1. The temperature at which the Fahrenheit and Celsius scales have the same reading is -40 degrees. 2. The change in diameter of a gold ring that falls into hot water is 4.6 x 10-6 m, as calculated using the coefficient of linear expansion for gold. 3. The original volume of a copper kettle that expands when heated is calculated to be 2.8 x 10-3 m3 based on the coefficient of volume expansion for copper.

1. Memo/EMH1501/101/03/2021
1
ASSIGNMENTS - 2021
(The assignments that follow are for students registered for the 2021 academic session)
Semester 1
Assignment: AS1 # Unique No:781080
Due date: 12th
May 2021
Total marks 100 (20 x 5 = 100)
1. There is one temperature at which the Fahrenheit and Celsius scales have the same
reading. Find the temperature.
Solution: Formula for Fahrenheit, F = 9/5 C + 32
Formula for Centigrade, C =5/9 (F-32)
Let F = C or C= F
F= C= 5/9(F-32)= (5F-160)/9
Or, 9F= 5F - 160
Or, 9F - 5F = -160
Or, 4F = -160
Or, F = -160/4
Or, F= -40 degree answer in Fahrenheit
Or, C= F= 9/5 C + 32
Or, C = (9C + 160)/5
Or, 5C = 9C + 160
Or, 5C - 9C = 160
Or, -4C= 160
Or, C=160/-4
Or, C= -40 degree answer in Centigrade
Conclusion: The temperature wherein degrees centigrade is the same in degrees Fahrenheit is
-40 degrees .

2. Memo/EMH1501/101/03/2021
2
2. A gold engagement ring has an inner diameter of 1.5 x 10-2
m and a temperature of 27
o
C. The ring falls into a sink of hot water whose temperature is 49 o
C. What is the
change in the diameter of the hole in the ring? [ = 14 x 10-6
(o
C)-1
is the coefficient
of linear expansion for gold].
Solution: The hole expands as it were filled with gold, so the change in the diameter is
given by L = L0T, where  = 14 x 10-6
(o
C)-1
is the coefficient of linear expansion for
gold, L0 is the original diameter and T is the change in temperature.
The change in the ring’s diameter is L = L0T
= [14 x 10-6
(o
C)-1
] (1.5 x 10-2
m) (49 o
C – 27 o
C) = 4.6 x 10-6
m.
3. A copper kettle contains water at 15 o
C. When the water is heated to its boiling point
of 100.0 o
C, the volume of the kettle expands by 1.2 x 10-5
m3
. Determine the volume
of the kettle at 15 o
C [ = 51 x 10-6
(o
C)-1
is the coefficient of volume expansion for
copper].
Solution: The volume expands, so the change in volume is given by V = V0T, where
 = 51x 10-6
(o
C)-1
is the coefficient of volume expansion for copper, V0 is the original
volume and T is the change in temperature.
The change in volume is V = V0T,
V0 = V/(T) = (1.2 x 10-5
m3
) / 51 x 10-6
(o
C)-1
x (100 o
C - 15 o
C) = 2.8 x 10-3
m3
.
4. A 7.5 g lead sphere is fired into a target at a speed of 240 m/s. If 75% of the kinetic
energy goes into heating the bullet as it is brought to rest, find the increase in
temperature of the sphere. [Specific heat capacity of lead = 128 J/(Kg.o
C)]
Solution: Kinetic energy of the lead = ½ (7.5 x 10-3
kg) (240 m/s)2
.
75% of kinetic energy = 0.75 x ½ x 7.5 x 10-3
x 240 x 240 kg. m2
s-2
= 162000 J = Q
Increase of temperature of the sphere, T= Q/cm
= (0.75 x ½ x 7.5 x 10-3
x 240 x 240) / (128 x 7.5 x 10-3
)
= (0.75 x 120 x 240) / 128 = 168.75 0
C
5. (a) Objects A and B have the same mass of 3.0 Kg. They melt when 3.0 x 104
J of heat
is added to object A and when 9.0 x 104
J is added to object B. Determine the latent
heat of fusion for the substance from which each object is made. (b) Find the heat
required to melt object A when its mass is 6.0 kg.
Solution: (a) The object A is melted when 3.0 x 104
J of heat is added i.e. LA = QA/m
LA = (3.0 x 104
J) / 3.0 kg
LA = 1.0 x 104
J/kg.
The object B is melted when 9.0 x 104
J of heat is added i.e. LB = QB/m
LB = (9.0 x 104
J) / 3.0 kg

3. Memo/EMH1501/101/03/2021
3
LB = 3.0 x 104
J/kg.
(b) Q = mA LA = 6.0 kg x 1.0 x 104
J/kg = 6.0 x 104
J.
6. A closed box is filled with dry ice at a temperature of -78.5 o
C, while the outside
temperature is 21.0 o
C. The box is cubical, measuring 0.35 m on a side, and the
thickness of the walls is 2.75 x 10-2
m. In one day, 3.10 x 106
J of heat is conducted
through the six walls. Find the thermal conductivity of the material from which the
box is made.
Solution: The energy Q conducted through one face (thickness L and surface area A) of
the cubical box in a time t is
( )
kA T t
Q
L

= , where ΔT is the temperature difference
between the outside and inside surface of the box and k is the thermal conductivity of the
material from which the box is made. With the aid of this equation, we can determine k.
Since the cube has six faces, the total heat conducted through all six faces is total 6
Q Q
= .
Using equation for Q, we have
total
6( )
kA T t
Q
L

=
Noting that the number of seconds in one day is ( ) 4
3600 s
24 h 8.64 10 s
1 h
 
= 
 
 
and
solving for the thermal conductivity k, we find that
( ) ( )
total
6 2
2
2 4
6
(3.10 10 J)(2.75 10 m)
1.35 10 J / (s m C )
6(0.350 m) 21.0 C 78.5 C 8.64 10 s
Q L
k
A Tt
−
−
=

 
= =    
 − −  
 
 
7. Two identically sized objects, one at 25 o
C with emissivity e1 and the other at 75 o
C
with emissivity e2, radiate the same power. Find the emissivity ratio e1/ e2.
Solution: The radiant energy Q radiated by an object is given by 4
Q e T At

= , where e is
the emissivity,  is the Stefan-Boltzmann constant, T is its temperature (in Kelvins), A is
the surface area of the object, and t is the time. The radiant energy emitted per second, or
power, is Q/t = 4
.
e T A
 Since both objects radiate the same power, we can set the
equation for each of them equal to one another and solve for the ratio 1 2
/
e e .
Setting the two equations equal to one another with appropriate subscripts and using the
fact that the objects are identical in size gives

4. Memo/EMH1501/101/03/2021
4
1 2
1
Q Q
t t
e 
=
4
1
T A 2
e 
= 4
2
T A
( )
( )
4
4
1 2
4 4
2 1
75 C
81
25 C
e T
e T
= = =
8. A solar collector is placed in direct sunlight where it absorbs energy at the rate of 880
J/s for each square meter of its surface. The emissivity of the solar collector is e =
0.78. What equilibrium temperature does the collector reach? Assume that the only
energy loss is due to the emission of radiation.
Solution: At equilibrium the temperature of the solar collector is not changing. Since the only
energy loss is due to the emission of radiation, we conclude that the temperature is not
changing because the collector is emitting the same amount of energy as it is absorbing.
Thus, it is emitting 880 J/s for each 1.0 m2
of its surface. The Stefan-Boltzmann law of
radiation specifies that the energy Q radiated by an object in a time t is 4
Q e T At

=
(Equation 13.2), where e is the object’s emissivity, –8 2 4
5.67 10 J / (s m K )
 =    is the
Stefan-Boltzmann constant, T is the Kelvin temperature, and A is the object’s surface
area. We can rearrange this equation as follows to give the energy per second Q/t:
4
Q
e T A
t

= (1)
This equation can be solved for the temperature T.
Solving Equation (1) for the temperature T, we find that
1/4
1/4
–8 2 4 2
/ 880 J/s
376 K
0.78[5.67 10 J / (s m K )](1.0 m )
Q t
T
e A

 
 
= = =
 
 
  
   
9. A certain element has a mass per mole of 196.967 g/mol. What is the mass of a single
atom in (i) atomic mass unit (amu) and kilograms? (c) How many moles of atoms are
in a 285-g sample?
Solution: The mass of one of its atoms (in atomic mass units) has the same numerical value
as the element’s mass per mole (in units of g/mol). Atomic mass units can be converted
into kilograms using the fact that 1 u = 1.6605  10–27
kg. Dividing the mass of the
sample by the mass per mole gives the number of moles of atoms in the sample.
(a) The mass per mole is 196.967 g/mol. Since the mass of one of its atoms (in atomic
mass units) has the same numerical value as the mass per mole, the mass of a single
atom is
m = 196.967 u .
(b) To convert the mass from atomic mass units to kilograms, we use the conversion
factor of 1 u = 1.6605  10–27
kg:

5. Memo/EMH1501/101/03/2021
5
( )
27
25
1.6605 10 kg
196.967 u 3.2706 10 kg
1 u
m
−
−
 

= = 
 
 
(c) The number n of moles of atoms is equal to the mass m divided by the mass per mole:
300 g
1.52 mol
Mass per mole 196.967 g/mol
m
n = = =
10. Find the number of moles of air there are in a FIFA World cup ball if it has an inner
diameter of 22.0 cm and is inflated to a pressure of 75.2 kPa at a temperature of 293
K.
Solution: We can rearrange the ideal gas law, /
P nRT V
= , where n is the number of moles of
gas, R is the universal gas constant, T is the Kelvin temperature, and V is the volume to
find n. The variables R and T, area known and V, can be calculated using the volume of a
sphere 3
4
3
V r

= , where r will be half the diameter given in the problem.
Rearranging the ideal gas law, we have
nRT PV
P n
V RT
= → =
Substituting in the volume of a sphere and solving we find the number of moles to be equal to
( )
( )
 ( )
3
3
3
4 0.22 m
4 75.2 10 Pa
3 2
3
0.18 mol
8.31 J/ mol K 273 K
P r
PV
n
RT RT


 
 
    
 
   
 
   
= = = =

11. Two ideal gases have the same mass density and the same absolute pressure. One of
the gases is helium (He), and its temperature is 180 K. The other gas is neon (Ne).
What is the temperature of the neon?
Solution: The ideal gas law is PV nRT
= . We need to put this in terms of the mass density
/
m V
 = . We can then set He Ne
P P
= , and solve the resulting expression for TNe, the
temperature of the neon.
We begin by writing Equation 14.1 in terms of the mass density  for an ideal gas.
Recall that the number n of moles of a substance is equal to its mass m in grams, divided
by its mass per mole M. Therefore, /
PV mRT M
= , and we have
( )
/ / /
P m V RT M RT M

