Chapter28 Solutions
P28.3 The total resistance is R =
P28.6
3.00 V
= 5.00 Ω .
0.600 A
(a)
R lamp = R − rbatteries = 5.00 Ω − 0.408 Ω = 4.59 Ω
(b)
Pbatteries ( 0.408 Ω ) I 2
=
= 0.081 6 = 8.16%
Ptotal
( 5.00 Ω) I 2
(a)
Rp =
1
(1 7.00 Ω) + ( 1 10.0 Ω)
FIG. P28.3
= 4.12 Ω
R s = R 1 + R2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 Ω
(b)
ΔV = IR
34.0 V = I (17.1 Ω)
FIG. P28.6
I = 1.99 A for 4.00 Ω , 9.00 Ω resistors.
(1.99 A) (4.12 Ω) = 8.18 V
8.18 V = I (7.00 Ω)
Applying ΔV = IR ,
so
I = 1.17 A for 7.00 Ω resistor
8.18 V = I (10.0 Ω)
so
I = 0.818 A for 10.0 Ω resistor.
1
P28.9
If we turn the given diagram on its side, we find that it is the same as
figure (a). The 20.0 Ω and 5.00 Ω resistors are in series, so the first
reduction is shown in (b). In addition, since the 10.0 Ω , 5.00 Ω , and
25.0 Ω resistors are then in parallel, we can solve for their equivalent
resistance as:
R eq =
1
(
1
10.0 Ω
+ 5.001 Ω +
1
25.0 Ω
)
= 2.94 Ω .
This is shown in figure (c), which in turn reduces to the circuit
shown in figure (d).
ΔV
and
R
The 12.94 Ω
Next, we work backwards through the diagrams applying I =
ΔV = IR alternately to every resistor, real and equivalent.
resistor is connected across 25.0 V, so the current through the battery
in every diagram is
ΔV 25.0 V
=
= 1.93 A .
R
12.94 Ω
goes through the 2.94 Ω
I=
In figure (c), this 1.93 A
to give a potential difference of:
equivalent resistor
ΔV = IR = ( 1.93 A )(2.94 Ω ) = 5.68 V .
From figure (b), we see that this potential difference is the same
across ΔVab , the 10 Ω resistor, and the 5.00 Ω resistor.
€
(b)
Therefore, ΔVab = 5.68 V .
(a)
Since the current through the 20.0 Ω resistor is also the current
through the 25.0 Ω line ab,
ΔVab 5.68 V
I=
=
= 0.227 A = 227 mA .
R ab
25.0 Ω
(a)
Since all the current in the circuit must pass through the series
100 Ω resistor, P = I 2 R
€
P28.11
FIG. P28.9
Pmax = RI2max
so
€
I max =
P
25.0 W
=
= 0.500 A
R
100 Ω
# 1
1 &
R eq = 100 Ω + %
+
(
$ 100 100 '
€
−1
Ω = 150 Ω
ΔVmax = R eqI max = 75.0 V
(b)
€
P = IΔV = ( 0.500 A )(75.0 V ) = 37.5 W total power
P1 = 25.0 W
2
P2 = P3 = RI2 ( 100 Ω)( 0.250 A) = 6.25 W
€
2
FIG. P28.11
−1
P28.15
" 1
1 %
Rp = $
+
' = 0.750 Ω
# 3.00 1.00 &
R s = (2.00 + 0.750 + 4.00) Ω = 6.75 Ω
ΔV 18.0 V
I battery =
=
= 2.67 A
Rs
6.75 Ω
2
P = I 2R :
P2 = (2.67 A) (2.00 Ω )
P2 = 14.2 W in 2.00 Ω
2
P4 = (2.67 A) ( 4.00 A) = 28.4 W in 4.00 Ω
ΔV2 = ( 2.67 A)( 2.00 Ω) = 5.33 V,
ΔV4 = ( 2.67 A)( 4.00 Ω ) = 10.67 V
ΔVp = 18.0 V − ΔV2 − ΔV4 = 2.00 V( = ΔV3 = ΔV1 )
P3 =
P1 =
P28.19
(ΔV3 )2 (2.00 V )2
R3
=
3.00 Ω
(ΔV1) ( 2.00 V) 2
R1
=
1.00 Ω
= 1.33 W in 3.00 Ω
FIG. P28.15
= 4.00 W in 1.00 Ω
(a)
The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series
with resistor 1, with resistance R, so the equivalent resistance of the whole
circuit is 3R. In series, potential difference is shared in proportion to the
1
resistance, so resistor 1 gets
of the battery voltage and the 2-3-4 parallel
3
2
combination get
of the battery voltage. This is the potential difference across
3
1
2
resistor 4, but resistors 2 and 3 must share this voltage.
