15.
The equivalent resistance of a series combination of resistors is the algebraic sum of the individual
resistances and is always greater than any individual resistance. Therefore, choices (a) and (d) are
true statements and all others are false.
ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS
Conceptual 2.
4.
86
A short circuit can develop when the last bit of insulation frays away between the two conductors
in a lamp cord. Then the two conductors touch each other, creating a low resistance path in parallel
with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A
very large current will be produced in the power source, the house wiring, and the wire in the lamp
cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before
considerable
heating and sparking is produced in the short-circuit path.
Chapter
18
6.
68719_18_ch18_p083-119.indd 85
8.
A wire or cable in a transmission line is thick and made of material with very low resistivity.
Only when its length is very large does its resistance become significant. To transmit power over
a long distance it is most efficient to use low current at high voltage. The power loss per unit
length of the transmission line is Ploss L = I 2 ( R L ), where R L is the resistance per unit length
of the line. Thus, a low current is clearly desirable, but to transmit a significant amount of power
P = ( ∆V ) I with low current, a high voltage must be used.
The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one
bulb produces an open circuit with infinite resistance and zero current.
10.
(a) ii. The power delivered may be expressed as P = I 2 R, and while resistors connected in series
Problems: have the same current in each, they may have different values of resistance.
(b) ii. The power delivered may also be expressed as P = (∆V )2 / R, and while resistors
connected in parallel have the same potential difference across them, they may have different
values of resistance.
12.
Compare two runs in series to two resistors connected in series. Compare three runs in parallel to
three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a
battery followed immediately by two resistors in parallel.
The junction rule for ski resorts says that the number of skiers coming into a junction must be
equal to the number of skiers leaving. The loop rule would be stated as the total change in altitude
must be zero for any skier completing a closed path.
14.
Because water is a good conductor, if you should become part of a short circuit when fumbling
with any electrical circuit while in a bathtub, the current would follow a pathway through you,
the water, and to ground. Electrocution would be the obvious result.
ANSWERS TO EVEN NUMBERED PROBLEMS
27 Ω
2.
(a)
4.
(a) 1.13 A
6.
(a)
(b) 0.44 A
(b)
(c)
3.0 Ω, 1.3 A
9.17 Ω
35.0 Ω 3 = 11.7 Ω
(b) 1.00 A in the 12.0-Ω and 8.00-Ω resistors, 2.00 A in the 6.00-Ω and 4.00-Ω resistors, and
3.00 A in the 5.00-Ω resistor
1/7/11 2:41:09 PM
18.1
From ∆V = I ( R + r ), the internal resistance is
r=
18.2
∆V
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)
∆V 12 V
=
= 0.44 A
Req 27 Ω
Direct-Current
CircuitsCircuits
89
Direct-Current
89
9.0-Ω
If the
resistors
are now are
connected
in parallel
with each
other,
equivalent
9.0-Ω resistors
(c)three
If the
three
now connected
in parallel
with
eachthe
other,
the equivalent
resistanceresistance
is
is
continued on next
13
3
1 page1 1
11
1
+
+=
=
=
+=
or
R
9.0
Ω
9.0
Ω
9.0
Ω
9.0
Ω
Req 9.0 Ω
9.0 Ω
9.0 Ω
eq
9.0 Ω 9.0 Ω
oreq = Req == 3.0 Ω
= 3.0 Ω
R
3
3
thiscombination
parallel combination
is connected
to the battery,
the potential
difference
When thisWhen
parallel
is connected
to the battery,
the potential
difference
across across
eachinresistor
in the combination
is ∆V
= 12
so the current
of the resistors
is
each resistor
the combination
is ∆V = 12
V, so
theV,current
throughthrough
each ofeach
the resistors
is
68719_18_ch18_p083-119.indd 88
1/7/11 2:41:16 PM
I=
18.3
∆V
12 V
∆V I 12
V
= = R == 9.0
1.3 Ω
A = 1.3 A
R
9.0 Ω
18.3
acts as resistor
a 192-Ω (see
resistor
(see below),
so thediagram
circuit diagram
is:
(a) The(a)
bulb The
acts bulb
as a 192-Ω
below),
so the circuit
is:
(b) For the bulb in use as intended, Rbulb2 = ( ∆V ) P = (2120 V ) 75.0 W = 192 Ω .
For the bulb in use as intended, Rbulb = ( ∆V ) P = (120 V ) 75.0 W = 192 Ω .
