Homework for the week of October 27. 5th week of classes.
Ch. 25: 12, 23, 32, 36, 39, 45, 49, 57
Ch. 26: 12, 17, 19, 34, 38, 40, 46, 50
Ch. 25
12.
Use Eq. 25-3 to calculate the resistance, with the area as A = π r 2 = π d 2 4 .
R=ρ
l
A
=ρ
4l
πd
2
(
= 1.68 × 10−8 Ω m
)
4 ( 4.5m )
π (1.5 × 10 m )
−3
2
= 4.3 × 10−2 Ω
23.
The original resistance is R0 = V I 0 , and the high temperature resistance is R = V I ,
where the two
voltages are the same. The two resistances are related by Eq. 25-5, multiplied by l A so
that it expresses resistance instead of resistivity.
R = R0 [1 + α ( T − T0 ) ] → T = T0 +
1 R


1V I
1I

− 1 = T0 + 
− 1 = T0 +  0 − 1

α  R0 
α  V I0 
α I

= 20.0°C +
32.
0.00429 ( C°)
−1
 0.4212 A 
 0.3818 A − 1 = 44.1°C


Use Eq. 25-7b to find the resistance from the voltage and the power.
P=
36.
1
( a)
V2
→ R=
R
Since P =
V2
R
V2
P
=
( 240 V )2
→ R=
3300 W
V2
P
= 17 Ω
says that the resistance is inversely proportional to
the power for a
constant voltage, we predict that the 850 W setting has the higher resistance.
( b)
R=
P
V2
=
=
(120 V )2
850 W
(120 V )2
= 17 Ω
= 12 Ω
P
1250 W
39. To find the kWh of energy, multiply the kilowatts of power consumption by the number of
hours in operation.
 1kW 
 1h 
Energy = P ( in kW ) t ( in h ) = ( 550 W) 
( 6.0 min ) 

 = 0.055kWh
 1000 W 
 60 min 
To find the cost of the energy used in a month, multiply times 4 days per week of usage,
times 4 weeks per month, times the cost per kWh.
kWh   4d  4 week   9.0cents 

Cost =  0.055

 1month   kWh  = 7.9cents month
d   1week 



( c)
R=
V2
45.
Find the current used to deliver the power in each case, and then find the power
dissipated in the
resistance at the given current.
P
P2
P = IV → I =
Pdissipated = I 2 R = 2 R
V
V
( 7.5 × 10 W ) ( 3.0 Ω ) = 11719 W
=
(1.2 × 10 V )
Pdissipated
12,000 V
50,000 V
49.
57.
value.
2
4
2
( 7.5 × 10 W ) ( 3.0 Ω ) = 675 W
=
(5 × 10 V )
2
5
Pdissipated
5
4
difference = 11719 W − 675 W = 1.1 × 104 W
2
Use Ohm’s law and the relationship between peak and rms values.
V
220 V
I peak = 2 I rms = 2 rms = 2
= 0.12 A
R
2700 Ω
(a)
We follow the derivation in Example 25-14. Start with Eq. 25-14, in absolute
j = nevd → vd =
vd =
j
ne
=
I
neA
=
(
I
 N (1 mole ) 
2
1
 m 1 mole ρ D  e π ( 2 d ) 
) 
 (
)(
m )(1.60 × 10
4 2.3 × 10−6 A 63.5 × 10−3 kg
( 6.02 × 10 )(8.9 × 10
23
3
kg
3
−19
=
4I m
N ρ D eπ d 2
)
) (
−3
C π 0.65 × 10 m
)
2
= 5.1 × 10 −10 m s
(b)
Calculate the current density from Eq. 25-11.
4 ( 2.3 × 10 −6 A )
4I
I
I
=
= 6.931A m 2 ≈ 6.9 A m 2
j= = 2 =
A πr
π d 2 π ( 6.5 × 10 −4 m )2
(c)
The electric field is calculated from Eq. 25-17.
Ch. 26
12.
(a)
Each bulb should get one-eighth of the total voltage, but let us prove that instead
of assuming it.
Since the bulbs are identical, the net resistance is Req = 8R . The current flowing
through the
bulbs is then Vtot = IReq → I =
Vtot
Req
=
Vtot
8R
. The voltage across one bulb is found from
Ohm’s law.
V = IR =
(b)
I=
Vtot
8R
Vtot
8R
R=
Vtot
=
8
→ R=
Vtot
8I
110 V
8
=
= 13.75 V ≈ 14 V
110 V
8 ( 0.42 A )
= 32.74 Ω ≈ 33 Ω
P = I 2 R = ( 0.42 A ) ( 32.74 Ω ) = 5.775 W ≈ 5.8 W
2
17.
The resistance of each bulb can be found by using Eq. 25-7b, P = V 2 R . The two
individual
resistances are combined in parallel. We label the bulbs by their wattage.
1 P
=
P =V2 R →
R V2
−1
−1
 75W
 1
1 
25W 
+
=
+
= 121Ω ≈ 120 Ω
Req = 

