F ( 30 V) = 1.5 × 10 −4 C , and
τ = R C = (1.0 × 10 6 Ω ) ( 5.0 × 10 −6 F ) = 5.0 s
Thus, at t = 10 s = 2 τ
)
)
)
Q = Qmax (1 − e−t τ = (1.5 × 10 −4 C (1 − e−2 = 1.3 × 10 −4 C
18.34
)
)
The charge on the capacitor at time t is Q = Qmax (1 − e−t τ , where
Q = C ( ΔV ) and Qmax = Cε . Thus, ΔV = ε (1 − e−t τ
We are given that
ε = 12 V, and at
Therefore, e−1.0 s τ = 1 −
or e−t τ = 1 − ( ΔV )
ε
t = 1.0 s, ΔV = 10 V
10 12 − 10
1
=
=
or e+1.0 s τ = 6.0
12
12
6.0
Taking the natural logarithm of each side of the equation gives
1.0 s
= ln ( 6.0 )
τ
or
τ=
1.0 s
= 0.56 s
ln ( 6.0 )
Since the time constant is τ = RC , we have
C=
56157_18_ch18_p110-170.indd 155
τ
0.56 s
=
= 4.7 × 10 −5 F = 47 μ F
R 12 × 10 3 Ω
3/19/08 1:35:48 AM
156
18.35
Chapter 18
(a)
The charge remaining on the capacitor after time t is q = Qe−t τ .
Thus, if q = 0.75Q , then e−t τ = 0.75 and −t τ = ln ( 0.75 ),
t = −τ ln ( 0.75 ) = − (1.5 s ) ln ( 0.75 ) = 0.43 s
or
18.36
(b)
τ = RC , so
(a)
I max =
ε
R
C=
τ
1.5 s
=
= 6.0 × 10 −6 F = 6.0 μ F
R 250 × 10 3 Ω
, so the resistance is
R=
ε
I max
=
48.0 V
= 9.60 × 10 4 Ω
0.500 × 10 −3 A
The time constant is τ = R C, so the capacitance is found to be
C=
(b)
τ
0.960 s
=
= 1.00 × 10 −5 F = 10.0 μ F
R 9.60 × 10 4 Ω
Qmax = Cε = (10.0 μ F ) ( 48.0 V) = 480 μC, so the charge stored in the capacitor at
t = 1.92 s is
1.92 s
−
⎛
⎞
Q = Qmax (1 − e−t τ = ( 480 μC ) ⎜ 1 − e 0.960 s ⎟ = ( 480 μC ) (1 − e−2 = 415 μC
⎠
⎝
)
18.37
The current drawn by a single 75-W bulb connected to a 120-V source is
I1 = P ΔV = 75 W 120 V. Thus, the number of such bulbs that can be connected in parallel with
this source before the total current drawn will equal 30.0 A is
n=
18.38
)
(a)
30.0 A
⎛ 120 V ⎞
= 48
= ( 30.0 A ) ⎜
⎝ 75 W ⎟⎠
I1
The equivalent resistance of the parallel combination is
⎛ 1
1
1⎞
Req = ⎜ +
+ ⎟
⎝ R1 R2 R3 ⎠
−1
1
1 ⎞
⎛ 1
=⎜
+
+
⎝ 150 Ω 25 Ω 50 Ω ⎟⎠
−1
= 15 Ω
so the total current supplied to the circuit is
I total =
(b)
ΔV 120 V
=
= 8.0 A
15 Ω
R
Since the appliances are connected in parallel,
the voltage across each one is ΔV = 120 V .
56157_18_ch18_p110-170.indd 156
ΔV
120 V
=
= 0.80 A
Rlamp 150 Ω
( ΔV )2 (120 V)2
=
=
= 5.8 × 10 2 W
Rheater
25 Ω
(c)
I lamp =
(d)
Pheater
3/19/08 1:35:49 AM
Direct-Current Circuits
18.39
From
157
P = ( ΔV )2 R, the resistance of the element is
R=
( ΔV )2 ( 240 V)2
P
=
3 000 W
= 19.2 Ω
When the element is connected to a 120-V source, we find that
18.40
18.41
ΔV 120 V
=
= 6.25 A , and
R 19.2 Ω
(a)
I=
(b)
P = ( ΔV ) I = (120 V) ( 6.25 A ) = 750 W
(a)
The current drawn by each appliance operating separately is
Coffee Maker:
I=
Toaster:
I=
Waffle Maker:
I=
P
ΔV
P
ΔV
P
ΔV
=
1 200 W
= 10 A
120 V
=
1100 W
= 9.2 A
120 V
=
1 400 W
= 12 A
120 V
(b)
If the three appliances are operated simultaneously, they will draw a total current
of I total = (10 + 9.2 + 12 ) A = 31 A.