Now, assuming the bulb resistance is unchanged, the current in the circuit shown is
Now, assuming the bulb resistance is unchanged, the current in the circuit shown is
∆V
120 V
= 0.620 A
∆V I = R = 0.800
120 V
Ω
+
192
Ω + 0.800
Ω A
eq
I=
=
= 0.620
Req 0.800 Ω + 192 Ω + 0.800 Ω
and
the actual power dissipated in the bulb is
and the actual power dissipated in the bulb is
2
P = I 2 Rbulb = ( 0.620 A ) (192 Ω ) = 73.8 W
2
2
P = I Rbulb = ( 0.620 A ) (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
current
both
the resistor
and the
batterypotential
is
(a) When thebattery,
8.00-Ωthe
resistor
is through
connected
across
the 9.00-V
terminal
difference of the
battery, the current through both the resistor and the battery is
∆V 9.00 V
I=
=
= 1.13 A
∆V 9.00RV 8.00 Ω
I=
=
= 1.13 A
R
8.00 Ω
(b) The relation between the emf and the terminal potential difference of a battery supplying
current I is ∆V = e − Ir, where r is the internal resistance of the battery. Thus, if the
(b) The relation
between the emf and the terminal potential difference of a battery supplying
battery has r = 0.15 Ω and maintains a terminal potential difference of ∆V = 9.00 V while
current I is ∆V = e − Ir, where r is the internal resistance of the battery. Thus, if the
supplying the current found above, the emf of this battery must be
battery has r = 0.15 Ω and maintains a terminal potential difference of ∆V = 9.00 V while
supplying the current found above, the emf of this battery must be
e = ∆V + Ir = 9.00 V + (1.13 A )( 0.15 Ω ) = ( 9.00 + 0.17 ) Ω = 9.17 Ω
2
2
(b)
18.4
18.5
18.5
(a)
∆V +equivalent
Ir = 9.00 resistance
V + (1.13 A
0.15
) = ( 9.00 + 0.17 ) Ω = 9.17 Ω
(a)e = The
of)(the
twoΩparallel
resistors is
The equivalent resistance of the two parallel
−1
1 
 1
resistors is
Rp = 
+
= 4.12 Ω

 7.00 Ω 10.0
Ω
−1
1
1
(b)
The relation between the emf and the terminal potential difference of a battery supplying
current I is ∆V = e − 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 ∆V = 9.00 V while
supplying the current found above, the emf of this battery must be
e = ∆V + Ir = 9.00 V + (1.13 A )( 0.15 Ω ) = ( 9.00 + 0.17 ) Ω = 9.17 Ω
18.5
(a)
The equivalent resistance of the two parallel
resistors is
1 
 1
Rp = 
+
 7.00 Ω 10.0 Ω 
−1
= 4.12 Ω
Thus,
90
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
Chapter 18
90
Chapter 18
(b)
( ∆V )ab
continued on next page
34.0 V
I ab =
=V1.99 A, so I 4 = I 9 = 1.99 A
( ∆V=)ab17.134.0
R
(b) I ab =ab
= Ω
= 1.99 A, so I 4 = I 9 = 1.99 A
17.1 Ω
Rab
Also,
A )( 4.12 Ω ) = 8.20 V
( ∆V )( p∆V= )I ab=RIp =R(1.99
= (1.99 A )( 4.12 Ω ) = 8.20 V
Also,
ab
p
p
( ∆V()∆V
VV
p
) 8.208.20
68719_18_ch18_p083-119.indd 89
1/7/11 2:41:18 PM
Then,Then,
I7 = I =
=p =
= 1.17 A
7
R7 R7 7.00
Ω Ω = 1.17 A
7.00
) p 8.208.20
( ∆V()∆V
VV
p
0.820AA
and and I10 =I10 =
= =
= =0.820
R
10.0
R10 10 10.0 Ω Ω
18.6 18.6(a)
(a)
The parallel
combination
of the
6.00-Ωand
and12.0-Ω
12.0-Ωresistors
resistors has
of of
The parallel
combination
of the
6.00-Ω
has an
anequivalent
equivalentresistance
resistance
1
1
2 +1
1
1
=1
+1
=2 + 1
=R
6.00+Ω 12.0 =
Ω 12.0 Ω
Rp1 p16.00 Ω 12.0 Ω 12.0 Ω
12.0 Ω
R = 12.0 Ω = 4.00 Ω
= 4.00 Ω
3
or
Rp1p1 =
or
3
Similarly, the equivalent resistance of the 4.00-Ω and 8.00-Ω parallel combination is
Similarly, the equivalent resistance of the 4.00-Ω and 8.00-Ω parallel combination is
1
1
2 +1
1