goes to 2 and
to 3.
3
3
The ranking by potential difference is ΔV4 > ΔV3 > ΔV1 > ΔV2 .
(b)
Based on the reasoning above the potential differences are
ε
2ε
4ε
2ε
.
ΔV1 = , ΔV2 =
, ΔV3 =
, ΔV4 =
3
9
9
3
(c)
All the current goes through resistor 1, so it gets the most. The current then
splits at the parallel combination. Resistor 4 gets more than half, because the
resistance in that branch is less than in the other branch. Resistors 2 and 3 have
equal currents because they are in series. The ranking by current is
I1 > I4 > I 2 = I3 .
(d)
Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice
that of resistor 4, twice as much current goes through 4 as through 2 and 3. The
I
2I
current through the resistors are I1 = I, I2 = I 3 = , I 4 =
.
3
3
continued on next page
3
€
€
€
€
(e)
(f)
P28.24 We
Increasing resistor 3 increases the equivalent resistance of the entire circuit. The
current in the circuit, which is the current through resistor 1, decreases. This
decreases the potential difference across resistor 1, increasing the potential
difference across the parallel combination. With a larger potential difference
the current through resistor 4 is increased. With more current through 4, and
less in the circuit to start with, the current through resistors 2 and 3 must
decrease. To summarize, I4 increases and I1 , I 2 , and I3 decrease .
If resistor 3 has an infinite resistance it blocks any current from passing
through that branch, and the circuit effectively is just resistor 1 and resistor 4 in
series with the battery. The circuit now has an equivalent resistance of 4R. The
3
current in the circuit drops to
of the original current because the resistance
4
4
has increased by . All this current passes through resistors 1 and 4, and none
3
3I
3I
passes through 2 or 3. Therefore I1 = , I 2 = I3 = 0, I4 =
.
4
4
name the currents I1, I2 , and I3 as shown.
[1]
70.0 − 60.0 − I2 ( 3.00 kΩ) − I 1( 2.00 kΩ) = 0
[2]
80.0 − I3 ( 4.00 kΩ) − 60.0 − I2 ( 3.00 kΩ) = 0
[3]
I2 = I 1 + I3
(a)
Substituting for I2 and solving the resulting simultaneous
equations yields
I1 = 0.385 mA
I3 = 2.69 mA
I2 = 3.08 mA
(b)
FIG. P28.24
(through R1 )
(through R3 )
( through R2 )
ΔVcf = −60.0 V − ( 3.08 mA)( 3.00 kΩ) = −69.2 V
Point c is at higher potential.
P28.26 Name the currents as shown in the figure to the right. Then w + x + z = y .
Loop equations are
−200w − 40.0 + 80.0x = 0
−80.0x + 40.0 + 360 − 20.0y = 0
+360 − 20.0y − 70.0z + 80.0 = 0
FIG. P28.26
Eliminate y by substitution.
#x = 2.50w + 0.500
%
$400 − 100x − 20.0w − 20.0z = 0
%440 − 20.0w − 20.0x − 90.0z = 0
&
Eliminate x.