(c)
No. The total current required exceeds the limit of the circuit breaker, so they cannot
be operated simultaneously. In fact, with a 15 A limit, no two of these appliances could be
operated at the same time without tripping the breaker.
(a)
The area of each surface of this axon membrane is
)
A = L ( 2π r ) = ( 0.10 m ) ⎡⎣ 2π (10 × 10 −6 m ⎤⎦ = 2π × 10 −6 m 2
and the capacitance is
C = κ ∈0
⎛ 2π × 10 −6 m 2 ⎞
A
= 1.67 × 10 −8 F
= 3.0 ( 8.85 × 10 −12 C2 N ⋅ m 2 ⎜
d
⎝ 1.0 × 10 −8 m ⎟⎠
)
In the resting state, the charge on the outer surface of the membrane is
)
)
Qi = C ( ΔV )i = (1.67 × 10 −8 F ( 70 × 10 −3 V = 1.17 × 10 −9 C → 1.2 × 10 −9 C
The number of potassium ions required to produce this charge is
N K+ =
Qi 1.17 × 10 −9 C
=
= 7.3 × 10 9 K + ions
e 1.6 × 10 −19 C
and the charge per unit area on this surface is
σ=
−20
2
Qi 1.17 × 10 −9 C ⎛
1e
1e
⎞ ⎛ 10 m ⎞
=
=
=
⎟
2 ⎜
−6
−19
2
⎟
⎜
A 2π × 10 m ⎝ 1.6 × 10 C ⎠ ⎝ 1 Å ⎠ 8.6 × 10 4 Å 2
1e
( 290 Å)
2
This corresponds to a low charge density of one electronic charge per square of side 290 Å,
compared to a normal atomic spacing of one atom per several Å 2.
continued on next page
56157_18_ch18_p110-170.indd 157
3/19/08 3:30:39 AM
158
Chapter 18
(b)
In the resting state, the net charge on the inner surface of the membrane is
− Qi = − 1.17 × 10 −9 C, and the net positive charge on this surface in the excited state is
)
)
Q f = C ( ΔV ) f = (1.67 × 10 −8 F ( +30 × 10 −3 V = + 5.0 × 10 −10 C
The total positive charge which must pass through the membrane to produce the excited
state is therefore
ΔQ = Q f − Qi
)
= + 5.0 × 10 −10 C − ( − 1.17 × 10 −9 C = 1.67 × 10 −9 C → 1.7 × 10 −9 C
corresponding to
N Na + =
(c)
If the sodium ions enter the axon in a time of Δt = 2 .0 ms, the average current is
I=
(d)
ΔQ 1.67 × 10 −9 C
=
= 8.3 × 10 −7 A = 0.83 μ A
Δt
2 .0 × 10 −3 s
When the membrane becomes permeable to sodium ions, the initial influx of sodium ions
neutralizes the capacitor with no required energy input. The energy input required to charge
the now neutral capacitor to the potential difference of the excited state is
W =
18.42
ΔQ
1.67 × 10 −9 C
=
= 1.0 × 1010 Na + ions
1.6 × 10 −19 C Na + ion
e
)
1
1
2
C ( ΔV ) f = (1.67 × 10 −8 F ( 30 × 10 −3 V
2
2
)
2
= 7.5 × 10 −12 J
The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10 −8 F in the solution of
Problem 18.41.