1 Rp 2 =14.00 Ω + 18.00 Ω =28.00
+1 Ω
=
+
=
Rp 2 4.00 Ω 8.00 Ω 8.00 Ω
Rp 2 =
or
8.00 Ω
8.00 Ω
Rp 2 = 3
3
or
The total resistance of the series combination between points a and b is then
The total resistance of the series combination between points a and b is then
8.00
35.0
Ω=
Ω
3
3
8.00
35.0
Rab = Rp1 + 5.00 Ω + Rp 2 = 4.00 Ω + 5.00 Ω +
Rab = Rp1 + 5.00 Ω + Rp 2 = 4.00 Ω + 5.00 Ω +
(b)
(b)
If ∆Vab = 35.0 V, the total current from a to b is
3
Ω=
3
Ω
If ∆Vab = 35.0
total current from a to b is
I ab = V,
∆Vthe
ab Rab = 35.0 V ( 35.0 Ω 3 ) = 3.00 A
I ab =the∆V
35.0 V ( 35.0
3) two
= 3.00
A combinations are
and
potential
differences
acrossΩthe
parallel
ab Rab =
∆Vp1 = Idifferences
A )( 4.00
V, and
) = 12.0
and the potential
theΩtwo
parallel
combinations are
ab R p1 = ( 3.00across
8.00 
∆Vp1 ∆V
= Ip ab
R I ab=Rp(23.00
A )(A4.00
V, and
= ( 3.00
V
)  Ω ) =Ω12.0
 = 8.00
2 = p1
 3

 8.00
 3


∆Vp individual
= ( 3.00 through
A )  theΩ
V are:
The
various
resistors
 = 8.00
2 = I ab R p 2 currents
I12 = ∆Vp1 12.0 Ω = 1.00 A ;
I 6 = ∆Vp1 6.00 Ω = 2.00 A ;
The individual currents through the various resistors are:
I 5 = I ab = 3.00 A ;
I 6 = ∆Vp1 6.00 Ω = 2.00 A ;
I 5 = I ab = 3.00 A ;
I 8 = ∆Vp 2 8.00 Ω = 1.00 A ;
and
18.7
I 8 = ∆Vp 2 8.00 Ω = 1.00 A ;
I12 = ∆Vp1 12.0 Ω = 1.00 A ;
I 4 = ∆Vp 2 4.00 Ω = 2.00 A
When connected in series, we have R1 + R2 = 690 Ω
and
I = ∆V
4.00 Ω = 2.00 A
4
p2
which we may
rewrite
as R2 = 690 Ω − R1
18.7
[1a]
When connected in series,
1 we
1 have1R1 + R2 = 690 Ω R1 R2
When in parallel,
R1
+
R2
=
150 Ω
which we may rewrite as R2 = 690 Ω − R1
[1]
or
R1 + R2
= 150 Ω
[1]
[2]
[1a]
with two solutions of
R1 = 470 Ω
and R1 = 220 Ω
Then Equation [1a] yields
R2 = 220 Ω
or
R2 = 470 Ω
Thus, the two resistors have resistances of 220 Ω and 470 Ω .
18.8
(a)
The equivalent resistance of this first parallel
combination is
1
1
1
=
+
Rp1 10.0 Ω 5.00 Ω
(b)
Rp1 = 3.33 Ω
or
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)
or
Rp 2 = 2.13 Ω
For the second series combination (and hence the entire resistor network)
(e)
The total current supplied by the battery is
 e 
and that across R4 is ∆V4 = ∆Vab = IR234 = 
( 2R ) = 2e 3
 3R 
∆V
8.00 V
I total =
=
= 1.94 A
Rtotal 4.13 Ω
The current through
R2 and R3 is I 23 = ∆Vab R23 = ( 2e 3R ) 6R = e 9R
(f )
The potential drop across the 2.00 Ω resistor is
e 
so the potential difference across R2 is ∆V2 = I 23 R2 = 
( 2R ) = 2e 9
 9R 
∆V2 = R2 I total = ( 2.00 Ω )(1.94 A ) = 3.88 V
 e 
The
drop
across
and potential
that across
R3 is
∆V3 =the
I 23second
R3 =  parallel
( 4Rcombination
) = 4e 9 must be
 9R 
(g)
∆Vp 2 = ∆V − ∆V2 = 8.00 V − 3.88 V = 4.12 V
From above, we have I1 = I and I 2 = I 3 = I 23 = e 9R = I 3
∆Vp 2 4.12 V
So the current through the 3.00 Ω resistor is I total =
=
= 1.37 A
The current through R4 is I 4 = ∆V4 R4 = ( 2e 3) 3R = 2eR3 9R =3.00
2I Ω
3
(b)
(h)
18.11
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 
Thus,
68719_18_ch18_p083-119.indd 91
93
Direct-Current Circuits
Rtotal = 2.00 Ω + Rp 2 = 2.00 Ω + 2.13 Ω = 4.13 Ω
75 Ω = R +
−1
1
 1

=
+
 30 Ω R + 5.0 Ω 
−1
=
( 30 Ω )( R + 5.0 Ω )
R + 35 Ω
( 30 Ω )( R + 5.0 Ω ) R 2 + ( 65 Ω ) R + 150 Ω2
R + 35 Ω
=
R + 35 Ω
which reduces to R 2 − (10 Ω ) R − 2 475 Ω2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0.
1/7/11 2:41:23 PM
Only the positive solution is physically acceptable, so R = 55 Ω .