= = , where  is the same for each gas. The two gases have
the same absolute pressures, so that He Ne
P P
= , and it follows that
He Ne
He Ne
RT RT
M M
 
=
The term R can be eliminated algebraically from this result. Solving for the
temperature of the neon TNe and using the mass per mole for helium (4.0026 g/mol) and
neon (20.180 g/mol) from the periodic table on the inside of the back cover of the text,
we find

6. Memo/EMH1501/101/03/2021
6
Ne
Ne He
He
20.180 g/mol
(175 K) 882 K
4.0026 g/mol
M
T T
M
   
= = =
   
 
 
12. Calculate the rms speed of an electron near the Sun’s surface if the temperature and
pressure are 2.01 x 106
K and 0.295 Pa, respectively. Assume the electrons behave
like an ideal gas.
Solution: The electrons will have an average translational kinetic energy
based on the temperature of the surface of the sun, namely, 2 3
1
rms
2 2
mv kT
= . In this
expression m is the mass of an electron, vrms is the rms speed of an electron, k is
Boltzmann’s constant, and T is the Kelvin temperature. The pressure is not needed
for the calculation.
Solving equation the rms
v of an electron, we find
( )( )
31
23 6
6
rms
3 1.38 10 J/K 2.01 10
9.11 10
K
3
9.6 10 m/s
kg
kT
v
m −
−
= = 

 
=
13. A gas is diffusing from a container through a 2.5 cm long tube. Once the higher
concentration has been reduced to ½ its original value, calculate the change in
diameter of tubing that would be needed to maintain the original diffusion rate,
assuming the gas is diffusing to open air (essentially none of this gas present).
Solution: The diffusion of the gas is driven by the concentration difference ΔC = C2 – C1
between the ends of the tube. Since we are assuming diffusion to open air, C1 = 0, so
ΔC = C2. As stated in the problem we will be addressing the situation when the higher
concentration has been reduced to ½ its original value, or C’
= 0.5C2 and since the gas is
still diffusing to open air, ΔC’
= 0.5C2. The concentration difference is related to the
mass rate of diffusion by Fick’s law:
m DA C
t L

= , where D is the diffusion constant, and
A and L are, respectively, the cross-sectional area and length of the tube. If the mass
diffusion rate is to stay constant, then we can set to two situations equal to one another
' ' '
'
DA C D A C
L L
 
=
Noting that the length of the tubing will remain the same as well as the diffusion constant we
find that
' '
A C A C
 = 
Meaning the cross-sectional area of the tubing will be inversely proportional to the change in
the concentration difference. Knowing we are using a circular tube we can replace the area
formula with the area of a circle to arrive at
2 2
' '
r C r C
 

 = 
2
r C 
 = 2
2
2
2
' '
'
'
r C
r C
r C
r
C


= =
 2
0.5C
2r
=

7. Memo/EMH1501/101/03/2021
7
So the radius (and thus the diameter) will need to be increased by a factor of 2
increased by a factor of 2 to maintain to flow rate.
14. When 2330 J of work is done by an ideal monoatomic gas and 924 J of heat is
removed, the temperature goes from 345 K to 475 K. Find the quantity of gas (in
moles) in the system.
Solution: The internal energy U of a monatomic ideal gas is given by 3
2
U nRT
= , where n is
the number of moles, R is the universal gas constant, and T is the Kelvin temperature.
When the temperature changes to a final value of Tf from an initial value of Ti, the
internal energy changes by an amount
( )
3
f i f i
2
U U nR T T
U
− = −

Solving this equation for the number of moles yields
2
3
U
n
R T

=

.We are given the change
in temperature, but must determine U. The change U in the internal energy of the gas
is related to the heat Q and the work W by the first law of thermodynamics, U = Q − W.
Using these two relations will allow us to find the final temperature of the gas.
Substituting U = Q − W into the expression for the number of moles of gas gives
( ) ( ) ( )
( )( )
2 924 J 2330 J
2
2
3 3 3 8.31 J/(mol K) 457 K 345 K
2.33 mol
Q W
U
n
R T R T
n
 
− −
−
  
= = =
  −
=
Note that the heat is negative (Q = -924 J) since the system (the gas) loses heat, and the
work is positive (W = +2330 J), since it is done by the system.
15. An ideal gas expands at a constant pressure from a volume of 750 L to 2300 L while
doing 91 J of work. What is the pressure during the process?
Solution: W = PV, the pressure of the gas can be found using the work W done by the gas
and the amount the volume changes.
Using the values given in the problem and the fact that 1 L = 1.0 x 10-3
m3
( )
3 3
91 J
59 Pa
2.3 m 0.75 m
W
P
V
+
= = =
 −
16. Three moles of neon expand isothermally to 0.250 m3
from 0.100 m3
, into the gas
flows 4.75 x 103
J of heat. Assuming that neon is an ideal gas, find its temperature.
Solution: When n moles of an ideal gas change quasi statically to a final volume Vf from an
initial volume Vi at a constant temperature T, the work W done is,
f
i f
i
ln or
ln
V W
W nRT T
V V
nR
V
 
= =
 
 
 
 
 
(1)
where R is the universal gas constant. To determine T from Equation (1), we need a
value for the work, which we do not have. However, we do have a value for the heat Q.
To take advantage of this value, the internal energy of an ideal gas is directly

8. Memo/EMH1501/101/03/2021
8
proportional to its Kelvin temperature. Since the temperature is constant (the neon
expands isothermally), the internal energy remains constant. According to the first law of
thermodynamics, the change U in the internal energy is given by U = Q – W. Since
the internal energy U is constant, U = 0, so that W = Q.
Substituting W = Q into the expression for T in Equation (1), we find that the temperature of
the gas during the isothermal expansion is
( ) ( )
3
3
f
3
i
3.25 10 J
142 K
0.250 m
3.00 mol 8.31 J/ mol K ln
ln
0.100 m
Q
T
V
nR
V

= = =
   

 
   
 
 
 
17. Find the change in (a) temperature and (b) pressure if two moles of a monoatomic
ideal gas at a constant volume of 2.54 L have 576 J of heat removed.
Solution: The gas undergoes heating at constant volume. For a constant-volume process
involving an ideal monatomic gas, the amount Q of heat transferred is given by
V
Q C n T
=  , where 3
V 2
C R
= is the molar specific heat capacity at constant volume, R
is the universal gas constant, n is the number of moles of gas, and T is the change in
temperature. The equation can be solved for the change in temperature. Since the
pressure is directly proportional to the temperature, we can use the ideal gas law to show
that for a fixed volume
nR T nR
P T
V V
  
 = = 
 
 
(a) Solving V
Q C n T
=  for the change in temperature we obtain
V
Q
T
nC
 = (1)
Substituting 3
V 2
C R
= into Equation (1) and realizing the quantity of heat was
removed yields
( ) ( ) ( )
( )
3 3
2 2
576 J
14 K
2.0 mol 8.31 J/(mol K)
Q
T
n R
−
 = = = −

(b) Using the ideal gas law and converting L to m3
( )( )
( ) 4
3 3
2.0 mol 8.31 J/(mol K)
14 K 9.2 10 Pa
2.54 10 m
nR
P T
V −

 
 =  = − = − 
 

 
18. An engine receives 680 J of heat from a hot reservoir and gives off 420 J of heat to a
cold reservoir. Calculate (a) the work done and (b) the efficiency of the engine.
Solution: The efficiency of the engine is H
/
e W Q
= , where W is the magnitude of the
work and H
Q is the magnitude of the input heat. In addition, energy conservation
requires that H C
Q W Q
= + (Equation 15.12), where C
Q is the rejected heat.
(a) Solving for W, we can find the magnitude of the rejected heat:
H C H C
or 680 J 420 J 260 J
Q W Q W Q Q
= + = − = − =

9. Memo/EMH1501/101/03/2021
9
(b) The efficiency of the heat pump is:
H
260 J
0.38
680 J
W
e
Q
= = =
19. An engine has a hot-reservoir temperature of 950 K and a cold-reservoir temperature
of 605 K. The engine operates at three-fifths maximum efficiency. What is the
efficiency of the engine?
Solution: The maximum efficiency of the engine is the efficiency that a Carnot engine
would have operating with the same hot and cold reservoirs. Thus, the maximum
efficiency is C
Carnot
H
1
T
e
T
= − (Equation 15.15), where TC and TH are the Kelvin
temperatures of the cold and the hot reservoir, respectively.
Using Equation for eCarnot and recognizing that the engine has an efficiency e that is
three-fifths the maximum or Carnot efficiency, we obtain
3 3 3
C
Carnot
5 5 5
H
605 K
1 1 0.22
950 K
T
e e
T
   
= = − = − =
   
   
 
20. Calculate the change in entropy as a 3.50 kg box slides to a stop from a speed of 2.25
m/s. Assume a temperature of 293 K for all the objects involved.
Solution: The change in entropy S of a system for a process in which heat Q enters or
leaves the system reversibly at a constant temperature T is given by
equation R
( / )
S Q T
 = . In this case, all of the kinetic energy that was present when the
box was moving has been converted into heat, so 2
1
2
Q KE mv
= = . Substituting this
expression into the above equation, we find the change in entropy is
( )( )
( )
2
2 2
2
R
1 1
3.50 kg 2.25 m/s
2 2
( / ) 3.02 10 J/K
293 K
mv
S Q T
T
−
 = = = = 
Since heat is lost from the hot reservoir (inside the house), the change in entropy is
negative: SH = −QH/TH. Since heat is gained by the cold reservoir (the outdoors), the
change in entropy is positive: SC = +QC/TC. Here we are using the symbols QH and QC
to denote the magnitudes of the heats. The change in the entropy of the universe is
C H
universe C H
C H
28 800 J 28 800 J
+ 13.7 J/K
258 K 294 K
Q Q
S S S
T T
 =   = − = − =
In this calculation we have used the fact that TC = 273 − 15 C = 258 K and
TH = 273 + 21 C = 294 K.