(a)
When the membrane becomes permeable to potassium ions, these ions flow out of the axon
with no energy input required until the capacitor is neutralized. To maintain this outflow of
potassium ions and charge the now neutral capacitor to the resting action potential requires
an energy input of
W =
(b)
)
1
1
2
C ( ΔV ) = (1.67 × 10 −8 F ( 70 × 10 −3 V
2
2
)
2
= 4.1 × 10 −11 J
As found in the solution of Problem 18.41, the charge on the inner surface of the membrane
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 which must flow out of the axon as it goes from
the excited state to the resting state is
ΔQ = 5.0 × 10 −10 C + 1.17 × 10 −9 C = 1.67 × 10 −9 C
and the average current during the 3.0 ms required to return to the resting state is
I=
18.43
From Figure 18.28, the duration of an action potential pulse is 4.5 ms. From the solution of
Problem 18.41, the energy input required to reach the excited state is W1 = 7.5 × 10 −12 J. The
energy input required during the return to the resting state is found in Problem 18.42 to be
W2 = 4.1 × 10 −11 J. Therefore, the average power input required during an action potential pulse is
P=
56157_18_ch18_p110-170.indd 158
ΔQ 1.67 × 10 −9 C
=
= 5.6 × 10 −7 A = 0.56 μ A
Δt
3.0 × 10 −3 s
Wtotal W1 + W2 7.5 × 10 −12 J + 4.1 × 10 −11 J
= 1.1 × 10 −8 W = 11 nW
=
=
4..5 × 10 −3 s
Δt
Δt
3/19/08 1:35:50 AM
Direct-Current Circuits
18.44
159
Using a single resistor → 3 distinct values: R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 6.0 Ω
2 resistors in Series → 2 additional distinct values: R4 = 2.0 Ω + 6.0 Ω = 8.0 Ω, and
R5 = 4.0 Ω + 6.0 Ω = 10 Ω. Note: 2.0 Ω and 4.0 Ω in series duplicates R3 above.
2 resistors in Parallel → 3 additional distinct values:
R6 = 2.0 Ω and 4.0 Ω in parallel = 1.3 Ω
R7 = 2.0 Ω and 6.0 Ω in parallel = 1.5 Ω
R8 = 4.0 Ω and 6.0 Ω in parallel = 2.4 Ω
3 resistors in Series → 1 additional distinct value:
R9 = 2.0 Ω + 4.0 Ω + 6.0 Ω = 12 Ω
3 resistors in Parallel → 1 additional distinct value:
R10 = 2.0 Ω, 4.0 Ω, and 6.0 Ω in parallel = 1.1 Ω
1 resistor in Parallel with Series combination of the other 2: → 3 additional values:
(
= (R
= (R
)
= 4.0 Ω, 2.0 Ω, and 6.0 Ω in series ) = 2.7 Ω
= 6.0 Ω, 2.0 Ω, and 4.0 Ω in series ) = 3.0 Ω
R11 = Rp = 2.0 Ω, 4.0 Ω, and 6.0 Ω in series = 1.7 Ω
R12
R12
p
p
1 resistor in Series with Parallel combination of the other 2: → 3 additional values:
R14 = ( Rs = 2.0 Ω, 4.0 Ω and 6.0 Ω in parallel ) = 4.4 Ω
R15 = ( Rs = 4.0 Ω, 2.0 Ω, and 6.0 Ω in parallel ) = 5.5 Ω
R16 = ( Rs = 6.0 Ω, 2.0 Ω, and 4.0 Ω in Parallel ) = 7.3 Ω
Thus, 16 distinct values of resistance are possible using these three resistors.
18.45
The resistive network between a an b reduces, in the stages shown below, to an equivalent
resistance of Req = 7.5 Ω .
18.46
(a)
R=
(b)
The resistance in the circuit consists of a series combination
with an equivalent resistance of Req = 2.0 kΩ + 3.0 kΩ = 5.0 kΩ.
The emf of the battery is then
ΔV
6.0 V
=
= 2.0 × 10 3 Ω = 2.0 kΩ
I
3.0 × 10 −3 A
ε = IReq = ( 3.0 × 10 −3 A ) ( 5.0 × 10 3 Ω ) = 15 V
continued on next page
56157_18_ch18_p110-170.indd 159
3/19/08 3:30:39 AM
160
Chapter 18
(c)
18.47
)
)
ΔV3 = IR3 = ( 3.0 × 10 −3 A ( 3.0 × 10 3 Ω = 9.0 V
(d)
In this solution, we have assumed that we have ideal devices in the circuit . In particular,
we have assumed that the battery has negligible internal resistance, the voltmeter has an
extremely large resistance and draws negligible current, and the ammeter has an extremely
low resistance and a negligible voltage drop across it.
(a)
The resistors combine to an equivalent resistance of Req = 15 Ω as shown.