18.12
The sketch at the right shows the equivalent circuit
when the switch is in the open position. For this simple
series circuit,
R1 + R2 + R3 =
or
e
When the switch is closed in position a, the equivalent
circuit is shown in Figure 2. The equivalent resistance
of the two parallel resistors, R2, is Rp = R2 2 and the
total resistance of the circuit is Ra = R1 + ( R2 2 ) + R3.
− R1
+
R2
Io
e
6.00 V
=
I o 1.00 × 10 −3 A
R1 + R2 + R3 = 6.00 kΩ
Figure 1
[1]
R2
− R1
R3
of the parallel
combination
of the 3.00 Ω
I 2resistance
=
=
= 1.7 A
ΩΩ
+ 2.0
Ω 5.0
and3.0
6.00
resistors
is Ω
so
18.14
−
Ω = 5.1 V
( ∆V )be =1 I 2 Rbe =1(1.7 A )( 3.0
1 )
3
4.00 Ω
3.00 Ω
=
+
=
or Rp = 2.00 Ω
Figure P18.14
Rp 3.00 Ω 6.00
6.00V Ω
( ∆V )Ω
5.1
be
Finally, from Figure 1, I12 =
=
= 0.43 A
R12
12 Ω
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 .
(a) The resistor network connected
to the battery in
Figure P18.14 can be reduced to a single equivalent
2.00 Ω
(b) resistance
The current
in the
2.00 Ωsteps.
resistor
the total current supplied by the battery and
6.00isΩequal to
in the
following
Theisequivalent
+
18.0
V
resistance of the parallel combination of the 3.00 Ω
3.00 Ω
−
∆V 18.0
V
and 6.00
I totalΩ=resistors
= is
= 2.25 A
4.00
Ω
R
8.00 Ω
eq
(c)
18.15
1
1
1
3
=
+
=
or R = 2.00 Ω
The power
the Ω
battery
to Ω
the circuitp is
Rp 3.00
6.00delivers
Ω 6.00
Figure P18.14
P = ( ∆V )isI total
= (18.0
V )the
40.5
( 2.25
) =and
This resistance
in series
with
4.00A Ω
theW
other 2.00 Ω resistor, giving a total
equivalent resistance of Req = 2.00 Ω + Rp + 4.00 Ω = 8.00 Ω .
(a)
Connect two 50-Ω resistors in parallel to get 25 Ω. Then connect that parallel
(b)
The current in the 2.00 Ω resistor is the total current supplied by the battery and is equal to
(b)
∆V 18.0 V
I total = two 50-Ω
=
= 2.25
Connect
resistors
in A
parallel to get 25 Ω.
Req 8.00 Ω
(c)
Also,
connect
two 20-Ω
resistors
parallel
The
power
the battery
delivers
to the in
circuit
is to get 10 Ω.
combination in series with a 20-Ω resistor for a total resistance of 45 Ω.
Then,P =connect
twoVparallel
= (18.0
( ∆V ) I totalthese
)( 2.25 Acombinations
) = 40.5 W in series to obtain 35 Ω.
18.16
18.15
(a)
(a)
The equivalent
resistance
parallel
Connect
two 50-Ω
resistorsofinthe
parallel
to get 25 Ω. Then connect that parallel
combination between points b and e is
combination in series with a 20-Ω resistor for a total resistance of 45 Ω.
(b)
96
1
1
Connect1 two
to
25 Ω.
= 50-Ω+resistors in parallel
or
Rbeget
= 8.0
Ω
Rbe 12 Ω 24 Ω
Also, connect two 20-Ω resistors in parallel to get 10 Ω.
The total resistance between points a and e is then
Then, connect these two parallel combinations in series to obtain 35 Ω.