10. Memo/EMH1501/101/03/2021
10
Semester 1
Assignment: AS2 # Unique No: 756789
Due date: 12th
Aug 2021
Total marks 100 (20 x 5 = 100)
1. Four identical metallic objects carry the following charges: +0.4, +6.2, -5.7, and -9.4
C. The objects are brought simultaneously into contact, so that each touches the
others. Then they are separated. (a) what is the final charge on each object? (b) How
many electrons (or protons) make up the final charge on each object?
Solution: Total charge = (+0.4 C) + (6.2 C) +(-5.7 C) +(-9.4 C) = -8.5 C.
(a) Charge of each objects (Q) = -8.5 C/4 = -2.1 C.
(b) Number of electron / proton = Q/e = (-2.1 x 10-6
)/1.6 x 10-19
= 1.3 x 1013
electron
2. Two equally positively charged particles are placed 7.6 x 10-2
m apart and released.
One of the particles, with a mass of 2.8 x 10-7
kg, experiences an acceleration of 5.5
m/s2
. Find (a) the amount of charge on each particle and (b) the acceleration of the
other particle if it has a mass of 4.5 x 10-7
kg.
Solution: Let us look at particle one first. From Newton’s second law, we know that F = ma,
where F is force, m is mass, and a is acceleration
Substitute the values we know for particle one, F = ma = 2.8 x 10-7
x 5.5 = 15.4 x 10-7
N
We can now apply this to Coulomb’s law. Coulomb’s law states, F = k (q1)(q2)/r2
; where F is
electrostatic force, k is coulomb’s constant, q1 is the charge of first particle, q2 is the charge of
second particle, and r is the distance between the two charges.
We know the value of F, and so, let us again, substitute the values into this equation.
15.4 x 10-7
= (8.99 x 109
)(q2
) / (7.6 x 10-2
)2
[q1 = q2 =q]
q = [(15.4 x 7.6 x 7.6 / 8.99) x10-7
x10-4
]1/2
= 9.9 x 10-19
C
(b) The mass of the second particle can be calculated by using Newton’s second law again.
Since the force is the same for both particles, we can isolate for mass.
F = ma; a = 15.4 x 10-7
/ 4.5 x 10-7
m/s2
= 3.4 m/s2
.
3. (a) Calculate the magnitude of the acceleration experienced by an electron that is in an
electric field with a strength of 825 N/C.
(b) A proton and an electron are moving due east in a constant electric field that also
points due east. The electric field has a magnitude of 8.0 x 104
N/C. Determine the
magnitude of the acceleration of the proton and the electron.
Solution: (a) we know the electric field, E = F/q. For an electron the force, Fe = meae.

11. Memo/EMH1501/101/03/2021
11
So, in an electric field the electron experienced a force in an electric field, E is
E = me.ae / e
825 = 9.1 x 10-31
kg x ae /1.6 x 10-19
C
[me = 9.1 x 10-31
kg and charge of electron =1.6 x 10-19
C]
ae = 145 x 1012
m/s2
.
(b) For electron, ae = E. e / me = (8 x 104
N/C x 1.6 x 10-19
C) / 9.1 x 10-31
kg
= 1.4 x 1016
m/s2
.
For proton, ap = E. e / mp = (8 x 104
N/C x 1.6 x 10-19
C) / 1.67 x 10-27
kg
= 7.7 x 1012
m/s2
.
4. A proton and an electron, starting from rest, are accelerated through an electric
potential of the same magnitude. In the process, the electron acquires a speed ve, while
the proton acquires a speed vp. Find the ratio ve/vp.
Solution: The proton accelerates from point A to point B. According to energy conservation
principle only kinetic and electric potential energies. We have
KEP1B + EPEP1B = KEP1A + EPEP1A. Or, KEPB = EPEP1A- EPEP1B
Final total energy Initial total energy
of proton at point B. of proton at point A
The initial kinetic energy is zero since proton starts from rest.
The electric potential energy of a charge q0 is EPE = q0v1
Where v is potential experienced by charge.
Final kinetic energy of proton as follows
KEP1B = EPEP1A - EPEP1B = qp (VA-VB) --- (1)
But an electron has a negative charge that is equal in magnitude to charge on a proton,
so qe = (-qp)
The kinetic energy of electron as
KEe1A = EPEe1B – EPEe1A = -qp(VB-VA) = qp(VA-VB) --- (2)
Comparing equation (1) and (2) we obtain
KEP1B = KEe1A, or, ½ mpVp
2
= ½ meVe
2
Solving for ratio, Ve/Vp = (mp/me)1/2
= (1.67 x 10-27
kg / 9.11 x 10-31
kg)1/2
= 42.8.
5. A charge of +125 C is fixed at the corner of a square that is 0.75 m on a side. How
much work is done by the electric force as a charge of +7.0 C is moved from one
corner of the square to any other empty corner? Explain.
Solution:
Since this a square the distance from the center of the square to any of the four corners is the
same. Therefore the electric potential at each of the four corners is the same. So moving the
charge from one corner to any other corner results in the same electrical potential
energy. Therefore the work done is zero!
But

12. Memo/EMH1501/101/03/2021
12
6. Calculate the final speed of a free electron accelerated from rest through a potential
difference of 100 V.
Solution: We have a system with only conservative forces. Assuming the electron is
accelerated in a vacuum, and neglecting the gravitational force (we will check on this
assumption later), all of the electrical potential energy is converted into kinetic energy. We
can identify the initial and final forms of energy to be KEi = 0, KEf = ½(mv2
), PEi=qV, and
PEf =0.
Conservation of energy states that KEi + PE i = KE f + PE f .
Entering the forms identified above, we obtain qV = mv2
/2.
We solve this for v:
V = [(2qV)/m]1/2
Entering values for q, V, and m gives
v=[2(−1.60×10−19
C)(−100 J/C) / 9.11×10−31kg]1/2
=5.93×106
m/s
7. A resistor is connected across the terminals of a 9.0-V battery, which delivers 9.8 x
104
J of energy to the resistor in 6.5 hours. What is the resistance of the resistor?
Solution: According to Ohm’s law, the resistance is the voltage of the battery divided by the
current that the battery delivers. The current is the charge divided by the time during
which it flows. We know the time but are not given the charge directly. However, we can
determine the charge from the energy delivered to the resistor, because this energy comes
from the battery, and the potential difference between the battery terminals is the
difference in electric potential energy per unit charge, according to Equation 19.4. Thus, a
9.0-V battery supplies 9.0 J of energy to each coulomb of charge passing through it. To
calculate the charge, then, we need only divide the energy from the battery by the 9.0 V
potential difference.
Ohm’s law indicates that the resistance R is the voltage V of the battery divided by the
current I, or R = V/I. According to Equation 20.1, the current I is the amount of charge q
divided by the time t, or I = q/t. Using these two equations, we have
/
V V V t
R
I q t q

= = =
  
The potential difference V is the difference (EPE) in the electric potential energy
divided by the charge q, or
( )
EPE
V
q

 =

. However, it is customary to denote the
potential difference across a battery by V, rather than V, so
( )
EPE
V
q

=

. Solving this
expression for the charge gives
( )
EPE
q
V

 = . Using this result in the expression for the
resistance, we find that
( ) ( )
( ) ( )
2
2
4
9.0 V 6.5 3600 s
19
EPE / EPE 9.8 10 J
V t V t V t
R
q V

  
= = = = = 
   

13. Memo/EMH1501/101/03/2021
13
8. An incandescent light bulb has resistance of about 11  when it is operating. Find the
operating temperature if the average temperature coefficient of resistivity is 0.0060
(Co
)-1
.
Solution: The resistance as a function of temperature as follows:
( )
0 0
1
R R T T

 
= + −
 
where α is the temperature coefficient of resistivity. Rearranging for the temperature T
gives
0
0
1
R
R
T T

 
−
 
 
 
= + (1)
Applying Equation (1) with the values given in the problem gives
0 3
0 1
130
1 1
11
20 C 1.8 10 C
0.0060 C
R
R
T T
 −
  
 
−
  −
   

   
= + = + = 
9. A piece of Nichrome wire has radius of 6.5 x 10-4
m. It is used in a laboratory to make
a heater that uses 4.00 x 102
W of power when connected to a voltage source of 120
V. Ignoring the effect of temperature on resistance, estimate the necessary length of
wire.
Solution: We know that the resistance of the wire can be obtained from
P = V
2
/R or R = V
2
/P
We also know that R = L/A. Solving for the length, noting that A =  r
2
, and using
 = 100  10
–8
.m, we find
( )( ) ( ) ( )
( )( )
2
2
2 2 –4
2 2
–8 2
/ 120 V 6.5 10 m
50 m
100 10 m 4.00 10 W
V P r
RA V r
L
P
 

  

= = = = =
  

10. The rms current in a copy machine is 6.00 A, and the resistance of the machine is 22.5
. What are (a) the average power and (b) the peak power delivered to the machine?
Solution: (a) The average power P delivered to the copy machine is equal to the square of
the rms-current Irms times the resistance R, or 2
rms
P I R
= . Both Irms and R are known.
The average power is
( ) ( )
2
2
rms 6.00 A 22.5 810 W
P I R
= =  = (20.15b)
(b) The peak power peak
P is twice the average power, or peak 2
P P
= .

14. Memo/EMH1501/101/03/2021
14
The peak power is twice the average power, so
( )
peak 2 2 810 W 1620 W
P P
= = =
11. The current in a series circuit is 15.0 A. When an additional 6.00-  resistor is
inserted in series, the current drops to 10.0 A. What is the resistance in the original
circuit?
Solution: Using Ohm's law, we can write an expression for the voltage across the original
circuit as 0 0
V I R
= . When the additional resistor R is inserted in series, assuming that the
battery remains the same, the voltage across the new combination is given by 0
( )
V I R R
= + .
Since V is the same in both cases, we can write 0 0 0
( )
I R I R R
= + . This expression can be
solved for 0
R Solving for 0
R , we have
0 0 0 0 0
– or ( – )
I R IR IR R I I IR
= =
Therefore, we find that
0
0
(10.0 A)(6.00 )
12.0 A
– (15.0 A -10.0 A)
IR
R
I I

= = =
12. Two resistors have resistance R1 and R2. When the resistors are connected in series to
a 12.0-V battery, the current from the battery is 2.00 A. When the resistors are
connected in parallel to the battery, the total current from the battery is 10.0 A.
Determine R1 and R2.
Solution: The series combination has an equivalent resistance of S 1 2
R R R
= + . The parallel
combination has an equivalent resistance that can be determined from 1 1 1
P 1 2
R R R
− − −
= + .
In each case the equivalent resistance can be used in Ohm’s law with the given voltage
and current. Thus, we can obtain two equations that each contain the unknown
resistances. These equations will be solved simultaneously to obtain R1 and R2.
For the series case, Ohm’s law is ( )
S 1 2
V I R R
= + . Solving for sum of the resistances, we
have
1 2
S
12.0 V
6.00
2.00 A
V
R R
I
+ = = =  (1)
For the parallel case, Ohm’s law is P P
V I R
= , where 1 1 1
P 1 2
R R R
− − −
= + . Thus, we have
1
P
P 1 2
1 1 1 10.00 A
0.833
12.0 V
I
R R R V
−
= + = = =  (2)
Solving Equation (1) for R2 and substituting the result into Equation (2) gives
2
1 1
1 1
1 1
0.833 or 6.00 7.20 0
6.00
R R
R R
+ = − + =
−
In this result we have suppressed the units in the interest of clarity. Solving the quadratic
equation (see Appendix C.4 for the quadratic formula) gives