Figure 1
Figure 2
Figure 3
(b)
From Figure 5, I1 =
Figure 4
Figure 5
ΔVab 15 V
=
= 1.0 A
Req
15 Ω
Then, from Figure 4,
ΔVac = ΔVdb = I1 ( 6.0 Ω ) = 6.0 V and ΔVcd = I1 ( 3.0 Ω ) = 3.0 V
I2 = I3 =
From Figure 2,
ΔVed = I 3 ( 3.6 Ω ) = 1.8 V
ΔVed
1.8 V
=
= 0.30 A
6.0 Ω 6.0 Ω
ΔV fd
1.8 V
ΔVed
I5 =
=
=
= 0.20 A
9.0 Ω 9.0 Ω 9.0 Ω
Then, from Figure 1,
and
(c)
ΔVcd
3.0 V
=
= 0.50 A
6.0 Ω 6.0 Ω
From Figure 3,
I4 =
From Figure 2, ΔVce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences were
calculated above in part (b). The results were
ΔVac = ΔVdb = 6.0 V ; ΔVcd = 3.0 V ; and ΔV fd = ΔVed = 1.8 V
continued on next page
56157_18_ch18_p110-170.indd 160
3/19/08 3:30:40 AM
Direct-Current Circuits
(d)
The power dissipated in each resistor is found from
results:
Pac =
Ped =
Pcd =
18.48
(a)
( ΔV )ac2
=
Rac
( ΔV )ed2
=
Red
( ΔV )cd2
=
Rcd
( 6.0 V)2
6.0 Ω
(1.8 V)2
= 0.54 W
6.0 Ω
( 3.0 V)2
6.0 Ω
= 6.0 W
= 1.5 W
Pce =
Pdb =
P = ( ΔV )2 R with the following
( ΔV )ce2
Rce
P fd =
161
=
( ΔV )2fd
Rdb
= 0.60 W
2.4 Ω
=
R fd
( ΔV )db2
(1.2 V)2
=
(1.8 V)2
9.0 Ω
( 6.0 V)2
6.0 Ω
= 0.36 W
= 6.0 W
From P = ( ΔV ) R, the resistance of each of the
three bulbs is given by
2
( ΔV )2
R=
P
=
(120 V)2
60.0 W
= 240 Ω
As connected, the parallel combination of
R2 and R3 is in series with R1. Thus, the
equivalent resistance of the circuit is
⎛ 1
1⎞
Req = R1 + ⎜
+
⎝ R2 R3 ⎟⎠
−1
1 ⎞
⎛ 1
= 240 Ω + ⎜
+
⎝ 240 Ω 240 Ω ⎟⎠
−1
= 360 Ω
The total power delivered to the circuit is
P=
(b)
( ΔV )2
Req
=
(120 V)2
360 Ω
= 40.0 W
ΔV 120 V 1
=
= A . Thus, the potential differThe current supplied by the source is I =
Req 360 Ω 3
ence across R1 is
( ΔV )1 = I R1 = ⎛⎜⎝
1 ⎞
A⎟ ( 240 Ω ) = 80.0 V
3 ⎠
The potential difference across the parallel combination of R2 and R3 is then
( ΔV )2 = ( ΔV )3 = ( ΔV )source − ( ΔV ) 1 = 120 V − 80.0 V =
18.49
(a)
From
40.0 V
ε = I ( r + Rload ), the current supplied when the headlights are the entire load is
I=
ε
r + Rload
=
12 .6 V
= 2 .48 A
( 0.080 + 5.00 ) Ω
The potential difference across the headlights is then
ΔV = I Rload = ( 2.48 A ) ( 5.00 Ω ) = 12.4 V
continued on next page
56157_18_ch18_p110-170.indd 161
3/19/08 3:30:41 AM
162
Chapter 18
(b)
The starter motor connects in parallel with the headlights. If I hl is the current supplied to
the headlights, the total current delivered by the battery is I = I hl + 35.0 A.
The terminal potential difference of the battery is ΔV = ε − I r , so the total current
is I = (ε − ΔV ) r while the current to the headlights is I hl = ΔV 5.00 Ω.
Thus, I = I hl + 35.0 A becomes
ε − ΔV
r
=
ΔV
+ 35.0 A
5.00 Ω
which yields
ΔV =
18.50
(a)
ε − ( 35.0 A ) r 12.6 V − ( 35.0 A ) ( 0.080 Ω )
=
= 9.65 V
1 + r ( 5.00 Ω )
1 + ( 0.080 Ω ) ( 5.00 Ω )
After steady-state conditions have been
reached, there is no current in the branch
containing the capacitor.