Rae = Rab + Rbe = 6.0 Ω + 8.0 Ω = 14 Ω
Chapter 18
18.16
(a) The equivalent resistance of the parallel
combination between points b and e is
The total current supplied by the battery (and also the current in the 6.0-Ω resistor) is
1
1
1
=
+
or
Rbe = 8.0 Ω
Rbe 12∆V
Ω 24 42
Ω V
continued on next page
ae
I total = I 6 =
=
= 3.0 A
14 Ω
Rae
The total resistance
between points a and e is then
The potential
between
and e is
R = difference
R + R = 6.0
Ω + 8.0points
Ω = 14b Ω
ae
ab
be
∆Vbe = Rbe I total = (8.0 Ω )( 3.0 A ) = 24 V
68719_18_ch18_p083-119.indd 95
so
(b)
68719_18_ch18_p083-119.indd 95
I12 =
∆Vbe 24 V
=
= 2.0 A
Rbce
12 Ω
1/7/11 2:41:32 PM
I 24 =
and
Applying the junction rule at point b yields
∆Vbe 24 V
=
= 1.0 A
Rbde
24 Ω continued on next page
I 6 − I12 − I 24 = 0
Using the loop rule on loop abdea gives +42 − 6I 6 − 24I 24 = 0
or
I 6 = 7.0 − 4I 24
[2]
and using the loop rule on loop bcedb gives −12I12 + 24I 24 = 0
or
I12 = 2I 24
[3]
Substituting Equations [2] and [3] into [1] yields 7I 24 = 7.0
Then, Equations [2] and [3] yield
18.17
[1]
I 6 = 3.0 A
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
1/7/11 2:41:32 PM
e
−
Applying the loop rule around loop abca gives
and using the loop rule on loop bcedb gives −12I12 + 24I 24 = 0
I1 − I
or
18.17
4R
I1
−
3R
I12 = 2I 24
or
e − R ( I1 − I ) − 4RI1 = 0
Substituting Equations [2] and [3] into [1] yields 7I 24 = 7.0
or I 24 = I1.0 A
a
1 e

I1Then,
=  Equations
+ I  [2] and [3] yield I 6 = 3.0 A and I12 = 2.0 A
5 R 
[3]
I2 + I
I2
e
[1]
Going counterclockwise around the upper loop,
Next,
applying
the loop
rule
around
applying
Kirchhoff’s
loop
rule,
givesloop cedc gives
+15.02 V
− ( 7.00
−3RI
+ 2e
− 2R)(II12−+( 5.00
I ) = )0( 2.00 A
or) = 0 I 2 =
2 e

 − I
5 R 
[2]
15.0 V − 10.0 V
or I1 =applying the loop =rule0.714
A loop caec gives
Finally,
around
7.00 Ω
e
+ 3RI 2 junction
=0
or I +4I
3I 2 A = 0
1
From−4RI
Kirchhoff’s
rule,
I 21 −=2.00
1
[3]
so I 2 = 2.00
A − I1 = [1]
2.00and
A −[2]0.714
= 1.29 [3]
A yields
Substituting
Equations
into A
Equation
I=
e
5R
Going around the lower loop in a clockwise direction gives
Thus, if e = 250 V and R = 1.00 kΩ = 1.00 × 10 3 Ω, the current in the wire between a and e is
+ e − ( 2.00 ) I 2 − ( 5.00 )( 2.00 A ) = 0
250 V
I=
= 50.0 × 10 −3 A = 50.0 mA flowing from a toward e.
3
or e = (52.00
ΩA
(1.00Ω×)(101.29
) ) + (5.00 Ω)( 2.00 A ) = 12.6 V
18.18
18.20
98
(a) Applying
Kirchhoff’s
rulerecording
at a gives
Following
the path
of I1 fromjunction
a to b and
changes in potential gives
I2 = I4 − I6
[1]
Vb − Va = + 24 V − ( 6.0 Ω )( 3.0 A ) = + 6.0 V
Going counterclockwise around the lower loop,
and applying
looparule,
Now, following
the Kirchhoff’s
path of I 2 from
to b we
andobtain
recording changes in potential gives
+8.00 V − ( 6.00 Ω ) I 6 + ( 2.00 Ω ) I 2 = 0
Vb − Va = − ( 3.0 Ω ) I 2 = + 6.0 V, or I 2 = − 2.0 A
4 1
Chapter 18
or I 6 = + I 2
3 3
12.0 V
+ −
4.00 Ω
I4
I6
b
I2
a
2.00 Ω
− +
6.00 Ω
8.00 V
[2]
Thus, I 2 is directed from b toward a and has magnitude of 2.0 A.
continued on next page
Applying Kirchhoff’s junction rule at point a gives
continued on next page
I 3 = I1 + I 2 = 3.0 A + ( −2.0 A ) = 1.0 A
68719_18_ch18_p083-119.indd 97
18.21
(a)
68719_18_ch18_p083-119.indd 96
a
The circuit diagram at the
right shows the assumed
directions of the current in
each resistor. Note that the
total current flowing out of
the section of wire connecting
points g and f must equal the
current flowing into that
section. Thus,
b
c
40.0 V
+
− 360 V
−
80.0 V
+
200 Ω
80.0 Ω
20.0 Ω
I2
I3
I1
h
g
I 3 = I1 + I 2 + I 4
f
1/7/11 2:41:37 PM
d
+
−
1/7/11 2:41:35 PM
70.0 Ω
I4
e
[1]
Applying the loop rule around loop abgha gives
−200I1 − 40.0 + 80.0I 2 = 0
or
I2 =
1
(5I1 + 1.00 )
2
[2]
Next, applying the loop rule around loop bcfgb gives
+360 − 20.0I 3 − 80.0I 2 + 40.0 = 0
or
I 3 = 20.0 − 4I 2
[3]
Finally, applying the loop rule around the outer loop abcdefgha yields
−80.0 + 70I 4 − 200I1 = 0
or
I4 =
1
( 20I1 + 8.00 )
7
[4]
100
Chapter 18
99
Direct-Current Circuits
18.22
18.24
 10.0 mA 
Equation
[2], I 3 resistors
= 5.00 mA
− 0.750
 branch  = 2.69 mA
(a) and,
Thefrom
in the
upper
30.0-Ω
and 50.0-Ω
3.25
are in series, and add to give a total
resistance
Ruppergo=to
80.0
Ω for
this path.changes in potential to obtain
(b) Start
at pointofc and
point
f, recording
This 80.0-Ω resistance is in parallel with the 80.0-Ω
resistance
and
V f − Vc of
= −the
e 2 middle
− R2 I 2 =branch,
−60.0 V
− ( 3.00 × 10 3 Ω ) ( 3.08 × 10 −3 A ) = −69.2 V
the rule for combining resistors in parallel
resistance
Rab c=is40.0
Ω higher potential .