15. Memo/EMH1501/101/03/2021
15
( ) ( ) ( )( )
( )
2
1
6.00 6.00 4 1.00 7.20 6.00 36.0 28.8
4.34 or 1.66
2 1.00 2.00
R
− −  − −  −
= = =
Substituting these values for R1 into Equation (1) reveals that
2 1
6.00 1.66 or 4.34
R R
= − =
Thus, the values for the two resistances are 1.66 Ω and 4.34 Ω .
13. Determine the equivalent resistance between the points A and B for the group of
resistors in the drawing.
Solution: When two or more resistors are in series, the equivalent resistance is given by
Rs = R1 + R2 + R3 + . . . . Likewise, when resistors are in parallel, the expression to be solved
to find the equivalent resistance is given by
1
Rp
=
1
R1
+
1
R2
+
1
R3
+....
Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the
combination of those two resistors is 10.0 . The 9.0- and 8.0- resistors are in parallel;
their equivalent resistance is 4.24 . The equivalent resistances of the parallel combination
(9.0  and 8.0 ) and the series combination (4.0  and the 6.0 ) are in parallel; therefore,
their equivalent resistance is 2.98 . The 2.98- combination is in series with the 3.0-
resistor, so that equivalent resistance is 5.98 . Finally, the 5.98- combination and the
20.0- resistor are in parallel, so the equivalent resistance between the points A and B is
4 6
.  .
14. A battery delivering a current of 55.0 A to a circuit has terminal voltage of 23.4 V.
The electric power being dissipated by the internal resistance of the battery is 34.4 V.
Find the emf of the battery.
Solution: The voltage V across the terminals of a battery is equal to the emf of the battery
minus the voltage Vr across the internal resistance of the battery: V = Emf − Vr.
Therefore, we have that
r
Emf V V
= + (1)
The power P being dissipated by the internal resistance is equal to the product of the
voltage Vr across the internal resistance and the current I: r
P IV
= . Therefore, we can
express the voltage Vr across the internal resistance as
r
P
V
I
= (2)

16. Memo/EMH1501/101/03/2021
16
Substituting Equation (2) into Equation (1), we obtain
r
34.0 W
Emf 23.4 V 24.0 V
55.0 A
P
V V V
I
= + = + = + =
15. Find the magnitude and direction of the current in the 2.0- resistor in the drawing.
Solution: Label the currents with the resistor values. Take I3 to the right, I2 to the left and I1
to the right. Applying the loop rule to the top loop (suppressing the units) gives
I1 + 2.0 I2 = 1.0 (1)
and to the bottom loop gives
2.0 I2 + 3.0 I3 = 5.0 (2)
Applying the junction rule to the left junction gives
I2 = I1 + I3 (3)
Solving Equations (1), (2) and (3) simultaneously, we find I2 = 0.73 A .
The positive sign shows that the assumed direction is correct. That is, to the left .
16. A galvanometer with a coil resistance of 9.50  is used with a shunt resistor to make
a nondigital ammeter that has an equivalent resistance of 0.40 . The current in the
shunt resistor is 3.00 mA when the galvanometer reads full scale. Find the full-scale
current of the galvanometer.
Solution: The drawing at the right shows the galvanometer (G), the coil resistance RC
and the shunt resistance R. The full-scale current IG through the galvanometer and
the current IS in the shunt resistor are also shown. Note that RC and R are in parallel,
so that the voltage across each of them is the same. Our solution is based on this fact.
The equivalent resistance of the ammeter is the parallel equivalent resistance of RC
and R.
59.9 mA
0.100 mA
60.0 mA
G
Shunt
resistor R
Rc = 50.0 
A B IG
RC
= 9.50 
IS
= 3.00 mA

17. Memo/EMH1501/101/03/2021
17
Expressing voltage as the product of current and resistance and noting that the voltages across
RC and R are the same, we have
G C S
Voltage across Voltage across
coil resistance shunt resistance
I R I R
=
Solving this equation for the full-scale current IG through the galvanometer gives
S
G
C
I R
I
R
=
In this result IS and RC are given, but the shunt resistance R is unknown. However, the
equivalent resistance is given as P 0.40
R = , and it is the parallel equivalent resistance
of RC and R:
P C
1 1 1
R R R
= + . This equation can be solved for R as follows:
C P P C
P C P C P C C P
1 1 1 1 1 1
or = or
R R R R
R
R R R R R R R R R R
−
= + = − =
−
Substituting this result for R into the expression for IG gives
( ) ( ) ( )
S S P C P
G S
C C C P C P
3 4
0.40
3.00 10 A 1.3 10 A
9.50 0.40
I R I R R R
I I
R R R R R R
− −
   
= = =
   
   
− −
   
 

=  = 
 
 − 
 
 
17. Two capacitors are connected to a battery. The battery voltage is V = 60.0 V, and the
capacitances are C1 = 2.00 mF and C2 = 4.00 mF. Determine the total energy stored
by the two capacitors when they are wired (a) in parallel and (b) in series.
Solution: When capacitors are connected in parallel, each receives the entire voltage V of the
battery. Thus, the total energy stored in the two capacitors is 2 2
1 1
2 2
1 2
CV C V
+ . When
the capacitors are connected in series, the sum of the voltages across each capacitor
equals the battery voltage: 1 2
V V V
+ = . Thus, the voltage across each capacitor is series
is less than the battery voltage, so the total energy, 2 2
1 1
2 2
1 1 2 2
CV C V
+ , is less than when
the capacitors are wired in parallel.
(a) The voltage across each capacitor is the battery voltage, or 60.0 V. The energy stored
in both capacitors is
( )
( )( )
2 2 2
1 1 1
2 2 2
1 2 1 2
2
6 6 2
1
2
Total energy
2.00 10 F + 4.00 10 F 60.0 V 1.08 10 J
C V C V C C V
− − −
= + = +
=   = 
(b) According to the discussion in Section 20.12, the total energy stored by capacitors in
series is 2
1
2 S
Total energy = C V , where CS is the equivalent capacitance of the series
combination:

18. Memo/EMH1501/101/03/2021
18
6 6
S 1 2
1 1 1 1 1
2.00 10 F 4.00 10 F
C C C − −
= + = +
 
(20.19)
Solving this equation yields CS = 1.33  10
−6
F. The total energy is
( )( )2
6 3
1
2
Total energy 1.33 10 F 60.0 V 2.39 10 J
− −
=  = 
18. A 5.00-F and a 7.00 F capacitor are connected in series across a 30.0-V battery.
A9.00- F capacitor is then connected in parallel across the 3.00- F capacitor.
Determine the voltage across the 9.00- F capacitor.
Solution: The 9.00 and 5.00-F capacitors are in parallel. According to Equation 20.18, the
equivalent capacitance of the two is 9.00 F + 5.00 F = 14.0 F. This 14.0-F capacitance
is in series with the 7.00-F capacitance. According to Equation 20.19, the equivalent
capacitance of the complete arrangement can be obtained as follows:
( )
( )
–1
–1
1 1 1 1
= 0.214 F or 4.67 F
14.0 F 7.00 F 0.214 F
C
C
 
  
= + = =
The battery separates an amount of charge
Q = CV = (4.67  10
–6
F) (30.0 V) = 1.4  10
–4
C
This amount of charge resides on the 7.00 µF capacitor, so its voltage is
V7 = (1.4  10
–4
C)/(7.00  10
–6
F) = 20.0 V
The loop rule gives the voltage across the 5.00 µF capacitor to be
V5 = 30.0 V – 20.0 V = 10.0 V
This is also the voltage across the 9.00 µF capacitor, since it is in parallel, so V9 = 10.0 V .
19. Four identical capacitors are connected with a resistor in two different ways. When
they are connected as in part a of the drawing, the time constant to charge up this
circuit is 0.72 s. What is the time constant when they are connected with the same
resistor, as in part b?
Solution: In either part of the drawing the time constant τ of the circuit is eq
RC
 = ,
according to Equation 20.21, where R is the resistance and Ceq is the equivalent
capacitance of the capacitor combination. We will apply this equation to both circuits. To
obtain the equivalent capacitance, we will analyze the capacitor combination in parts. For

19. Memo/EMH1501/101/03/2021
19
the parallel capacitors P 1 2 3 ...
C C C C
= + + + applies (Equation 20.18), while for the
series capacitors 1 1 1 1
S 1 2 3 ...
C C C C
− − − −
= + + + .
we write the time constant of each circuit as follows:
a eq, a b eq, b
and
RC RC
 
= =
Dividing these two equations allows us to eliminate the unknown resistance algebraically:
eq, b eq, b
b
b a
a eq, a eq, a
or
RC C
RC C

 

 
 
= =
 
 
(1)
To obtain the equivalent capacitance in part a of the drawing, we note that the two
capacitors in series in each branch of the parallel combination have an equivalent
capacitance CS that can be determined using Equation
1
S 2
S
1 1 1
or C C
C C C
= + = (2)
Using Equation, we find that the parallel combination in part a of the drawing has an
equivalent capacitance of
1 1
eq, a 2 2
C C C C
= + = (3)
To obtain the equivalent capacitance in part b of the drawing, we note that the two
capacitors in series have an equivalent capacitance of 1
2
C , according to Equation (2).
The two capacitors in parallel have an equivalent capacitance of 2C. Finally, then, we
have a series combination of 1
2
C and 2C,
2
eq, b 5
1
eq, b 2
1 1 1 5
or
2 2
C C
C C C C
= + = = (4)
Using Equations (3) and (4) in Equation (1), we find that
( )
2
eq, b 5
b a
eq, a
0.72 s 0.29 s
C C
C C
 
 
 
= = =
 
 
20. An 86- resistor and a 67- resistor are connected in series across a battery. The
voltage across the 86- resistor is 72 V. What is the voltage across the 67- resistor?
Solution: Ohm’s law relates the resistance R of either resistor to the current I in it and the
voltage V across it:
V
R
I
= (20.2)
Because the two resistors are in series, they must have the same current I. We will,
therefore, apply Equation 20.2 to the 86- resistor to determine the current I. Following
(b)
R C C
C
C
+ −
(a)
C
C
C C
R
+ −

20. Memo/EMH1501/101/03/2021
20
that, we will use Equation 20.2 again, to obtain the potential difference across the 67−
resistor.
Let R1 = 86  be the resistance of the first resistor, which has a potential difference of
V1 = 27 V across it. The current I in this resistor is
1
1
V
I
R
= (1)
Let R2 = 67  be the resistance of the second resistor. Again employing Equation 20.2,
the potential difference V2 across this resistor is given by
2 2
V IR
= (2)
Since the current in both resistors is the same, substituting Equation (1) into Equation (2)
yields
( )
( )
( )
1
2 2 2
1
27 V
67 21 V
86
V
V IR R
R
 
= = =  =
 
  
 
Unisa ©2021

21. Memo/EMH1501/101/03/2021
21
Semester 2
Assignment: AS1 # Unique No: 756789
Due date: 12th May 2021
Total marks 100 (20 x 5 = 100)
1. There is one temperature at which the Fahrenheit and Celsius scales have the same
reading. Find the temperature.
Solution: Formula for Fahrenheit, F = 9/5 C + 32
Formula for Centigrade, C =5/9 (F-32)
Let F = C or C= F
F= C= 5/9(F-32)= (5F-160)/9
Or, 9F= 5F - 160
Or, 9F - 5F = -160
Or, 4F = -160
Or, F = -160/4
Or, F= -40 degree answer in Fahrenheit
Or, C= F= 9/5 C + 32
Or, C = (9C + 160)/5
Or, 5C = 9C + 160
Or, 5C - 9C = 160
Or, -4C= 160
Or, C=160/-4
Or, C= -40 degree answer in Centigrade
Conclusion: The temperature wherein degrees centigrade is the same in degrees Fahrenheit is
-40 degrees .