2
2 
110
V
110
V
(
)
(
)
 R75 R40 


19.
The resistors have been numbered in the accompanying diagram to
help in
the analysis. R1 and R2 are in series with an equivalent resistance of
R12 = R + R = 2 R . This combination is in parallel with R3 , with an
−1
1 1 
2
equivalent resistance of R123 =  +
 = 3 R . This combination is in
 R 2R 
R2
R1
R3
series with R4 , with an equivalent resistance of R1234 = 23 R + R = 53 R . This
R4
R5
combination is in parallel with R5 , with an equivalent resistance of
R6
−1
R12345
1 3 
5
= +
 = 8 R . Finally, this combination is in series with R6 ,
 R 5R 
and we calculate the final equivalent resistance.
Req = 58 R + R =
13
8
R
34. (a) There are three currents involved, and so there must be three
independent equations to determine those three currents. One comes
from Kirchhoff’s junction rule applied to the junction of the three
branches on the right of the circuit.
I 2 = I1 + I 3 → I1 = I 2 − I 3
I1
R1
e1
R2
I2
R3
e2
I3
Another equation comes from Kirchhoff’s loop rule applied to the top loop, starting at
the negative terminal of the battery and progressing clockwise.
e1 − I1R1 − I 2 R2 = 0 → 9 = 25I1 + 48 I 2
The final equation comes from Kirchhoff’s loop rule applied to the bottom loop,
starting at the negative terminal of the battery and progressing counterclockwise.
e 2 − I 3 R3 − I 2 R2 = 0 → 12 = 35I 3 + 48I 2
Substitute I1 = I 2 − I 3 into the top loop equation, so that there are two equations with
two unknowns.
9 = 25I1 + 48 I 2 = 25 ( I 2 − I 3 ) + 48 I 2 = 73I 2 − 25I 3 ; 12 = 35I 3 + 48 I 2
Solve the bottom loop equation for I 2 and substitute into the top loop equation,
resulting in an equation with only one unknown, which can be solved.
12 − 35I 3
12 = 35I 3 + 48I 2 → I 2 =
48
 12 − 35I 3  − 25I → 432 = 876 − 2555I − 1200 I →
9 = 73I 2 − 25I 3 = 73 

3
3
3
 48 
I3 =
444
3755
= 0.1182 A ≈ 0.12 A , up ; I 2 =
12 − 35I 3
48
= 0.1638 A ≈ 0.16 A , left
I1 = I 2 − I 3 = 0.0456 A ≈ 0.046 A , right
(b) We can include the internal resistances simply by adding 1.0Ω to R1 and R3. So let
R1 = 26 Ω and let R3 = 36 Ω. Now re-work the problem exactly as in part (a).
I 2 = I1 + I 3 → I1 = I 2 − I 3
e1 − I1R1 − I 2 R2 = 0 → 9 = 26 I1 + 48 I 2
e 2 − I 3 R3 − I 2 R2 = 0 → 12 = 36 I 3 + 48 I 2
9 = 26 I1 + 48I 2 = 26 ( I 2 − I 3 ) + 48 I 2 = 74 I 2 − 26 I 3 ; 12 = 36 I 3 + 48 I 2
12 = 36 I 3 + 48 I 2 → I 2 =
12 − 36 I 3
48
=
1 − 3I 3
4
 1 − 3I 3 
9 = 74 I 2 − 26 I 3 = 74 

