Homework 4 Solutions
14. S1, S2 and S3 all open: ia 0.00 A.
S1 closed, S2 and S3 open: ia
VB 2R1
S2 closed, S1 and S3 open: ia
VB 2R1 R2
S3 closed, S1 and S2 open: ia
VB 2R1 2R2
120 V 40.0
3.00 A.
120 V 50.0
2.40 A.
120 V 60.0
2.00 A.
S1 open, S2 and S3 closed:
Req
so ia
VB Req
R1
1
R2
R1
120 V 42.0
1
1
R1
R2
20
1
30
10
1
1
20
42.0
,
2.86 A.
S2 open, S1 and S3 closed:
Req
so ia
VB Req
1
R1 2 R2
R1
120 V 33.3
1
R1
1
20
1
40
1
20
20
1
30
1
20
1
33.3 ,
3.60 A.
S3 open, S1 and S2 closed:
Req
so ia
VB Req
R1
1
R1
120 V 32.0
1
1
R1
R2
1
32.0
,
3.75 A.
S1, S2 and S3 all closed: Starting from the right and working leftward…The rightmost
pair of series resistors is in parallel with the resistor directly below S2. Call the
equivalent resistance of that trio R3. Then
1
R3
R1
1
1
R1
R2
1
30
1
20
1
12 .
From there,
Req
R1
1
R2
R3
1
R1
1
20
1
22
1
20
1
30.5
,
so ia
VB Req
120 V 30.5
3.94 A.
22. The two 4.00 resistors are in parallel, and that parallel combination is in series with
the 2.50 resistor. The equivalent resistance between points D and E is therefore
1
4.00
Req
1
1
4.00
2.50
31. The potential difference across the 6.0
potential difference across the 2.0
4.50 .
resistor is
V1
resistor must be
6.0 A 6.0
V2
VA VB
36 V.
V1
The
42 V. The
current through the 2.0 resistor is therefore i2 42 V 2.0
21 A. By the junction rule,
the current in the 4.0 resistor is i3 21 A 6.0 A 15 A. Using P i2 R for each resistor,
we find the total power dissipated by the resistors:
P used
21 A
2
2.0
15 A
2
4.0
6.0 A
2
6.0
1998 W.
By contrast, the power supplied (externally) to this section is
Psupplied
i2 VA VB
21 A 78 V
1638 W.
Therefore, the “Box” must be supplying energy at a rate of 1998 – 1638 = 360 W.
43. (a) To provide 800 W h of energy to the lamp would require 400 batteries. That
would cost $320.
(b) At 12¢ per kW h , the 800 W h needed would cost 9.6¢.
45. If P is the rate at which the battery delivers energy, then the battery delivers an
amount of energy E P t in a time t. At the same time, the chemical energy of the
battery is reduced by E q E , where q is the charge that passes through in time t
and E is the emf of the battery. Delivering power at a rate of P 100 W, the battery will
be completely discharged in
t
qE
P
(120 A h) (12 V)
100 W
14.4 h 14 h 24 min.
49. (a) With the battery’s emf denoted E and the internal resistance r, the equation
describing the terminal voltage as a function of the current drawn from the battery is
VT E ir. Doing a linear regression fit for the VT versus i values listed, we obtain
VT
(b) From the fit, the emf is E 13.6 V.
13.611 0.0599i.
(c) From the fit, the internal resistance is r 0.060 . .
54. (a) The emf of the battery must be greater than the terminal voltage by the amount of
VB ir 12 V 10 A 0.050
12.5 V.
voltage drop across the internal resistor: E
(b) We’re going to use the same equation to solve for the current as we used in part (a),
but both VB and i take on different values from before. Call the old values of the
terminal voltage and the current through the battery VB 1 and i1 respectively. The new
value of the current through the battery is i2 imotor ilights 2 imotor 8.0A. The new terminal
voltage VB 2 is enough potential difference to produce a current of ilights 2 in the lights. The
resistance of those lights must be Rlights VB 1 i1 12 V 10 A 1.2 , so the new terminal
9.6 V.
voltage is VB 2 ilights 2 Rlights 8.0 A 1.2
We can determine the current in the motor from E
imotor
E
VB 2
VB 2
imotor
ilights 2 r :
12.5 V 9.6 V
8.0 A 50 A.
0.050
ilights 2
r
VB 2 i2 r
Chapter 29
1. (a) We use Eq. 29-3:
F mag
q vB sin
3.2 10
19
C 550m s 0.045 T sin 52
(b) The acceleration has a magnitude of a
 
F mag
m
6.2 10 18 N
6.6 10 27 kg
6.2 10
18
N.
9.5 108 m s2 .
(c) Since it is perpendicular to v , F mag does not do any work on the particle. There is
therefore no change in the particle's kinetic energy, and the speed does not change.
3. (a) Eq. 29-3 leads to
v
F mag
eB sin
1.60 10
19
6.50 10 17 N
C 2.60 10 3 T sin 23.0
4.00 105 m s.
(b) The kinetic energy of the proton is
K
1
2
mv 2
In electron-volts, this is K
1
2
1.67 10
1.34 10
16
27
J
kg 4.00 105 m s
1.60 10
19
2
1.34 10
J eV =835eV.
16
J.
7. We solve Eq. 29-7 for the magnetic field to obtain
B
9. (a) From K
31
9.11 10
me v
er
kg 1.3 106 m s
1.60 10
19
C 0.35 m
2.1 10 5 T.
1
me v 2 we get
2
2 1.20 103 eV 1.60 10
2K
me
v
9.11 10
31
19
J eV
kg
2.05 107 m s.
(b) Solving Eq. 29-7 for the magnetic field strength yields
9.11 10
me v
qr
B
31
1.60 10
kg 2.05 107 m s
19
C 25.0 10
2
m
4.67 10 4 T.
(c) The frequency is
(d) The period is T 1 f
2.07 107 m s
2 25.0 10 2 m
v
f
2 r
1
1.31 107 Hz
7.63 10
1.31 107 Hz.
8
s.
15. The particles will travel in circular paths with a speed obtained from Eq. 29-7 as
v q Br m . The kinetic energy of each particle is
1
2
K
mv2
B2 r 2 q 2
.
2 m
In order for the other particles to circulate in the same path as the proton,
(a) Kalpha
qalpha
2
qproton
mproton
malpha
2
(b) Kdeuteron
qdeuteron
qproton
K proton
mproton
mdeuteron
2
K proton
2
1
4
1
1.0 MeV
2
1
2
1.0 MeV;
1.0 MeV
0.50 MeV.