22. Memo/EMH1501/101/03/2021
22
2. A gold engagement ring has an inner diameter of 1.5 x 10-2
m and a temperature of 27
o
C. The ring falls into a sink of hot water whose temperature is 49 o
C. What is the
change in the diameter of the hole in the ring? [ = 14 x 10-6
(o
C)-1
is the coefficient
of linear expansion for gold].
Solution: The hole expands as it were filled with gold, so the change in the diameter is
given by L = L0T, where  = 14 x 10-6
(o
C)-1
is the coefficient of linear expansion for
gold, L0 is the original diameter and T is the change in temperature.
The change in the ring’s diameter is L = L0T
= [14 x 10-6
(o
C)-1
] (1.5 x 10-2
m) (49 o
C – 27 o
C) = 4.6 x 10-6
m.
3. A copper kettle contains water at 15 o
C. When the water is heated to its boiling point
of 100.0 o
C, the volume of the kettle expands by 1.2 x 10-5
m3
. Determine the volume
of the kettle at 15 o
C [ = 51 x 10-6
(o
C)-1
is the coefficient of volume expansion for
copper].
Solution: The volume expands, so the change in volume is given by V = V0T, where
 = 51x 10-6
(o
C)-1
is the coefficient of volume expansion for copper, V0 is the original
volume and T is the change in temperature.
The change in volume is V = V0T,
V0 = V/(T) = (1.2 x 10-5
m3
) / 51 x 10-6
(o
C)-1
x (100 o
C - 15 o
C) = 2.8 x 10-3
m3
.
4. A 7.5 g lead sphere is fired into a target at a speed of 240 m/s. If 75% of the kinetic
energy goes into heating the bullet as it is brought to rest, find the increase in
temperature of the sphere. [Specific heat capacity of lead = 128 J/(Kg.o
C)]
Solution: Kinetic energy of the lead = ½ (7.5 x 10-3
kg) (240 m/s)2
.
75% of kinetic energy = 0.75 x ½ x 7.5 x 10-3
x 240 x 240 kg. m2
s-2
= 162000 J = Q
Increase of temperature of the sphere, T= Q/cm
= (0.75 x ½ x 7.5 x 10-3
x 240 x 240) / (128 x 7.5 x 10-3
)
= (0.75 x 120 x 240) / 128 = 168.75 0
C
5. (a) Objects A and B have the same mass of 3.0 Kg. They melt when 3.0 x 104
J of heat
is added to object A and when 9.0 x 104
J is added to object B. Determine the latent
heat of fusion for the substance from which each object is made. (b) Find the heat
required to melt object A when its mass is 6.0 kg.
Solution: (a) The object A is melted when 3.0 x 104
J of heat is added i.e. LA = QA/m
LA = (3.0 x 104
J) / 3.0 kg
LA = 1.0 x 104
J/kg.

23. Memo/EMH1501/101/03/2021
23
The object B is melted when 9.0 x 104
J of heat is added i.e. LB = QB/m
LB = (9.0 x 104
J) / 3.0 kg
LB = 3.0 x 104
J/kg.
(b) Q = mA LA = 6.0 kg x 1.0 x 104
J/kg = 6.0 x 104
J.
6. A closed box is filled with dry ice at a temperature of -78.5 o
C, while the outside
temperature is 21.0 o
C. The box is cubical, measuring 0.35 m on a side, and the
thickness of the walls is 2.75 x 10-2
m. In one day, 3.10 x 106
J of heat is conducted
through the six walls. Find the thermal conductivity of the material from which the
box is made.
Solution: The energy Q conducted through one face (thickness L and surface area A) of
the cubical box in a time t is
( )
kA T t
Q
L

= , where ΔT is the temperature difference
between the outside and inside surface of the box and k is the thermal conductivity of the
material from which the box is made. With the aid of this equation, we can determine k.
Since the cube has six faces, the total heat conducted through all six faces is total 6
Q Q
= .
Using equation for Q, we have
total
6( )
kA T t
Q
L

=
Noting that the number of seconds in one day is ( ) 4
3600 s
24 h 8.64 10 s
1 h
 
= 
 
 
and
solving for the thermal conductivity k, we find that
( ) ( )
total
6 2
2
2 4
6
(3.10 10 J)(2.75 10 m)
1.35 10 J / (s m C )
6(0.350 m) 21.0 C 78.5 C 8.64 10 s
Q L
k
A Tt
−
−
=

 
= =    
 − −  
 
 
7. Two identically sized objects, one at 25 o
C with emissivity e1 and the other at 75 o
C
with emissivity e2, radiate the same power. Find the emissivity ratio e1/ e2.
Solution: The radiant energy Q radiated by an object is given by 4
Q e T At

= , where e is
the emissivity,  is the Stefan-Boltzmann constant, T is its temperature (in Kelvins), A is
the surface area of the object, and t is the time. The radiant energy emitted per second, or
power, is Q/t = 4
.
e T A
 Since both objects radiate the same power, we can set the
equation for each of them equal to one another and solve for the ratio 1 2
/
e e .
Setting the two equations equal to one another with appropriate subscripts and using the
fact that the objects are identical in size gives

24. Memo/EMH1501/101/03/2021
24
1 2
1
Q Q
t t
e 
=
4
1
T A 2
e 
= 4
2
T A
( )
( )
4
4
1 2
4 4
2 1
75 C
81
25 C
e T
e T
= = =
8. A solar collector is placed in direct sunlight where it absorbs energy at the rate of 880
J/s for each square meter of its surface. The emissivity of the solar collector is e =
0.78. What equilibrium temperature does the collector reach? Assume that the only
energy loss is due to the emission of radiation.
Solution: At equilibrium the temperature of the solar collector is not changing. Since the only
energy loss is due to the emission of radiation, we conclude that the temperature is not
changing because the collector is emitting the same amount of energy as it is absorbing.
Thus, it is emitting 880 J/s for each 1.0 m2
of its surface. The Stefan-Boltzmann law of
radiation specifies that the energy Q radiated by an object in a time t is 4
Q e T At

=
(Equation 13.2), where e is the object’s emissivity, –8 2 4
5.67 10 J / (s m K )
 =    is the
Stefan-Boltzmann constant, T is the Kelvin temperature, and A is the object’s surface
area. We can rearrange this equation as follows to give the energy per second Q/t:
4
Q
e T A
t

= (1)
This equation can be solved for the temperature T.
Solving Equation (1) for the temperature T, we find that
1/4
1/4
–8 2 4 2
/ 880 J/s
376 K
0.78[5.67 10 J / (s m K )](1.0 m )
Q t
T
e A

 
 
= = =
 
 
  
   
9. A certain element has a mass per mole of 196.967 g/mol. What is the mass of a single
atom in (i) atomic mass unit (amu) and kilograms? (c) How many moles of atoms are
in a 285-g sample?
Solution: The mass of one of its atoms (in atomic mass units) has the same numerical value
as the element’s mass per mole (in units of g/mol). Atomic mass units can be converted
into kilograms using the fact that 1 u = 1.6605  10–27
kg. Dividing the mass of the
sample by the mass per mole gives the number of moles of atoms in the sample.
(d) The mass per mole is 196.967 g/mol. Since the mass of one of its atoms (in atomic
mass units) has the same numerical value as the mass per mole, the mass of a single
atom is
m = 196.967 u .
(e) To convert the mass from atomic mass units to kilograms, we use the conversion
factor of 1 u = 1.6605  10–27
kg:

25. Memo/EMH1501/101/03/2021
25
( )
27
25
1.6605 10 kg
196.967 u 3.2706 10 kg
1 u
m
−
−
 

= = 
 
 
(f) The number n of moles of atoms is equal to the mass m divided by the mass per mole:
300 g
1.52 mol
Mass per mole 196.967 g/mol
m
n = = =
10. Find the number of moles of air there are in a FIFA World cup ball if it has an inner
diameter of 22.0 cm and is inflated to a pressure of 75.2 kPa at a temperature of 293
K.
Solution: We can rearrange the ideal gas law, /
P nRT V
= , where n is the number of moles of
gas, R is the universal gas constant, T is the Kelvin temperature, and V is the volume to
find n. The variables R and T, area known and V, can be calculated using the volume of a
sphere 3
4
3
V r

= , where r will be half the diameter given in the problem.
Rearranging the ideal gas law, we have
nRT PV
P n
V RT
= → =
Substituting in the volume of a sphere and solving we find the number of moles to be equal to
( )
( )
 ( )
3
3
3
4 0.22 m
4 75.2 10 Pa
3 2
3
0.18 mol
8.31 J/ mol K 273 K
P r
PV
n
RT RT


 
 
    
 
   
 
   
= = = =

11. Two ideal gases have the same mass density and the same absolute pressure. One of
the gases is helium (He), and its temperature is 180 K. The other gas is neon (Ne).
What is the temperature of the neon?
Solution: The ideal gas law is PV nRT
= . We need to put this in terms of the mass density
/
m V
 = . We can then set He Ne
P P
= , and solve the resulting expression for TNe, the
temperature of the neon.
We begin by writing Equation 14.1 in terms of the mass density  for an ideal gas.
Recall that the number n of moles of a substance is equal to its mass m in grams, divided
by its mass per mole M. Therefore, /
PV mRT M
= , and we have
( )
/ / /
P m V RT M RT M

= = , where  is the same for each gas. The two gases have
the same absolute pressures, so that He Ne
P P
= , and it follows that
He Ne
He Ne
RT RT
M M
 