oryields
∆V acf total
= 69.2
V and of
point
at the
between points a and b. This resistance is in
series with
the 20.0-Ω
resistor,
total gives
(a) Applying
Kirchhoff’s
loop
rule toso
thethe
circuit
equivalent resistance of the circuit is
+ 3.00 V − ( 0.255 Ω + 0.153 Ω + R )( 0.600 A ) = 0
Req = 20.0 Ω + Rab = 20.0 Ω + 40.0 Ω = 60.0 Ω
3.00 V
or R =
− ( 0.255 Ω + 0.153 Ω ) = 4.59 Ω
0.600 A
∆V
12 V
(b) The current supplied to this circuit by the battery is I total =
=
= 0.20 A .
(b) The total power input to the circuit is
Req 60.0 Ω
(c)
2
+ 1.50isVP)total
= 1.80
W Ω )( 0.20 A )2 = 2.4 W .
(1.50
( 0.600
= ReqAI)total
= ( 60.0
ThePpower
delivered
the V
battery
input = ( e
1 + e 2 ) I =by
power
loss by
heating between
within thepoints
batteries
is b is
(d) The
The
potential
difference
a and
2
Ploss
r1 ab+Irtotal
A))( 0.20
0.153
( 0.255
∆V=abI =(R
=((0.600
40.0 Ω
A )Ω
= +8.0
V Ω ) = 0.147 W
2)=
2
Thus, the fraction of the power input that is dissipated
∆Vab internally
8.0 V is
and the current in the upper branch is I upper =
=
= 0.10 A, so the power
Rupper 80.0 Ω
delivered
to
the
50.0
Ω
resistor
is
Ploss
0.147 W
=
= 0.081 6 or 8.16%
Pinput
1.802 W
2
P50 = R50 I upper = ( 50.0 Ω )( 0.10 A ) = 0.50 W
18.25
18.23
(a)
No. Some simplification could be made by recognizing that the 2.0-Ω and 4.0-Ω resistors
(a) areWe
currents
I1 , Ia2 ,total
and of
I 3 as
in name
series,the
adding
to give
6.0shown.
Ω; and the 5.0-Ω and 1.0-Ω resistors form a
e cannot beesimplified any e
series combination with a total resistance of 6.0 Ω. The circuit
Applying
Kirchhoff’s rules
loop rule
further,
and Kirchhoff’s
musttobeloop
usedabcfa
to analyze the circuit.
gives + e 1 − e 2 − R2 I 2 − R1 I1 = 0
(b)
Applying Kirchhoff’s junction rule at junction a gives
or 3I 2 + 2I1 = 10.0 mA
I1 = I 2 + I 3
and I1 = 5.00 mA − 1.50I 2
[1]
[1]
Using Kirchhoff’s loop rule on the upper loop yields
Applying the loop rule to loop edcfe yields
+ 24 V − ( 2.0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0
+ e 3 − R3 I 3 − e 2 − R2 I 2 = 0 or 3I 2 + 4I 3 = 20.0 mA
or I 3 = 8.0 A − 2 I1
[2]
and I 3 = 5.00 mA − 0.750I 2
For the lower loop: +12 V + ( 3.0 ) I 3 − (1.0 + 5.0 ) I 2 = 0
Finally, applying Kirchhoff’s junction rule at junction c gives
Using Equation [2], this reduces to
I 2 = I1 + I 3
12 V + 3.0 (8.0 A − 2 I1 )
I2 =
or
I 2 = 6.0 A − I1
Substituting
Equations
6.0 [1] and [2] into [3] yields
[2]
[3]
[3]
I 2 = 5.00
mA − 1.50I
5.00
mA
0.750II2 = 3.5
or A3.25I
2 +[3]
2 = 10.0 mA
Substituting
Equations
[2] and
into
[1]−gives
.
1
ThisEquation
gives I 2 =[3](10.0
mA)I 3.25
= 3.08 mA . Then, Equation [1] yields
Then,
gives
2 = 2.5 A , and Equation [2] yields I 3 = 1.0 A .