I R3 = 0 ( steady-state )
Thus, for R3 :
ε
For the other two resistors, the steady-state
current is simply determined by the 9.00-V emf
across the 12.0-kΩ and 15.0-kΩ resistors in series:
For R1 and R2 :
I ( R1 + R2 ) =
(b)
ε
R1 + R2
=
9.00 V
= 333 μ A ( steady-state )
(12.0 k Ω + 15.0 k Ω )
When the steady state has been reached, the potential difference across C is the same as the
potential difference across R2 because there is no change in potential across R3. Therefore,
the charge on the capacitor is
Q = C ( ΔV )R2
)
)
= C ( I R2 ) = (10.0 μ F ) ( 333 × 10 −6 A (15.0 × 10 3 Ω = 50.0 μC
18.51
(a)
When switch S is open, all three bulbs are
in series and the equivalent resistance is
Reqopen = R + R + R = 3R .
When the switch is closed, bulb C is shorted
across and no current will flow through that
bulb. This leaves bulbs A and B in series with
an equivalent resistance of
Reqclosed = R + R = 2 R .
(b)
With the switch open, the power delivered by the battery is
the switch closed,
(c)
Pclosed = ε 2 Reqclosed = ε 2 2 R .
Popen =
ε2
open
eq
R
=
ε2
3R
, and with
When the switch 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 when
the switch was open.
56157_18_ch18_p110-170.indd 162
3/19/08 1:35:54 AM
Direct-Current Circuits
18.52
163
With the switch open, the circuit may be reduced as follows:
With the switch closed, the circuit reduces as shown below:
Since the equivalent resistance with the switch closed is one-half that when the switch is open,
we have
1
R + 18 Ω = ( R + 50 Ω ), which yields R = 14 Ω
2
18.53
When a generator with emf
ΔV = ε − I r.
ε
and internal resistance r supplies current I, its terminal voltage is
If ΔV = 110 V when I = 10.0 A , then
110 V = ε − (10.0 A ) r
[1]
Given that ΔV = 106 V when I = 30.0 A, yields
106 V = ε − ( 30.0 A ) r
[2]
Subtracting Equation [2] from [1] gives 4.0 V = ( 20.0 A ) r , or r = 0.20 Ω . Then, Equation [1]
yields
18.54
ε = 112 V
.
)
At time t, the charge on the capacitor will be Q = Qmax (1 − e−t τ , where
τ = RC = ( 2 .0 × 10 6 Ω ) ( 3.0 × 10 −6 F ) = 6.0 s
When Q = 0.90 Qmax , this gives 0.90 = 1 − e−t τ
or
e−t τ = 0.10
Thus, −
t
= ln ( 0.10 )
τ
giving t = − ( 6.0 s ) ln ( 0.10 ) = 14 s
18.55
(a)
For the first measurement, the equivalent circuit is
as shown in Figure 1. From this,
Rab = R1 = Ry + Ry = 2 Ry
1
R1
[1]
2
For the second measurement, the equivalent circuit
is shown in Figure 2. This gives
so
Ry =
Rac = R2 =
1
Ry + Rx
2
Figure 1
[2]
Substitute [1] into [2] to obtain
R2 =
1⎛1 ⎞
1
⎜ R 1 ⎟ + Rx, or Rx = R2 − R1
2⎝2 ⎠
4
Figure 2
continued on next page
56157_18_ch18_p110-170.indd 163
3/19/08 3:30:42 AM
164
Chapter 18
(b)
If R1 = 13 Ω and R2 = 6.0 Ω , then Rx = 2 .8 Ω .
Since this exceeds the limit of 2 .0 Ω , the antenna is inadequately grounded .