=
The term R can be eliminated algebraically from this result. Solving for the
temperature of the neon TNe and using the mass per mole for helium (4.0026 g/mol) and
neon (20.180 g/mol) from the periodic table on the inside of the back cover of the text,
we find

26. Memo/EMH1501/101/03/2021
26
Ne
Ne He
He
20.180 g/mol
(175 K) 882 K
4.0026 g/mol
M
T T
M
   
= = =
   
 
 
12. Calculate the rms speed of an electron near the Sun’s surface if the temperature and
pressure are 2.01 x 106
K and 0.295 Pa, respectively. Assume the electrons behave
like an ideal gas.
Solution: The electrons will have an average translational kinetic energy
based on the temperature of the surface of the sun, namely, 2 3
1
rms
2 2
mv kT
= . In this
expression m is the mass of an electron, vrms is the rms speed of an electron, k is
Boltzmann’s constant, and T is the Kelvin temperature. The pressure is not needed
for the calculation.
Solving equation the rms
v of an electron, we find
( )( )
31
23 6
6
rms
3 1.38 10 J/K 2.01 10
9.11 10
K
3
9.6 10 m/s
kg
kT
v
m −
−
= = 

 
=
13. A gas is diffusing from a container through a 2.5 cm long tube. Once the higher
concentration has been reduced to ½ its original value, calculate the change in
diameter of tubing that would be needed to maintain the original diffusion rate,
assuming the gas is diffusing to open air (essentially none of this gas present).
Solution: The diffusion of the gas is driven by the concentration difference ΔC = C2 – C1
between the ends of the tube. Since we are assuming diffusion to open air, C1 = 0, so
ΔC = C2. As stated in the problem we will be addressing the situation when the higher
concentration has been reduced to ½ its original value, or C’
= 0.5C2 and since the gas is
still diffusing to open air, ΔC’
= 0.5C2. The concentration difference is related to the
mass rate of diffusion by Fick’s law:
m DA C
t L

= , where D is the diffusion constant, and
A and L are, respectively, the cross-sectional area and length of the tube. If the mass
diffusion rate is to stay constant, then we can set to two situations equal to one another
' ' '
'
DA C D A C
L L
 
=
Noting that the length of the tubing will remain the same as well as the diffusion constant we
find that
' '
A C A C
 = 
Meaning the cross-sectional area of the tubing will be inversely proportional to the change in
the concentration difference. Knowing we are using a circular tube we can replace the area
formula with the area of a circle to arrive at

27. Memo/EMH1501/101/03/2021
27
2 2
' '
r C r C
 

 = 
2
r C 
 = 2
2
2
2
' '
'
'
r C
r C
r C
r
C


= =
 2
0.5C
2r
=
So the radius (and thus the diameter) will need to be increased by a factor of 2
increased by a factor of 2 to maintain to flow rate.
14. When 2330 J of work is done by an ideal monoatomic gas and 924 J of heat is
removed, the temperature goes from 345 K to 475 K. Find the quantity of gas (in
moles) in the system.
Solution: The internal energy U of a monatomic ideal gas is given by 3
2
U nRT
= , where n is
the number of moles, R is the universal gas constant, and T is the Kelvin temperature.
When the temperature changes to a final value of Tf from an initial value of Ti, the
internal energy changes by an amount
( )
3
f i f i
2
U U nR T T
U
− = −

Solving this equation for the number of moles yields
2
3
U
n
R T

=

.We are given the change
in temperature, but must determine U. The change U in the internal energy of the gas
is related to the heat Q and the work W by the first law of thermodynamics, U = Q − W.
Using these two relations will allow us to find the final temperature of the gas.
Substituting U = Q − W into the expression for the number of moles of gas gives
( ) ( ) ( )
( )( )
2 924 J 2330 J
2
2
3 3 3 8.31 J/(mol K) 457 K 345 K
2.33 mol
Q W
U
n
R T R T
n
 
− −
−
  
= = =
  −
=
Note that the heat is negative (Q = -924 J) since the system (the gas) loses heat, and the
work is positive (W = +2330 J), since it is done by the system.
15. An ideal gas expands at a constant pressure from a volume of 750 L to 2300 L while
doing 91 J of work. What is the pressure during the process?
Solution: W = PV, the pressure of the gas can be found using the work W done by the gas
and the amount the volume changes.
Using the values given in the problem and the fact that 1 L = 1.0 x 10-3
m3
( )
3 3
91 J
59 Pa
2.3 m 0.75 m
W
P
V
+
= = =
 −
16. Three moles of neon expand isothermally to 0.250 m3
from 0.100 m3
, into the gas
flows 4.75 x 103
J of heat. Assuming that neon is an ideal gas, find its temperature.
Solution: When n moles of an ideal gas change quasi statically to a final volume Vf from an
initial volume Vi at a constant temperature T, the work W done is,

28. Memo/EMH1501/101/03/2021
28
f
i f
i
ln or
ln
V W
W nRT T
V V
nR
V
 
= =
 
 
 
 
 
(1)
where R is the universal gas constant. To determine T from Equation (1), we need a
value for the work, which we do not have. However, we do have a value for the heat Q.
To take advantage of this value, the internal energy of an ideal gas is directly
proportional to its Kelvin temperature. Since the temperature is constant (the neon
expands isothermally), the internal energy remains constant. According to the first law of
thermodynamics, the change U in the internal energy is given by U = Q – W. Since
the internal energy U is constant, U = 0, so that W = Q.
Substituting W = Q into the expression for T in Equation (1), we find that the temperature of
the gas during the isothermal expansion is
( ) ( )
3
3
f
3
i
3.25 10 J
142 K
0.250 m
3.00 mol 8.31 J/ mol K ln
ln
0.100 m
Q
T
V
nR
V

= = =
   

 
   
 
 
 
17. Find the change in (a) temperature and (b) pressure if two moles of a monoatomic
ideal gas at a constant volume of 2.54 L have 576 J of heat removed.
Solution: The gas undergoes heating at constant volume. For a constant-volume process
involving an ideal monatomic gas, the amount Q of heat transferred is given by
V
Q C n T
=  , where 3
V 2
C R
= is the molar specific heat capacity at constant volume, R
is the universal gas constant, n is the number of moles of gas, and T is the change in
temperature. The equation can be solved for the change in temperature. Since the
pressure is directly proportional to the temperature, we can use the ideal gas law to show
that for a fixed volume
nR T nR
P T
V V
  
 = = 
 
 
(a) Solving V
Q C n T
=  for the change in temperature we obtain
V
Q
T
nC
 = (1)
Substituting 3
V 2
C R
= into Equation (1) and realizing the quantity of heat was
removed yields
( ) ( ) ( )
( )
3 3
2 2
576 J
14 K
2.0 mol 8.31 J/(mol K)
Q
T
n R
−
 = = = −

(b) Using the ideal gas law and converting L to m3
( )( )
( ) 4
3 3
2.0 mol 8.31 J/(mol K)
14 K 9.2 10 Pa
2.54 10 m
nR
P T
V −

 
 =  = − = − 
 

 
18. An engine receives 680 J of heat from a hot reservoir and gives off 420 J of heat to a
cold reservoir. Calculate (a) the work done and (b) the efficiency of the engine.
Solution: The efficiency of the engine is H
/
e W Q
= , where W is the magnitude of the
work and H
Q is the magnitude of the input heat. In addition, energy conservation
requires that H C
Q W Q
= + (Equation 15.12), where C
Q is the rejected heat.

29. Memo/EMH1501/101/03/2021
29
(a) Solving for W, we can find the magnitude of the rejected heat:
H C H C
or 680 J 420 J 260 J
Q W Q W Q Q
= + = − = − =
(b) The efficiency of the heat pump is:
H
260 J
0.38
680 J
W
e
Q
= = =
19. An engine has a hot-reservoir temperature of 950 K and a cold-reservoir temperature
of 605 K. The engine operates at three-fifths maximum efficiency. What is the
efficiency of the engine?
Solution: The maximum efficiency of the engine is the efficiency that a Carnot engine
would have operating with the same hot and cold reservoirs. Thus, the maximum
efficiency is C
Carnot
H
1
T
e
T
= − (Equation 15.15), where TC and TH are the Kelvin
temperatures of the cold and the hot reservoir, respectively.
Using Equation for eCarnot and recognizing that the engine has an efficiency e that is
three-fifths the maximum or Carnot efficiency, we obtain
3 3 3
C
Carnot
5 5 5
H
605 K
1 1 0.22
950 K
T
e e
T
   
= = − = − =
   
   
 
20. Calculate the change in entropy as a 3.50 kg box slides to a stop from a speed of 2.25
m/s. Assume a temperature of 293 K for all the objects involved.
Solution: The change in entropy S of a system for a process in which heat Q enters or
leaves the system reversibly at a constant temperature T is given by
equation R
( / )
S Q T
 = . In this case, all of the kinetic energy that was present when the
box was moving has been converted into heat, so 2
1
2
Q KE mv
= = . Substituting this
expression into the above equation, we find the change in entropy is
( )( )
( )
2
2 2
2
R
1 1
3.50 kg 2.25 m/s
2 2
( / ) 3.02 10 J/K
293 K
mv
S Q T
T
−
 = = = = 
Since heat is lost from the hot reservoir (inside the house), the change in entropy is
negative: SH = −QH/TH. Since heat is gained by the cold reservoir (the outdoors), the
change in entropy is positive: SC = +QC/TC. Here we are using the symbols QH and QC
to denote the magnitudes of the heats. The change in the entropy of the universe is
C H
universe C H
C H
28 800 J 28 800 J
+ 13.7 J/K
258 K 294 K
Q Q
S S S
T T
 =   = − = − =
In this calculation we have used the fact that TC = 273 − 15 C = 258 K and
TH = 273 + 21 C = 294 K.