 10.0 mA 
= 0.385 mA
I1 = 5.00 mA − 1.50 
 3.25 
continued on next page
68719_18_ch18_p083-119.indd 100
68719_18_ch18_p083-119.indd 99
1/7/11 2:41:45 PM
1/7/11 2:41:42 PM
Direct-Current Circuits
18.26
18.26
I1
Using Kirchhoff’s loop rule on the outer
perimeter
circuit
+12 ofVthe
− ( 0.01
− ( 0.06 ) I = 0
) I gives
1
V −×(10
0.01
3 ) I1 − ( 0.06 ) I 3 = 0
I+12
= 1.2
A − 6.0 I 3
1
or
I = 1.2 × 10 3 A − 6.0 I
1
For the
rightmost loop, the3 loop rule gives
0.01+Ω
[1]
[1]
For the rightmost loop, the loop rule gives
+10 V + (1.00 ) I 2 − ( 0.06 ) I 3 = 0
or
or
I2
Ω I
I0.01
1
2
3
or
101
Direct-Current Circuits
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
− 12 V
+
12 V
−Live
Battery
Live
Battery
101
I3
0.06 Ω
Starter
I31.00 Ω
1.00 +
Ω
− 10 V
0.06 Ω
Starter
+
V
Dead
− 10
Battery
Dead
Battery
+10 V + (1.00 ) I 2 − ( 0.06 ) I 3 = 0
I 2 = 0.06 I 3 − 10 A
[2]
I 2 = 0.06 I 3 − 10 A
[2]
Applying Kirchhoff’s junction rule at either junction gives
Applying Kirchhoff’s junction rule at either junction gives
I1 = I 2 + I 3
I1 = I 2 + I 3
(a)
(a)
[3]
[3]
Substituting Equations [1] and [2] into [3] yields
Substituting Equations [1] and [2] into [3] yields
7.1I = 1.2 × 10 3 A and I = 1.7 × 10 2 A ( in starter )
7.1I 3 3= 1.2 × 10 3 A and I 3 =3 1.7 × 10 2 A ( in starter )
(b)
Then, Equation [2] gives I 2 =2 0.20 A ( in (dead battery ) .
(a)
(a)
No. This
Thismulti-loop
multi-loopcircuit
circuit
does
contain
resistors
in series
(i.e., connected
No.
does
notnot
contain
anyany
resistors
in series
(i.e., connected
so all so all
the current
currentininone
onemust
mustpass
pass
through
other)
in parallel
(connected
the voltage
the
through
thethe
other)
nor nor
in parallel
(connected
so theso
voltage
dropacross
acrossone
oneisisalways
always
same
as that
across
the other).
this circuit
be
drop
thethe
same
as that
across
the other).
Thus,Thus,
this circuit
cannotcannot
be
simplified
edany
anyfurther,
further,and
and
Kirchhoff’s
rules
must
be used
to analyze
simplifi
Kirchhoff’s
rules
must
be used
to analyze
it. it.
(b)
Assume
I 3 in
thethe
directions
shown.
Assumecurrents
currentsI1I,1I,2 I, 2and
, and
I 3 in
directions
shown.
Then,
junction
rulerule
at junction
a gives
Then,using
usingKirchhoff’s
Kirchhoff’s
junction
at junction
a gives
(b)
18.27
18.27
Then, Equation [2] gives I = 0.20 A in dead battery ) .
II3 ==I1I ++I 2I
3
1
2
[1] [1]
Applying
rule
onon
thethe
lower
loop,
ApplyingKirchhoff’s
Kirchhoff’sloop
loop
rule
lower
loop,
+10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
+10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
or
or
I 2 = 2.00 A − 4 I 3
[2]
I 2 = 2.00 A − 4 I 3
and for the loop around the perimeter of the circuit,
and for the loop around the perimeter of the circuit,
or
or
I1 = 0.667 A − 0.667I 3
[2]
20.0 V − 30.0I1 − 20.0I 3 = 0
20.0 V − 30.0I1 − 20.0I 3 = 0
I1 = 0.667 A − 0.667I 3
Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667I 3 + 2.00 A − 4 I 3
[3]
[3]
Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667I 3 + 2.00 A − 4 I 3
which reduces to 5.67I 3 = 2.67 A and gives I 3 = 0.471 A .
which reduces to 5.67I 3 = 2.67 A and gives I 3 = 0.471 A .
Then, Equation [2] gives I 2 = 0.116 A , and from Equation [3], I1 = 0.353 A .
Then, Equation [2] gives I = 0.116 A , and from Equation [3], I = 0.353 A .
2
1
All currents are in the directions
indicated in the circuit diagram given above.
All currents are in the directions indicated in the circuit diagram given above.
68719_18_ch18_p083-119.indd 101
68719_18_ch18_p083-119.indd 101
1/7/11 2:41:48 PM
1/7/11 2:41:48 PM
18.29
(g)
Equation [4] gives I 36 = 2.88 A − ( −0.416 A ), or
(h)
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.