18.56
Assume a set of currents as shown in the circuit
diagram at the right. Applying Kirchhoff’s loop
rule to the leftmost loop gives
+ 75 − ( 5.0 ) I − ( 30 ) ( I − I1 ) = 0
or
7 I − 6 I1 = 15
[1]
For the rightmost loop, the loop rule gives
⎛7 R⎞
− ( 40 + R ) I1 + ( 30 ) ( I − I1 ) = 0, or I = ⎜ + ⎟ I1
⎝ 3 30 ⎠
[2]
Substituting Equation [1] into [2] and simplifying gives
310 I1 + 7 ( I1 R ) = 450
Also, it is known that
[3]
PR = I12 R = 20 W, so I1 R =
20 W
I1
[4]
Substitution of Equation [4] into [3] yields
310 I1 +
140
= 450
I1
or
Using the quadratic formula: I1 =
310 I12 − 450 I1 + 140 = 0
− ( − 450 ) ±
( − 450 )2 − 4 ( 310 ) (140 )
,
2 ( 310 )
yielding I1 = 1.0 A and I1 = 0.452 A. Then, from R =
20 W
, we find two possible values for
I12
the resistance R. These are: R = 20 Ω or R = 98 Ω
18.57
When connected in series, the equivalent resistance is Req = R1 + R2 + ⋅ ⋅ ⋅ + Rn = n R . Thus, the
current is I s = ( ΔV ) Req = ( ΔV ) n R, and the power consumed by the series configuration is
Ps = I s2 Req =
( ΔV )2
(n R)
2 (n R) =
( ΔV )2
nR
For the parallel connection, the power consumed by each individual resistor is
the total power consumption is
P p = nP1 =
Therefore,
56157_18_ch18_p110-170.indd 164
n ( ΔV )
R
P1 =
( ΔV )2
R
, and
2
1
Ps ( ΔV )2
1
R
or Ps = 2 P p
=
⋅
2 =
2
n
Pp
n R n ( ΔV )
n
3/19/08 1:35:57 AM
Direct-Current Circuits
18.58
Consider a battery of emf ε connected between
points a and b as shown. Applying Kirchhoff’s
loop rule to loop acbea gives
165
ε
− (1.0 ) I1 − (1.0 ) ( I1 − I 3 ) + ε = 0
or
I 3 = 2 I1 − ε
[1]
Applying the loop rule to loop adbea gives
− ( 3.0 ) I 2 − ( 5.0 ) ( I 2 + I 3 ) + ε = 0
or
8 I2 + 5 I3 = ε
[2]
For loop adca, the loop rule yields
I1 + I 3
3
[3]
Substituting Equation [1] into [3] gives I 2 = I1 − ε 3.
[4]
− ( 3.0 ) I 2 + (1.0 ) I 3 + (1.0 ) I1 = 0 or I 2 =
26
ε , which reduces to I1 = 13 ε .
3
27
4
=
ε , and [1] yields I 3 = − 1 ε .
27
27
Now, substitute Equations [1] and [4] into [2] to obtain 18 I1 =
⎡ 13 9 ⎤
Then, Equation [4] gives I 2 = ⎢ − ⎥ ε
⎣ 27 27 ⎦
Then, applying Kirchhoff’s junction rule at junction a gives
I = I1 + I 2 =
18.59
13
ε + 4 ε = 17 ε . Therefore, Rab = ε = ε = 27 Ω .
I (17 ε 27 )
27
27
27
17
(a) and (b). With R the value of the load resistor, the
current in a series circuit composed of a 12.0 V battery,
an internal resistance of 10.0 Ω, and a load resistor is
I=
12.0 V
R + 10.0 Ω
and the power delivered to the load resistor is
(144 V ) R
2
PL = I 2 R =
( R + 10.0 Ω )2
Some typical data values for the graph are
R (Ω)
PL (W)
1.00
1.19
5.00
3.20
10.0
3.60
15.0
3.46
20.0
3.20
25.0
2.94
30.0
2.70
PL
The curve peaks at PL = 3.60 W at a load resistance of R = 10.0 Ω.
56157_18_ch18_p110-170.indd 165
3/19/08 1:35:57 AM
166
18.60
Chapter 18
The total resistance in the circuit is
⎛ 1
1⎞
R=⎜ + ⎟
⎝ R1 R2 ⎠
−1
⎛ 1
1 ⎞
=⎜
+
⎝ 2 .0 kΩ 3.0 kΩ ⎟⎠
−1
= 1.2 kΩ
and the total capacitance is C = C1 + C2 = 2 .0 μ F + 3.0 μ F = 5.0 μ F.
Thus, Qmax = C ε = ( 5.0 μ F ) (120 V) = 600 μC
and
τ = RC = (1.2 × 10 3 Ω ) ( 5.0 × 10 −6 F ) = 6.0 × 10 −3 s =
6.0 s
1 000
The total stored charge at any time t is then
Q = Q1 + Q2 = Qmax (1 − e−t τ
)
or
Q1 + Q2 = ( 600 μC ) (1 − e−1 000 t
6.0 s
)
[1]
Since the capacitors are in parallel with each other, the same potential difference exists across
both at any time.
