Physics 102 Group Session for Chapter 20 &21 Key
1) There are approximately 110 million TVs in the United States. each one uses, on average, 75 W of power and is turned
on for 6 hrs per day. If electricity costs $0.10 per KWh, how much money is spent each day on running all the TVs in the
US?
REASONING The total cost of keeping all the TVs turned on is equal to the number of TVs times the cost to keep each
one on. The cost for one TV is equal to the energy it consumes times the cost per unit of energy ($0.10 per kWh). The
energy that a single set uses is, according to Equation 6.10b, the power it consumes times the time of use.
SOLUTION The total cost is
Total cost = 110 million sets  Cost per set 
 $0.10 
 110 million sets   Energy  in kW  h  used per set  

 1 kW  h 
The energy (in kWh) used per set is the product of the power and the time, where the power is expressed in kilowatts and
the time is in hours:
 1 kW 
Energy used per set = P t   75 W  
  6.0 h 
 1000 W 
(6.10b)
The total cost of operating the TV sets is

  $0.10 
 1 kW 
6
Total cost = 110 million sets   75 W  
  6.0 h   
  $5.0  10
 1000 W 

  1 kW  h 
2) REASONING AND SOLUTION Each resistor can tolerate a current of no more than
I
P
0.25 W

 0.073 A
R
47 
Ohm's law applied to a series circuit containing N such resistors gives V = IRs = INR, so
N
V
9.0 V

 2.6
IR 0.073 A47 
Only three resistors can be used.
3) REASONING AND SOLUTION REASONING In preparation for applying Kirchhoff’s rules, we now choose
the currents in each resistor. The directions of the currents are arbitrary, and should they be incorrect, the
currents will turn out to be negative quantities. Having chosen the currents, we also mark the ends of the
resistors with the plus and minus signs that indicate that the currents are directed from higher (+) toward
lower () potential. These plus and minus signs will guide us when we apply Kirchhoff’s loop rule.
A
+
2.00

Ω
I1
+

3.00
V
B + 8.00
Ω
I
 6.00 2
+ V

I3
 E
+

9.00
V
4.00
Ω
+
D
C
SOLUTION Applying the junction rule to junction B, we find
I1  I3 
Into junction
F
(1)
I2
Out of junction
Applying the loop rule to loop ABCD (going clockwise around the loop), we obtain
I1  2.00    6.00 V  I3  4.00    3.00 V
Potential drops
(2)
Potential rises
Applying the loop rule to loop BEFC (going clockwise around the loop), we obtain
I 2  8.00    9.00 V  I3  4.00    6.00 V 
Potential drops
0
(3)
Potential rises
Substituting I2 from Equation (1) into Equation (3) gives
 I1  I3  8.00    9.00 V  I3  4.00    6.00 V  0
I1  8.00    I3 12.00    15.00 V  0
(4)
Solving Equation (2) for I1 gives
I1  4.50 A  I3  2.00 
This result may be substituted into Equation (4) to show that
 4.50 A  I3  2.00    8.00    I3 12.00    15.00 V  0
I3  28.00    51.00 V  0
or
I3 
51.00 V
 1.82 A
28.00 
The minus sign indicates that the current in the 4.00- resistor is directed downward , rather than upward
as selected arbitrarily in the drawing.
4) REASONING
Equation 21.2 gives the radius r of the circular path as r = mv/( q B), where m, v, and q are, respectively,
the mass, speed, and charge magnitude of the particle, and B is the magnitude of the magnetic field. We
wish the radius to be the same for both the proton and the electron. The speed v and the charge magnitude
q are the same for the proton and the electron, but the mass of the electron is 9.11  10–31 kg, while that
of the proton is 1.67  10–27 kg. Therefore, to offset the effect of the smaller electron mass m in
Equation 21.2, the magnitude B of the field must be reduced for the electron.
SOLUTION Applying Equation 21.2 to the proton and the electron, both of which carry charges of the
same magnitude q = e, we obtain
r
mp v
eBp
r
and
mev
eBe
Electron
Proton
Dividing the proton-equation by the electron-equation gives
mp v
r eBp

r me v
eBe
1
or
mp Be
me Bp
Solving for Be, we obtain
Be 
me Bp
9.111031 kg   0.50 T 



1.67 10
mp
27
2.7 104 T
kg
5) REASONING AND SOLUTION The current associated with the lightning bolt is
I 
15 C
q

 1.0  10 4 A
t
1.5  10 3 s
The magnetic field near this current is given by



4  107 T  m/A 1.0  104 A
0 I
B 

 8.0  105T
2 r
2  25 m 