I3 =
38
326
4
 − 26 I 3 → 36 = 74 − 222 I 3 − 104 I 3 →

= 0.1166 A ≈ 0.12 A , up ; I 2 =
1 − 3I 3
4
= 0.1626 A ≈ 0.16 A, left
I1 = I 2 − I 3 = 0.046 A , right
The currents are unchanged to 2 significant figures by the inclusion of the internal
resistances.
38. The circuit diagram has been labeled with six different currents.
b
We apply the junction rule to junctions a, b, and c. We apply the
R3
loop rule to the three loops labeled in the diagram.
R1
I1
I
3
1) I = I1 + I 2 ; 2 ) I1 = I 3 + I 5 ; 3) I 3 + I 4 = I
4 ) − I1R1 − I 5 R5 + I 2 R2 = 0 ; 5) − I 3 R3 + I 4 R4 + I 5 R5 = 0
a
I5
1
6) e − I 2 R2 − I 4 R4 = 0
Eliminate I using equations 1) and 3).
R5
I4
I2
R2
R4
d
3
I
2
å
+
–
c
1) I 3 + I 4 = I1 + I 2 ; 2 ) I1 = I 3 + I 5
4 ) − I1R1 − I 5 R5 + I 2 R2 = 0 ; 5) − I 3 R3 + I 4 R4 + I 5 R5 = 0
6) e − I 2 R2 − I 4 R4 = 0
Eliminate I1 using equation 2.
1) I 3 + I 4 = I 3 + I 5 + I 2 → I 4 = I 5 + I 2
4 ) − ( I 3 + I 5 ) R1 − I 5 R5 + I 2 R2 = 0 → − I 3 R1 − I 5 ( R1 + R5 ) + I 2 R2 = 0
5) − I 3 R3 + I 4 R4 + I 5 R5 = 0
6) e − I 2 R2 − I 4 R4 = 0
Eliminate I 4 using equation 1.
4 ) − I 3 R1 − I 5 ( R1 + R5 ) + I 2 R2 = 0
5) − I 3 R3 + ( I 5 + I 2 ) R4 + I 5 R5 = 0 → − I 3 R3 + I 5 ( R4 + R5 ) + I 2 R4 = 0
6) e − I 2 R2 − ( I 5 + I 2 ) R4 = 0 → e − I 2 ( R2 + R4 ) − I 5 R4 = 0
Eliminate I 2 using equation 4: I 2 =
5) − I 3 R3 + I 5 ( R4 + R5 ) +
1
R2
1
R2
[I R
3
1
[I R
3
1
+ I 5 ( R1 + R5 ) ] .
+ I 5 ( R1 + R5 ) ] R4 = 0 →
I 3 ( R1R4 − R2 R3 ) + I 5 ( R2 R4 + R2 R5 + R1R4 + R5 R4 ) = 0
6) e −
1
R2
[I R
3
1
+ I 5 ( R1 + R5 ) ] ( R2 + R4 ) − I 5 R4 = 0 →
e R2 − I 3 R1 ( R2 + R4 ) − I 5 ( R1R2 + R1 R4 + R5 R2 + R5 R4 + R2 R4 ) = 0
Eliminate I 3 using equation 5: I 3 = − I 5
( R2 R4 + R2 R5 + R1R4 + R5 R4 )
( R1R4 − R2 R3 )
( R2 R4 + R2 R5 + R1R4 + R5R4 ) 
 R1 ( R2 + R4 ) − I 5 ( R1R2 + R1R4 + R5 R2 + R5 R4 + R2 R4 ) = 0
( R1R4 − R2 R3 )



I  ( R R + R2 R5 + R1R4 + R5 R4 ) 
e = − 5  2 4
+
−
+
+
+
+
R
R
R
R
R
R
R
R
R
R
R
R
R
(
)
(
)