4
#350 − 270w − 20.0z = 0
$
%430 − 70.0 w − 90.0z = 0
Eliminate z = 17.5 − 13.5w to obtain
430 − 70.0w − 1 575 + 1 215w = 0
w=
Now
70.0
= 1.00 A upward in 200 Ω
70.0
.
z = 4.00 A upward in 70.0 Ω
x = 3.00 A upward in 80.0 Ω
y = 8.00 A downward in 20.0 Ω
and for the 200 Ω,
ΔV = IR = ( 1.00 A )(200 Ω) = 200 V .
P28.28
ΔVab = (1.00 )I 1 + ( 1.00)(I 1 − I 2 )
ΔVab = (1.00 )I 1 + ( 1.00) I2 + ( 5.00)( I − I 1 + I2 )
ΔVab = ( 3.00)( I − I 1) + ( 5.00)(I − I1 + I 2 )
Let I = 1.00 A , I1 = x , and I2 = y .
Then, the three equations become:
ΔVab = 2.00x − y ,
FIG. P28.28
or y = 2.00x − ΔVab
ΔVab = −4.00x + 6.00y + 5.00
and ΔVab = 8.00 − 8.00x + 5.00y .
€
Substituting the first into the last two gives:
7.00ΔVab = 8.00x + 5.00
€
Solving these simultaneously yields ΔVab =
€
Then, Rab
€
and 6.00ΔVab = 2.00x + 8.00 .
27
V
ΔVab
=
= 17
I
1.00 A
or
27
V.
17
R ab =
27
Ω .
17
P28.30 We apply Kirchhoff’s rules to the second diagram.
€
50.0 − 2.00I1 − 2.00I 2 = 0
(1)
20.0 − 2.00I 3 + 2.00 I2 = 0
(2)
(3)
I1 = I2 + I3
Substitute (3) into (1), and solve for I1, I2 , and I3
I1 = 20.0 A ; I2 = 5.00 A ; I3 = 15.0 A .
Then apply P = I 2 R to each resistor:
(2.00 Ω)1 :
2
P = I12 (2.00 Ω ) = (20.0 A) ( 2.00 Ω) =
2
( 4.00 Ω) :
800 W
" 5.00 %
P =$
A ' ( 4.00 Ω) = 25.0 W
# 2
&
5
FIG. P28.30
(Half of I2 goes through each)
(2.00 Ω)3 :
2
P = I32 (2.00 Ω ) = (15.0 A) ( 2.00 Ω) =
P28.36
(
)(
450 W
)
(a)
τ = RC = 1.50 × 105 Ω 10.0 × 10 −6 F = 1.50 s
(b)
τ = 1.00 × 105 Ω 10.0 × 10−6 F = 1.00 s
(c)
The battery carries current
(
.
)(
)
10.0 V
= 200 µA .
50.0 × 103 Ω
The 100 kΩ carries current of magnitude
% 10.0 V ( −t 1.00 s
.
I = I 0 e −t RC = '
*e
& 100× 103 Ω )
200 µA + (100 µA )e − t 1.00 s .
Call the potential at the left junction VL and at the right VR . After a
“long” time, the capacitor is fully charged.
So the switch carries downward current
P28.37
(a)
VL = 8.00 V because of voltage divider:
10.0 V
= 2.00 A
5.00 Ω
VL = 10.0 V − (2.00 A)(1.00 Ω ) = 8.00 V
IL =
Likewise,
#
2.00 Ω
&
VR = %
( (10.0 V ) = 2.00 V
$ 2.00 Ω + 8.00 Ω '
or
IR =
FIG. P28.37(a)
10.0 V
= 1.00 A
10.0 Ω
VR = ( 10.0 V) − (8.00 Ω )( 1.00 A ) = 2.00 V .
(b)
Therefore,
ΔV = VL − VR = 8.00 − 2.00 = 6.00 V .
Redraw the circuit
R=
€
and
so
1
= 3.60 Ω
(1 9.00 Ω) + (1 6.00 Ω)
RC = 3.60 × 10 −6 s
1
e −t RC =
10
t = RC ln 10 = 8.29 µs .
€
€
€
6
FIG. P28.37(b)