30. Memo/EMH1501/101/03/2021
30
Semester 2
Assignment: AS2 # Unique No: 756789
Due date: 12th
Aug 2021
Total marks 100 (10 x 10 = 100)
1. Four identical metallic objects carry the following charges: +0.4, +6.2, -5.7, and -9.4
C. The objects are brought simultaneously into contact, so that each touches the
others. Then they are separated. (a) what is the final charge on each object? (b) How
many electrons (or protons) make up the final charge on each object?
Solution: Total charge = (+0.4 C) + (6.2 C) +(-5.7 C) +(-9.4 C) = -8.5 C.
(c) Charge of each objects (Q) = -8.5 C/4 = -2.1 C.
(d) Number of electron / proton = Q/e = (-2.1 x 10-6
)/1.6 x 10-19
= 1.3 x 1013
electron
2. Two equally positively charged particles are placed 7.6 x 10-2
m apart and released.
One of the particles, with a mass of 2.8 x 10-7
kg, experiences an acceleration of 5.5
m/s2
. Find (a) the amount of charge on each particle and (b) the acceleration of the
other particle if it has a mass of 4.5 x 10-7
kg.
Solution: Let us look at particle one first. From Newton’s second law, we know that F = ma,
where F is force, m is mass, and a is acceleration
Substitute the values we know for particle one, F = ma = 2.8 x 10-7
x 5.5 = 15.4 x 10-7
N
We can now apply this to Coulomb’s law. Coulomb’s law states, F = k (q1)(q2)/r2
; where F is
electrostatic force, k is coulomb’s constant, q1 is the charge of first particle, q2 is the charge of
second particle, and r is the distance between the two charges.
We know the value of F, and so, let us again, substitute the values into this equation.
15.4 x 10-7
= (8.99 x 109
)(q2
) / (7.6 x 10-2
)2
[q1 = q2 =q]
q = [(15.4 x 7.6 x 7.6 / 8.99) x10-7
x10-4
]1/2
= 9.9 x 10-19
C
(b) The mass of the second particle can be calculated by using Newton’s second law again.
Since the force is the same for both particles, we can isolate for mass.
F = ma; a = 15.4 x 10-7
/ 4.5 x 10-7
m/s2
= 3.4 m/s2
.

31. Memo/EMH1501/101/03/2021
31
3. (a) Calculate the magnitude of the acceleration experienced by an electron that is in an
electric field with a strength of 825 N/C.
(b) A proton and an electron are moving due east in a constant electric field that also
points due east. The electric field has a magnitude of 8.0 x 104
N/C. Determine the
magnitude of the acceleration of the proton and the electron.
Solution: (a) we know the electric field, E = F/q. For an electron the force, Fe = meae.
So, in an electric field the electron experienced a force in an electric field, E is
E = me.ae / e
825 = 9.1 x 10-31
kg x ae /1.6 x 10-19
C
[me = 9.1 x 10-31
kg and charge of electron =1.6 x 10-19
C]
ae = 145 x 1012
m/s2
.
(b) For electron, ae = E. e / me = (8 x 104
N/C x 1.6 x 10-19
C) / 9.1 x 10-31
kg
= 1.4 x 1016
m/s2
.
For proton, ap = E. e / mp = (8 x 104
N/C x 1.6 x 10-19
C) / 1.67 x 10-27
kg
= 7.7 x 1012
m/s2
.
4. A proton and an electron, starting from rest, are accelerated through an electric
potential of the same magnitude. In the process, the electron acquires a speed ve, while
the proton acquires a speed vp. Find the ratio ve/vp.
Solution: The proton accelerates from point A to point B. According to energy conservation
principle only kinetic and electric potential energies. We have
KEP1B + EPEP1B = KEP1A + EPEP1A. Or, KEPB = EPEP1A- EPEP1B
Final total energy Initial total energy
of proton at point B. of proton at point A
The initial kinetic energy is zero since proton starts from rest.
The electric potential energy of a charge q0 is EPE = q0v1
Where v is potential experienced by charge.
Final kinetic energy of proton as follows
KEP1B = EPEP1A - EPEP1B = qp (VA-VB) --- (1)
But an electron has a negative charge that is equal in magnitude to charge on a proton,
so qe = (-qp)
The kinetic energy of electron as
KEe1A = EPEe1B – EPEe1A = -qp(VB-VA) = qp(VA-VB) --- (2)
Comparing equation (1) and (2) we obtain
KEP1B = KEe1A, or, ½ mpVp
2
= ½ meVe
2
Solving for ratio, Ve/Vp = (mp/me)1/2
= (1.67 x 10-27
kg / 9.11 x 10-31
kg)1/2
= 42.8.

32. Memo/EMH1501/101/03/2021
32
5. A charge of +125 C is fixed at the corner of a square that is 0.75 m on a side. How
much work is done by the electric force as a charge of +7.0 C is moved from one
corner of the square to any other empty corner? Explain.
Solution:
Since this a square the distance from the center of the square to any of the four corners is the
same. Therefore the electric potential at each of the four corners is the same. So moving the
charge from one corner to any other corner results in the same electrical potential
energy. Therefore the work done is zero!
But
6. Calculate the final speed of a free electron accelerated from rest through a potential
difference of 100 V.
Solution: We have a system with only conservative forces. Assuming the electron is
accelerated in a vacuum, and neglecting the gravitational force (we will check on this
assumption later), all of the electrical potential energy is converted into kinetic energy. We
can identify the initial and final forms of energy to be KEi = 0, KEf = ½(mv2
), PEi=qV, and
PEf =0.
Conservation of energy states that KEi + PE i = KE f + PE f .
Entering the forms identified above, we obtain qV = mv2
/2.
We solve this for v:
V = [(2qV)/m]1/2
Entering values for q, V, and m gives
v=[2(−1.60×10−19
C)(−100 J/C) / 9.11×10−31kg]1/2
=5.93×106
m/s
7. A resistor is connected across the terminals of a 9.0-V battery, which delivers 9.8 x
104
J of energy to the resistor in 6.5 hours. What is the resistance of the resistor?
Solution: According to Ohm’s law, the resistance is the voltage of the battery divided by the
current that the battery delivers. The current is the charge divided by the time during
which it flows. We know the time but are not given the charge directly. However, we can
determine the charge from the energy delivered to the resistor, because this energy comes
from the battery, and the potential difference between the battery terminals is the
difference in electric potential energy per unit charge, according to Equation 19.4. Thus, a
9.0-V battery supplies 9.0 J of energy to each coulomb of charge passing through it. To
calculate the charge, then, we need only divide the energy from the battery by the 9.0 V
potential difference.

33. Memo/EMH1501/101/03/2021
33
Ohm’s law indicates that the resistance R is the voltage V of the battery divided by the
current I, or R = V/I. According to Equation 20.1, the current I is the amount of charge q
divided by the time t, or I = q/t. Using these two equations, we have
/
V V V t
R
I q t q

= = =
  
The potential difference V is the difference (EPE) in the electric potential energy
divided by the charge q, or
( )
EPE
V
q

 =

. However, it is customary to denote the
potential difference across a battery by V, rather than V, so
( )
EPE
V
q

=

. Solving this
expression for the charge gives
( )
EPE
q
V

 = . Using this result in the expression for the
resistance, we find that
( ) ( )
( ) ( )
2
2
4
9.0 V 6.5 3600 s
19
EPE / EPE 9.8 10 J
V t V t V t
R
q V

  
= = = = = 
   
8. An incandescent light bulb has resistance of about 11  when it is operating. Find the
operating temperature if the average temperature coefficient of resistivity is 0.0060
(Co
)-1
.
Solution: The resistance as a function of temperature as follows:
( )
0 0
1
R R T T

 
= + −
 
where α is the temperature coefficient of resistivity. Rearranging for the temperature T
gives
0
0
1
R
R
T T

 
−
 
 
 
= + (1)
Applying Equation (1) with the values given in the problem gives
0 3
0 1
130
1 1
11
20 C 1.8 10 C
0.0060 C
R
R
T T
 −
  
 
−
  −
   

   
= + = + = 
9. A piece of Nichrome wire has radius of 6.5 x 10-4
m. It is used in a laboratory to make
a heater that uses 4.00 x 102
W of power when connected to a voltage source of 120
V. Ignoring the effect of temperature on resistance, estimate the necessary length of
wire.
Solution: We know that the resistance of the wire can be obtained from
P = V2
/R or R = V2
/P

34. Memo/EMH1501/101/03/2021
34
We also know that R = L/A. Solving for the length, noting that A =  r
2
, and using
 = 100  10
–8
.m, we find
( )( ) ( ) ( )
( )( )
2
2
2 2 –4
2 2
–8 2
/ 120 V 6.5 10 m
50 m
100 10 m 4.00 10 W
V P r
RA V r
L
P
 

  

= = = = =
  

10. The rms current in a copy machine is 6.00 A, and the resistance of the machine is 22.5
. What are (a) the average power and (b) the peak power delivered to the machine?
Solution: (a) The average power P delivered to the copy machine is equal to the square of
the rms-current Irms times the resistance R, or 2
rms
P I R
= . Both Irms and R are known.
The average power is
( ) ( )
2
2
rms 6.00 A 22.5 810 W
P I R
= =  = (20.15b)
(b) The peak power peak
P is twice the average power, or peak 2
P P
= .
The peak power is twice the average power, so
( )
peak 2 2 810 W 1620 W
P P
= = =
11. The current in a series circuit is 15.0 A. When an additional 6.00-  resistor is
inserted in series, the current drops to 10.0 A. What is the resistance in the original
circuit?
Solution: Using Ohm's law, we can write an expression for the voltage across the original
circuit as 0 0
V I R
= . When the additional resistor R is inserted in series, assuming that the
battery remains the same, the voltage across the new combination is given by 0
( )
V I R R
= + .
Since V is the same in both cases, we can write 0 0 0
( )
I R I R R
= + . This expression can be
solved for 0
R Solving for 0
R , we have
0 0 0 0 0
– or ( – )
I R IR IR R I I IR
= =
Therefore, we find that
0
0
(10.0 A)(6.00 )
12.0 A
– (15.0 A -10.0 A)
IR
R
I I

= = =
12. Two resistors have resistance R1 and R2. When the resistors are connected in series to
a 12.0-V battery, the current from the battery is 2.00 A. When the resistors are
connected in parallel to the battery, the total current from the battery is 10.0 A.
Determine R1 and R2.
Solution: The series combination has an equivalent resistance of S 1 2
R R R
= + . The parallel
combination has an equivalent resistance that can be determined from 1 1 1
P 1 2
R R R
− − −
= + .
In each case the equivalent resistance can be used in Ohm’s law with the given voltage