1
2
4
1
2
1
4
5
2
5
4
2
4

R2 
( R1R4 − R2 R3 )



e R2 +  I 5
 ( 25 Ω)(14 Ω) + ( 25 Ω)(15 Ω) + ( 22 Ω)(14 Ω) + (15 Ω)(14 Ω) 

( 22 Ω)( 25 Ω + 14 Ω) 
I 5 

( 22 Ω)(14 Ω) − ( 25 Ω)(12 Ω)
=−



25 Ω 

− [( 22 Ω)( 25 Ω) + ( 22)(14) + (15 Ω)( 25 Ω) + (15 Ω)(14 Ω) + ( 25 Ω)(14 Ω)]

= − I 5 ( 5261Ω) → I5 = −
6.0V
5261Ω
= −1.140mA ( upwards )
( R2 R4 + R2 R5 + R1R4 + R5 R4 )
( R1R4 − R2 R3 )
( 25 Ω)(14 Ω ) + ( 25 Ω)(15 Ω) + ( 22 Ω)(14 Ω) + (15 Ω)(14 Ω) = 0.1771A
= − ( −1.140 mA )
( 22 Ω)(14 Ω) − ( 25 Ω)(12 Ω)
I3 = − I5
I2 =
1
R2
[I R
3
1
+ I 5 ( R1 + R5 ) ] =
1
25 Ω
[( 0.1771A)( 22 Ω) + ( −0.00114 A )( 37 Ω)] = 0.1542 A
I 4 = I 5 + I 2 = −0.00114 A + 0.1542 A = 0.1531A
I1 = I 3 + I 5 = 0.1771A − 0.00114 A = 0.1760 A
We keep an extra significant figure to show the slight difference in the currents.
I 22 Ω = 0.176A
40.
I 25 Ω = 0.154 A
I12 Ω = 0.177 A
I14 Ω = 0.153A
I15Ω = 0.001A, upwards
(a)
As shown in the diagram, we use symmetry to reduce the
number of independent currents to six. Using Kirchhoff’s junction rule, we write
equations for junctions a, c, and d. We then use Kirchhoff’s loop rule to write the loop
equations for loops afgba, hedch, and aba (through the voltage source).
I = 2 I1 + I 2 [1] ; I 3 + I 4 = I1 [2] ; I 5 = 2 I 4 [3]
0 = −2 I1R − I 3 R + I 2 R [4] ; 0 = −2 I 4 R − I 5 R + I 3 R [5]
0 = e − I 2 R [6]
We have six equations with six unknown currents. We use the method of substitution
to reduce the equations to a single equation relating the emf from the power source to
the current through the power source. This resulting ratio is the effective resistance
between points a and b. We insert Eqs. [2], [3], and [6] into the other three equations to
eliminate I1, I2, and I5.
e
e
I = 2 ( I 3 + I 4 ) + =2 I 3 + 2 I 4 +
[1*]
R
R
e
0 = −2 ( I 3 + I 4 ) R − I 3 R + R = − 2 I 4 R − 3I 3 R + e
[4*]
R
0 = −2 I 4 R − 2 I 4 R + I 3 R = −4 I 4 R + I 3 R
[5*]
We solve Eq. [5*] for I3 and insert that into Eq. [4*]. We then insert the two results
into Eq. [1*] and solve for the effective resistance.
e
I 3 = 4 I 4 ; 0 = −2 I 4 R − 3 ( 4 I 4 ) R + e → I 4 =
14 R
10e e 24e 12e
e
e
e
+ =
=
→ Req = = 127 R
I = 2 ( 4 I 4 ) + 2 I 4 + = 10 I 4 + =
R
R
14 R R 14 R 7 R
I
(b)
As shown in the diagram, we use symmetry to
I3
reduce the
g
f
I2
number of currents to four. We use Kirchhoff’s junction
I=0
I
rule at junctions a and d and the loop rule around loops
a
b
abca (through the voltage source) and afgdcha. This
I3
I3
I1
I1
results in four equations with four unknowns. We solve
I
I1
3
e
d
I=0
h
c
I1
I2
I
these equations for the ratio of the voltage source to current I, to obtain the effective
resistance.
I = 2 I1 + I 2 [1]
; 2 I 3 = I 2 [2]
0 = −2 I 2 R + e [3] ; 0 = −2 I 2 R − 2 I 3 R + 2 I1R [4]
We solve Eq. [3] for I2 and Eq. [2] for I3. These results are inserted into Eq. [4] to
determine I1. Using these results and Eq. [1] we solve for the effective resistance.
e
e
e
e
3e
I
; I3 = 2 =
; I1 = I 2 + I 3 =
I2 =
+
=
2R
2 4R
2R 4R 4R
2e
e
 3e  e
=
I = 2 I1 + I 2 = 2 
; Req = = 12 R
+
4
2
R
R
R
I