( ΔV )C
Therefore,
=
Q1 Q2
=
,
C1 C2
⎛C ⎞
Q2 = ⎜ 2 ⎟ Q1 = 1.5 Q1
⎝ C1 ⎠
or
[2]
Substituting Equation [2] into [1] gives
2.5Q1 = ( 600 μC ) (1 − e−1 000 t
6.0 s
)
6.0 s
) = ( 360 μC) (1 − e
Q1 =
and
( 240 μC) (1 − e−1 000 t 6.0 s )
Then, Equation [2] yields
Q2 = 1.5 ( 240 μC ) (1 − e−1 000 t
18.61
(a)
−1 000 t 6.0 s
)
Using the rules for combining resistors in series and parallel, the circuit reduces as shown
below:
From the figure of Step 3, observe that
I=
(b)
56157_18_ch18_p110-170.indd 166
25.0 V
= 1.93 A
12.94 A
and
From the figure of Step 1, observe that
ΔVab = I ( 2.94 Ω ) = (1.93 A ) ( 2.94 Ω ) = 5.68 V
I1 =
ΔVab
5.68 V
=
= 0.227 A
25.0 Ω 25.0 Ω
3/19/08 1:35:59 AM
Direct-Current Circuits
18.62
(a)
When the power supply is connected to points A and B, the circuit reduces as shown below
to an equivalent resistance of Req = 0.099 9 Ω .
From the center figure above, observe that I R1 = I1 =
and
(b)
I R2 = I R3 = I100 =
5.00 V
= 50.0 A
0.100 Ω
5.00 V
= 0.045 0 A = 45.0 mA
111 Ω
When the power supply is connected to points A and C, the circuit reduces as shown below
to an equivalent resistance of Req = 1.09 Ω .
From the center figure above, observe that I R1 = I R2 = I1 =
and
(c)
167
I R3 = I100 =
5.00 V
= 4.55 A
1.10 Ω
5.00 V
= 0.045 5 A = 45.5 mA
110 Ω
When the power supply is connected to points A and D, the circuit reduces as shown below
to an equivalent resistance of Req = 9.99 Ω .
5.00 V
= 0.450 A
From the center figure above, observe that I R1 = I R2 = I R3 = I1 =
11.1 Ω
5.00 V
and I100 =
= 0.050 0 A = 50.0 mA
100 Ω
56157_18_ch18_p110-170.indd 167
3/19/08 1:35:59 AM
168
18.63
Chapter 18
In the circuit diagram at the right, note that all points
labeled a are at the same potential and equivalent
to each other. Also, all points labeled c are equivalent.
To determine the voltmeter reading, go from point e
to point d along the path ecd, keeping track of all
changes in potential to find:
ΔVed = Vd − Ve = − 4.50 V + 6.00 V = + 1.50 V
Apply Kirchhoff’s loop rule around loop abcfa to find
− ( 6.00 Ω ) I + ( 6.00 Ω ) I 3 = 0
or
I3 = I
[1]
Apply Kirchhoff’s loop rule around loop abcda to find
− ( 6.00 Ω ) I + 6.00 V − (10.0 Ω ) I 2 = 0
or
I 2 = 0.600 A − 0.600 I
[2]
I1 = 0.900 A − 1.20 I
[3]
Apply Kirchhoff’s loop rule around loop abcea to find
− ( 6.00 Ω ) I + 4.50 V − ( 5.00 Ω ) I1 = 0
or
Finally, apply Kirchhoff’s junction rule at either point a or point c to obtain
I + I 3 = I1 + I 2
[4]
Substitute Equations [1], [2], and [3] into Equation [4] to obtain the current through the ammeter.
This gives
I + I = 0.900 A − 1.20 I + 0.600 A − 0.600 I
or
18.64
3.80 I = 1.50 A
and
I = 1.50 A 3.80 = 0.395 A
In the figure given below, note that all bulbs have the same resistance, R.
(a)
In the series situation, Case 1, the same current I1 flows through both bulbs. Thus, the
same power, P1 = I12 R, is supplied to each bulb. Since the brightness of a bulb is proportional to the power supplied to it, they will have the same brightness. We conclude that the
bulbs have the same current, power supplied,, and brightness .
(b)
In the parallel case, Case 2, the same potential difference ΔV is maintained across
each of the bulbs. Thus, the same current
I 2 = ΔV R will flow in each branch of this
parallel circuit. This means that, again, the same power
P2 = I 22 R is supplied to each
bulb, and the two bulbs will have equal brightness .
continued on next page
56157_18_ch18_p110-170.indd 168
3/19/08 3:30:42 AM
Direct-Current Circuits
18.65
169
(c)
The total resistance of the single branch of the series circuit (Case 1) is 2R. Thus, the
current in this case is I1 = ΔV 2 R. Note that this is one half of the current I 2 that flows
through each bulb in the parallel circuit (Case 2). Since the power supplied is proportional
to the square of the current, the power supplied to each bulb in Case 2 is four times that
supplied to each bulb in Case 1. Thus, the bulbs in Case 2 are much brighter than those
in Case 1.