I 36 = 3.30 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 − 4.00I 3 ) 5.00
or
I1 = 1.80 A − 0.800 I 3
[2]
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 − 4.00I 3 ) 5.00
or
I 2 = 3.00 A − 0.800 I 3
[3]
Direct-Current
Circuits
Direct-Current
Circuits103 103
2.60I2.60I
= 4.80
Substituting
Equations
[2] and
givesgives
and Iand
= 1.846
A. A.
continued
on next page
3
3
= 4.80
Substituting
Equations
[2] [3]
andinto
[3] [1]
intoand
[1] simplifying
and simplifying
I 3 = 1.846
3
ThenThen
Equations
[2] and
I1 = 0.323
A and
= 1.523
A. A.
2
Equations
[2] [3]
andyield
[3] yield
I1 = 0.323
A Iand
I 2 = 1.523
Therefore,
the potential
differences
across
the resistors
are are
Therefore,
the potential
differences
across
the resistors
68719_18_ch18_p083-119.indd 102
∆V2 =
Ω ) =Ω3.05
V , V
∆V,3 =∆VI 2 (=3.00
Ω ) =Ω4.57
V V
∆VI 22 (=2.00
I 2 ( 2.00
I 2 ( 3.00
) = 3.05
) = 4.57
3
1/7/11 2:41:50 PM
∆V4 =
Ω ) =Ω7.38
V , and
∆V5 =∆VI1 (=5.00
Ω ) =Ω1.62
V V
∆VI 43 (=4.00
I 3 ( 4.00
V , and
I1 ( 5.00
) = 7.38
) = 1.62
5
18.3018.30The time
constant
is t =isRC.
Considering
units,units,
we fiwe
nd find
The time
constant
t = RC.
Considering
 Volts   Coulombs
 
 Coulombs
  Coulombs
=  =  Coulombs
RC →
)( Farads
) =  ) =  Volts


RC( Ohms
→ ( Ohms
)( Farads

 

  Amperes
 Amperes
 Amperes
 Amperes
  Volts
 Volts


Coulombs

Coulombs = Second
= =
  = Second
 Coulombs
Second
 Coulombs
Second 
or or
t = RC
of time.
t = has
RC units
has units
of time.
3
−6
3
−6
18.3118.31(a) (a)
The time
constant
is: t is:
= RC
× 10 ×Ω
× 10 × 10
F ) = F1.88
s. s.
( 75.0
The time
constant
t ==RC
= ( 75.0
10) ( 25.0
Ω ) ( 25.0
) = 1.88
× 10 −6× 10
F )−6(12.0
V ) =V1.90
× 10 −4× 10
C −4
. C.
(b) (b)
At t =
(Ce )(=Ce0.632
( 25.0( 25.0
m ax = 0.632
q = 0.632Q
= 0.632
F )(12.0
Att t, =qt=, 0.632Q
) = 0.632
) = 1.90
m ax
18.3218.32(a) (a)t = RC
× 10 −6× 10
F )−6= F2.00
× 10 −3× 10
s =−3 2.00
ms ms
(100
)( 20.0
t = =RC
= (Ω
100
Ω )( 20.0
s = 2.00
) = 2.00
(b) (b)Qm ax Q
= C e==C(e20.0
× 10 −6× 10
F )−6( 9.00
V ) =V
1.80
× 10 −4× 10
C −4
= C
180
= ( 20.0
F )( 9.00
= mC
180 mC
) = 1.80
m ax
 mC
(c) (c)Q = Q
(1m−ax e(−1t −t )e=− t Qt )m=ax Q(1m−axe(−t1 −t )e−t= Qt )m=ax Q 1m−ax 1e1 −=1114
Qm=ax Q
 = 114 mC
e
18.3318.33(a) (a)
The time
constant
of anofRC
is t =isRC.
The time
constant
ancircuit
RC circuit
t = Thus,
RC. Thus,
t = (t1.00
× 10 6×Ω
× 10 −6× 10
F )−6= F5.00
s s
= (1.00
10)6( 5.00
Ω ) ( 5.00
) = 5.00
(b) (b)Qm ax Q
= Ce ==Ce
mF )(mF
30.0
V ) =V150
(5.00
= ( 5.00
150 mC
) = mC
)(30.0
m ax
(c) (c)
To obtain
the current
through
the resistor
at time
t after
the switch
To obtain
the current
through
the resistor
at time
t after
the switch
is closed,
recall
that
the
charge
on
the
capacitor
at
that
time
is is
is closed,
recall
that
the
charge
on
the
capacitor
at
that
time
+
−t t
q = Ce
− e(−1t −t )e and
difference
across
a capacitor
is ise −e
q =(1Ce
andpotential
the potential
difference
across
a capacitor
) the
+
R
R