(c)
As shown in the diagram, we again use symmetry
I2
g
to reduce
f
I1
the number of currents to three. We use Kirchhoff’s
I2
I
junction rule at points a and b and the loop rule around
a
I1
b
the loop abgda (through the power source) to write three
I2
I1
I1
equations for the three unknown currents. We solve
I1 I2
d
e
these equations for the ratio of the emf to the current
I
through the emf (I) to calculate the effective resistance.
I2
I
I = 3I1 [1] ; I1 = 2 I 2 [2]
1
h
c
I2
0 = −2 I1R − I 2 R + e [3]
We insert Eq. [2] into Eq. [3] and solve for I1. Inserting I1 into Eq. [1] enables us to
solve for the effective resistance.
e
2e
6e
→ Req = = 65 R
0 = −2 I1R − 12 I1 R + e → I1 =
; I = 3 I1 =
5R
5R
I
46.
Express the stored energy in terms of the charge on the capacitor, using Eq. 24-5. The
charge on the
capacitor is given by Eq. 26-6a.
U=
1
2
Q2
C
C e (1 − e −t τ )
2
2
=
= 12 C e 2 (1 − e −t τ ) = U max (1 − e −t τ ) ;
2
1
2
C
(
U = 0.75U max → U max 1 − e −t τ
(
)
2
= 0.75U max →
(1 − e )
−t τ
2
= 0.75 →
)
t = −τ ln 1 − 0.75 = 2.01τ
50.
(a)
With the currents and junctions labeled as in the diagram,
we use
point a for the junction rule and the right and left loops for the
loop rule. We set current I3 equal to the derivative of the charge
on the capacitor and combine the equations to obtain a single
differential equation in terms of the capacitor charge. Solving this
equation yields the charging time constant.
Q
I1 = I 2 + I 3 [1] ; e − I1R1 − I 2 R2 = 0 [2] ; − + I 2 R2 = 0 [3]
C
We use Eq. [1] to eliminate I1 in Eq. [2]. Then we use Eq. [3] to eliminate I2
from Eq. [2].
 Q 
0 = e − ( I 2 + I 3 ) R1 − I 2 R2 ; 0 = e − I 2 ( R1 + R2 ) − I 3 R1 ; 0 = e − 
 ( R1 + R2 ) − I 3 R1
 R2C 
We set I3 as the derivative of the charge on the capacitor and solve the differential
equation by separation of variables.
 Q 
dQ
0= e − 
R1 →
 ( R1 + R2 ) −
dt
 R2C 
Q
∫
0
( R1 + R2 )
R1 R2C
t
t
t −(R + R )
dQ ′
1
2
dt ′ →
=∫
0
R1 R2C
 R2C e 
Q′ − 

 R1 + R2 

 R2C e  
Q − 

 R1 + R2   = − ( R1 + R2 ) t →
→ ln 
  R2C e  
R1 R2C
 
 
  R1 + R2  

 R Ce 
( R1 + R2 ) t′
ln Q ′ −  2
 = −
R1R2C
 R1 + R2   0

0
−
R Ce 
1 − e
Q= 2
R1 + R2 

Q




From the exponential term we obtain the time constant, τ =
(b)
to infinity.
R1 R2C
.
R1 + R2
We obtain the maximum charge on the capacitor by taking the limit as time goes
−
R Ce 
1 − e
Qmax = lim 2
t →∞ R1 + R2 

( R1 + R2 )
R1 R2 C
t

R Ce
= 2
 R1 + R2