(d)
If either bulb goes out in Case 1, the only conducting 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 conducting path through the other bulb. Neglecting
any internal resistance of the battery, the battery continues to maintain the same potential
difference ΔV across this bulb as was present when both bulbs were lit. Thus, in the parallel
case, the second bulb remains lit with unchanged current and brightness when one
bulb fails.
(a)
The equivalent capacitance of this parallel combination is
Ceq = C1 + C2 = 3.00 μ F + 2.00 μ F = 5.00 μ F
When fully charged by a 12.0-V battery, the total stored charge
before the switch is closed is
Q0 = Ceq ( ΔV ) = ( 5.00 μ F ) (12.0 V) = 60.0 μC
Once the switch is closed, the time constant of the resulting RC
circuit is
τ = RCeq = ( 5.00 × 10 2 Ω ) ( 5.00 μ F ) = 2.50 × 10 −3 s = 2.50 ms
Thus, at t = 1.00 ms after closing the switch, the remaining total stored charge is
q = Q0 e−t τ = ( 60.0 μC ) e−1.00 ms 2.50 ms = ( 60.0 μC ) e−0.400 = 40.2 μC
The potential difference across the parallel combination of capacitors is then
ΔV =
40.2 μC
q
=
= 8.04 V
Ceq 5.00 μ F
and the charge remaining on the 3.00 μ F capacitor will be
q3 = C3 ( ΔV ) = ( 3.00 μ F ) ( 8.04 V) = 24.1 μC
(b)
The charge remaining on the 2.00 μ F at this time is
q2 = q − q3 = 40.2 μC − 24.1 μC = 16.1 μC
or, alternately,
(c)
Since the resistor is in parallel with the capacitors, it has the same potential difference
across it as do the capacitors at all times. Thus, Ohm’s law gives
I=
56157_18_ch18_p110-170.indd 169
q2 = C2 ( ΔV ) = ( 2.00 μ F ) ( 8.04 V) = 16.1 μC
ΔV
8.04 V
=
= 1.61 × 10 −2 A = 16.1 mA
R
5.00 × 10 2 Ω
3/19/08 1:36:01 AM
170
18.66
Chapter 18
The resistor network shown at the right does not contain any
obvious combinations of resistors in series or parallel. Thus,
the method of replacing such combinations by single resistors
to simplify the network and find the equivalent resistance is
not useful here. Still, the problem is easily solved if one takes
note of, and utilizes, the symmetry present in this network.
Imagine a battery of emf ε connected to points a and d, supplying current I which enters the network at a and exits at d.
When the current reaches a, it has two identical paths that it
could follow. Since the network is totally symmetric about a horizontal line drawn from a to d,
the current will split equally with I 2 flowing from a to b and I 2 flowing from a to c. This
means ΔVab = ΔVac = ( I 2 ) R, so the potential difference between points b and c will be zero.
Therefore, no current will flow through the vertical resistor between these two points, and currents of I 2 follow each of the paths b to d and c to d. There, they merge forming current I
again, which returns to the battery.
The potential difference between points a and d is ε = IReq . But it may also be written as
ε = ΔVac + ΔVcd = ⎛⎜⎝ I ⎞⎟⎠ R + ⎛⎜⎝ I ⎞⎟⎠ R = IR. Since both are equal to the emf of the battery, we see that
2
2
IReq = IR , and conclude that the equivalent resistance of this network is Req = R .
18.67
(a)
With 4.0 × 10 3 cells, each with an emf of 150 mV, connected in series, the total terminal
potential difference is
)
)
ΔV = ( 4.0 × 10 3 (150 × 10 −3 V = 6.0 × 10 2 V
When delivering a current of I = 1.0 A, the power output is
P = I ( ΔV ) = (1.0 A ) ( 6.0 × 10 2 V) = 6.0 × 10 2 W
(b)
The energy released in one shock is
)
)
E1 = P ( Δt )1 = ( 6.0 × 10 2 W ( 2.0 × 10 −3 s = 1.2 J
(c)
The energy released in 300 such shocks is Etotal = 300 E1 = 300 (1.2 J ) = 3.6 × 10 2 J . For
a 1.0-kg object to be given a gravitational potential energy of this magnitude, the height
the object must be lifted above the reference level is
h=
56157_18_ch18_p110-170.indd 170
PEg
mg
=
3.6 × 10 2 J
= 37 m
(1.0 kg ) ( 9.80 m s2
)
3/19/08 1:36:02 AM