35. Memo/EMH1501/101/03/2021
35
and current. Thus, we can obtain two equations that each contain the unknown
resistances. These equations will be solved simultaneously to obtain R1 and R2.
For the series case, Ohm’s law is ( )
S 1 2
V I R R
= + . Solving for sum of the resistances, we
have
1 2
S
12.0 V
6.00
2.00 A
V
R R
I
+ = = =  (1)
For the parallel case, Ohm’s law is P P
V I R
= , where 1 1 1
P 1 2
R R R
− − −
= + . Thus, we have
1
P
P 1 2
1 1 1 10.00 A
0.833
12.0 V
I
R R R V
−
= + = = =  (2)
Solving Equation (1) for R2 and substituting the result into Equation (2) gives
2
1 1
1 1
1 1
0.833 or 6.00 7.20 0
6.00
R R
R R
+ = − + =
−
In this result we have suppressed the units in the interest of clarity. Solving the quadratic
equation (see Appendix C.4 for the quadratic formula) gives
( ) ( ) ( )( )
( )
2
1
6.00 6.00 4 1.00 7.20 6.00 36.0 28.8
4.34 or 1.66
2 1.00 2.00
R
− −  − −  −
= = =
Substituting these values for R1 into Equation (1) reveals that
2 1
6.00 1.66 or 4.34
R R
= − =
Thus, the values for the two resistances are 1.66 Ω and 4.34 Ω .
13. Determine the equivalent resistance between the points A and B for the group of
resistors in the drawing.
Solution: When two or more resistors are in series, the equivalent resistance is given by
Rs = R1 + R2 + R3 + . . . . Likewise, when resistors are in parallel, the expression to be solved
to find the equivalent resistance is given by
1
Rp
=
1
R1
+
1
R2
+
1
R3
+....
Since the 4.0- and the 6.0- resistors are in series, the equivalent resistance of the
combination of those two resistors is 10.0 . The 9.0- and 8.0- resistors are in parallel;
their equivalent resistance is 4.24 . The equivalent resistances of the parallel combination
(9.0  and 8.0 ) and the series combination (4.0  and the 6.0 ) are in parallel; therefore,
their equivalent resistance is 2.98 . The 2.98- combination is in series with the 3.0-
resistor, so that equivalent resistance is 5.98 . Finally, the 5.98- combination and the

36. Memo/EMH1501/101/03/2021
36
20.0- resistor are in parallel, so the equivalent resistance between the points A and B is
4 6
.  .
14. A battery delivering a current of 55.0 A to a circuit has terminal voltage of 23.4 V.
The electric power being dissipated by the internal resistance of the battery is 34.4 V.
Find the emf of the battery.
Solution: The voltage V across the terminals of a battery is equal to the emf of the battery
minus the voltage Vr across the internal resistance of the battery: V = Emf − Vr.
Therefore, we have that
r
Emf V V
= + (1)
The power P being dissipated by the internal resistance is equal to the product of the
voltage Vr across the internal resistance and the current I: r
P IV
= . Therefore, we can
express the voltage Vr across the internal resistance as
r
P
V
I
= (2)
Substituting Equation (2) into Equation (1), we obtain
r
34.0 W
Emf 23.4 V 24.0 V
55.0 A
P
V V V
I
= + = + = + =
15. Find the magnitude and direction of the current in the 2.0- resistor in the drawing.
Solution: Label the currents with the resistor values. Take I3 to the right, I2 to the left and I1
to the right. Applying the loop rule to the top loop (suppressing the units) gives
I1 + 2.0 I2 = 1.0 (1)
and to the bottom loop gives
2.0 I2 + 3.0 I3 = 5.0 (2)
Applying the junction rule to the left junction gives
I2 = I1 + I3 (3)
Solving Equations (1), (2) and (3) simultaneously, we find I2 = 0.73 A .
The positive sign shows that the assumed direction is correct. That is, to the left .

37. Memo/EMH1501/101/03/2021
37
16. A galvanometer with a coil resistance of 9.50  is used with a shunt resistor to make
a nondigital ammeter that has an equivalent resistance of 0.40 . The current in the
shunt resistor is 3.00 mA when the galvanometer reads full scale. Find the full-scale
current of the galvanometer.
Solution: The drawing at the right shows the galvanometer (G), the coil resistance RC
and the shunt resistance R. The full-scale current IG through the galvanometer and
the current IS in the shunt resistor are also shown. Note that RC and R are in parallel,
so that the voltage across each of them is the same. Our solution is based on this fact.
The equivalent resistance of the ammeter is the parallel equivalent resistance of RC
and R.
59.9 mA
0.100 mA
60.0 mA
G
Shunt
resistor R
Rc = 50.0 
A B
Expressing voltage as the product of current and resistance and noting that the voltages across
RC and R are the same, we have
G C S
Voltage across Voltage across
coil resistance shunt resistance
I R I R
=
Solving this equation for the full-scale current IG through the galvanometer gives
S
G
C
I R
I
R
=
In this result IS and RC are given, but the shunt resistance R is unknown. However, the
equivalent resistance is given as P 0.40
R = , and it is the parallel equivalent resistance
of RC and R:
P C
1 1 1
R R R
= + . This equation can be solved for R as follows:
C P P C
P C P C P C C P
1 1 1 1 1 1
or = or
R R R R
R
R R R R R R R R R R
−
= + = − =
−
Substituting this result for R into the expression for IG gives
( ) ( ) ( )
S S P C P
G S
C C C P C P
3 4
0.40
3.00 10 A 1.3 10 A
9.50 0.40
I R I R R R
I I
R R R R R R
− −
   
= = =
   
   
− −
   
 

=  = 
 
 − 
 
 
IG
RC
= 9.50 
IS
= 3.00 mA

38. Memo/EMH1501/101/03/2021
38
17. Two capacitors are connected to a battery. The battery voltage is V = 60.0 V, and the
capacitances are C1 = 2.00 mF and C2 = 4.00 mF. Determine the total energy stored
by the two capacitors when they are wired (a) in parallel and (b) in series.
Solution: When capacitors are connected in parallel, each receives the entire voltage V of the
battery. Thus, the total energy stored in the two capacitors is 2 2
1 1
2 2
1 2
CV C V
+ . When
the capacitors are connected in series, the sum of the voltages across each capacitor
equals the battery voltage: 1 2
V V V
+ = . Thus, the voltage across each capacitor is series
is less than the battery voltage, so the total energy, 2 2
1 1
2 2
1 1 2 2
CV C V
+ , is less than when
the capacitors are wired in parallel.
(a) The voltage across each capacitor is the battery voltage, or 60.0 V. The energy stored
in both capacitors is
( )
( )( )
2 2 2
1 1 1
2 2 2
1 2 1 2
2
6 6 2
1
2
Total energy
2.00 10 F + 4.00 10 F 60.0 V 1.08 10 J
C V C V C C V
− − −
= + = +
=   = 
(b) According to the discussion in Section 20.12, the total energy stored by capacitors in
series is 2
1
2 S
Total energy = C V , where CS is the equivalent capacitance of the series
combination:
6 6
S 1 2
1 1 1 1 1
2.00 10 F 4.00 10 F
C C C − −
= + = +
 
(20.19)
Solving this equation yields CS = 1.33  10
−6
F. The total energy is
( )( )2
6 3
1
2
Total energy 1.33 10 F 60.0 V 2.39 10 J
− −
=  = 
18. A 5.00-F and a 7.00 F capacitor are connected in series across a 30.0-V battery.
A9.00- F capacitor is then connected in parallel across the 3.00- F capacitor.
Determine the voltage across the 9.00- F capacitor.
Solution: The 9.00 and 5.00-F capacitors are in parallel. According to Equation 20.18, the
equivalent capacitance of the two is 9.00 F + 5.00 F = 14.0 F. This 14.0-F capacitance
is in series with the 7.00-F capacitance. According to Equation 20.19, the equivalent
capacitance of the complete arrangement can be obtained as follows:
( )
( )
–1
–1
1 1 1 1
= 0.214 F or 4.67 F
14.0 F 7.00 F 0.214 F
C
C
 
  
= + = =
The battery separates an amount of charge
Q = CV = (4.67  10
–6
F) (30.0 V) = 1.4  10
–4
C
This amount of charge resides on the 7.00 µF capacitor, so its voltage is
V7 = (1.4  10
–4
C)/(7.00  10
–6
F) = 20.0 V
The loop rule gives the voltage across the 5.00 µF capacitor to be
V5 = 30.0 V – 20.0 V = 10.0 V

39. Memo/EMH1501/101/03/2021
39
This is also the voltage across the 9.00 µF capacitor, since it is in parallel, so V9 = 10.0 V .
19. Four identical capacitors are connected with a resistor in two different ways. When
they are connected as in part a of the drawing, the time constant to charge up this
circuit is 0.72 s. What is the time constant when they are connected with the same
resistor, as in part b?
Solution: In either part of the drawing the time constant τ of the circuit is eq
RC
 = ,
according to Equation 20.21, where R is the resistance and Ceq is the equivalent
capacitance of the capacitor combination. We will apply this equation to both circuits. To
obtain the equivalent capacitance, we will analyze the capacitor combination in parts. For
the parallel capacitors P 1 2 3 ...
C C C C
= + + + applies (Equation 20.18), while for the
series capacitors 1 1 1 1
S 1 2 3 ...
C C C C
− − − −
= + + + .
we write the time constant of each circuit as follows:
a eq, a b eq, b
and
RC RC
 
= =
Dividing these two equations allows us to eliminate the unknown resistance algebraically:
eq, b eq, b
b
b a
a eq, a eq, a
or
RC C
RC C

 

 
 
= =
 
 
(1)
To obtain the equivalent capacitance in part a of the drawing, we note that the two
capacitors in series in each branch of the parallel combination have an equivalent
capacitance CS that can be determined using Equation
1
S 2
S
1 1 1
or C C
C C C
= + = (2)
Using Equation, we find that the parallel combination in part a of the drawing has an
equivalent capacitance of
1 1
eq, a 2 2
C C C C
= + = (3)
(b)
R C C
C
C
+ −
(a)
C
C
C C
R
+ −

40. Memo/EMH1501/101/03/2021
40
To obtain the equivalent capacitance in part b of the drawing, we note that the two
capacitors in series have an equivalent capacitance of 1
2
C , according to Equation (2).
The two capacitors in parallel have an equivalent capacitance of 2C. Finally, then, we
have a series combination of 1
2
C and 2C,
2
eq, b 5
1
eq, b 2
1 1 1 5
or
2 2
C C
C C C C
= + = = (4)
Using Equations (3) and (4) in Equation (1), we find that
( )
2
eq, b 5
b a
eq, a
0.72 s 0.29 s
C C
C C
 
 
 
= = =
 
 
20. An 86- resistor and a 67- resistor are connected in series across a battery. The
voltage across the 86- resistor is 72 V. What is the voltage across the 67- resistor?
Solution: Ohm’s law relates the resistance R of either resistor to the current I in it and the
voltage V across it:
V
R
I
= (20.2)
Because the two resistors are in series, they must have the same current I. We will,
therefore, apply Equation 20.2 to the 86- resistor to determine the current I. Following
that, we will use Equation 20.2 again, to obtain the potential difference across the 67−
resistor.
Let R1 = 86  be the resistance of the first resistor, which has a potential difference of
V1 = 27 V across it. The current I in this resistor is
1
1
V
I
R
= (1)
Let R2 = 67  be the resistance of the second resistor. Again employing Equation 20.2,
the potential difference V2 across this resistor is given by
2 2
V IR
= (2)
Since the current in both resistors is the same, substituting Equation (1) into Equation (2)
yields
( )
( )
( )
1
2 2 2
1
27 V
67 21 V
86
V
V IR R
R
 
= = =  =
 
  
 
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