Chapter 21
Problem 40.
A dead battery is charged by connecting it to the live battery of another car with jumper cables (Fig. P21.40).
Determine the current in the starter and in the dead battery.
Let V
L
= 12 V and R dead battery, and R
S
L
= 0
= 0 .
.
01Ω be the parameters of the live battery,
06Ω be the resistance of the starter. Let I
L
V
D
= 10 V and R
D
= 1 .
00Ω be the parameters of the be the current going upward in the left branch, I
D be the current going upward in the middle branch, and I
S be the current going downward in the right branch.
Applying Kirchoff’s junction rule to the top node, we have
I
L
+ I
D
− I
S
= 0
Applying Kirchoff’s loop rule to the outer and right loops respectively, we have
(1)
V
L
− I
L
R
L
− I
S
R
S
= 0
V
D
− I
D
R
D
− I
S
R
S
= 0
(2)
(3)
Solving these using linear algebra (or your method of choice)

0


V
L 
V
D

1 1 − 1
 
I
L

=

R
L
0
0 R
D
R
R
S
S
 
I
D 
I
S

1 1 − 1
 
I
L

=

0 .
01 0 0 .
06
 
I
D 
0 1 .
00 0 .
06 I
S
 I
L


I
D 
I
S
=
 0 .
850 15 .
0 − 0 .
850  

0 .
0085 − 0 .
850 0 .
992
− 0 .
142 14 .
2 0 .
142

0

12
10


 172 
=

− 0 .
283

171
A
(4)
(5)
Where I
2
< 0 indicates that the current in the middle branch actually flows downward, recharging the dead battery.
Problem 46.
A C = 10 .
0 µ F capacitor is charged by a = 10 .
0 V battery through a resistance R . The capacitor reaches a potential difference of V
C
( t f
) = 4 .
00 V at the instant t f
= 3 .
00 s after the charging begins. Find R .
Applying Kirchoff’s loop rule,
− V
C
− V
R
C
R dq dq
− dt dq dt q q
= −
C q
= −
C
C − q
=
RC dt
=
RC
− IR = 0
Z dq Z
= dt
C − q RC
− ln( C − q ) = t/RC + A
C − q = Be
− t/RC q = C − Be
− t/RC q = C (1 − e
− t/RC
)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
Where A is a constant of integration, B = e
− A is another way of writing that constant, C = Q (because as t → ∞ , q → Q ), and B = Q = C (because at t = 0, q = 0).
Note: The book derives this on pages 709-710 if you want more details.
Now applying this to our particular problem,
V
C
( t f
) = q ( t f
)
C
=
C
(1 − e
− t f
C
/RC
)
V
C
( t f
)
= 1 − e
− t f
/RC
(15)
(16) e
− t f
/RC
= 1 −
V
C
( t f
)
− t f
RC
= ln 1 −
V
C
( t f
)
R =
− t f
C ln(1 − V
C
( t f
) / )
=
10 .
0 · 10 − 6
− 3 .
00 s
F · ln(1 − 4 .
00 V / 10 .
0 V)
= 587 kΩ
(17)
(18)
(19)

Problem 55.
Four V = 1 .
50 V AA batteries in series are used to power a transistor radio. If the batteries can move a charge of ∆ Q = 240 C, how long will they last if the radio has a resistance of R = 200Ω?
Using Kirchoff’s loop rule,
V + V + V + V − IR = 0
I =
∆ t =
4 V
=
R
∆ qR
4 V
∆
∆
= q t
240 C
4 · 1 .
· 200Ω
50 V
= 8000 s = 2 .
22 hours
(20)
(21)
(22)