PHYSICS 212
CHAPTER 18
DIRECT-CURRENT CIRCUITS
WORKBOOK
ANSWERS
_____________________________________________
STUDENT’S FULL NAME
(By placing your name above and submitting this for credit
you are affirming this to be predominantly your own work.)
_____ / _____ / _____
DATE DUE
INSTRUCTIONS
1. Turn this workbook in on time for credit, even if it is not complete. (No credit
if late.)
2. Complete this workbook neatly. Do not write in ink so that corrections can be
made. (Credit will be lost if this is turned in messy.)
3. Complete the chapter outline section as early as possible. Don’t wait for the due
date to be assigned to start.
4. Complete the sections in sequence.
5. Study and learn definitions of terms, physical quantities, units, principles, and
basic equations before attempting problems.
6. You may work on this with other students but do not copy another student’s
workbook or let a student copy this workbook. Do not copy from other sources
either.
7. Wherever possible, include diagrams in your solutions. Diagrams are required.
8. Keep this workbook after it is graded and returned to you.
9. Using the answer key, redo all questions and problems until you can answer
them all correctly by yourself without help.
10. Use the workbook to learn the general problem-solving strategy rather than
how individual problems are solved.
11. (NEW) Questions marked (Basic) should be answered by students without the
need for additional assistance.
-1-
PHYSICS 212 CHAPTER 18 OUTLINE
DIRECT-CURRENT CIRCUITS
18.1
SOURCES OF EMF
What does the abbreviation emf represent?
electromotive force
Explain the difference between a battery’s emf and its terminal voltage.
The battery emf is the voltage produced internally.
The terminal voltage is the accessible external voltage.
18.2
RESISTORS IN SERIES
Draw a simple circuit with two resistors in series.
How are the currents through the two resistors related?
The currents through the resistors is the same through both.
How are the voltages across the two resistors related?
The voltages add together to equal the voltage across their group, in this case the
battery voltage.
How is the equivalent resistor calculated?
The equivalent resistor is the sum of the individual resistors in series.
18.3
RESISTORS IN PARALLEL
Draw a simple circuit with two resistors in parallel.
-2-
How are the currents through the two resistors related?
The currents through the resistors adds together and equals the current entering the
group .
How are the voltages across the two resistors related?
The voltages across the resistors are equal.
How is the equivalent resistor calculated?
The reciprocal of the equivalent resistor is equal to the sum of the reciprocal of all
the resistors in parallel.
18.4
SKIP
18.5
RC CIRCUITS
Suppose a capacitor and resistor are connected in series. During the charging process I, Q, and VC
change. Write equations for these quantities.
I = I 0e
 t 
−

 RC 
 t 

−

Q = Q0 1 − e  RC  




 t 

−

V = V0 1 − e  RC  




What is the RC time constant?
It is the it takes the current from a discharging capacitor to fall to 1/e of its in initial
value.
When a capacitor discharges through a resistor I, Q, and VC change. Write equations for these
quantities.
I = I0e
 t 
−

 RC 
Q = Q0 e
V = V0 e
 t 
−

 RC 
 t 
−

 RC 
-3-
18.6
HOUSEHOLD CIRCUITS
How is 240 V produced in circuits for large appliances?
The appliance is connected between two wires, one is 120 V above ground and the
other is 120 V below ground.
18.7
ELECTRICAL SAFETY
List what may likely occur for each of the following currents if passing through a person.
5 mA - sensation of shock, usually no damage or harm
10 mA - muscle contraction and possible inability to let go of wire
100 mA - potentially fatal
What does the abbreviation GFI represent and how does the device work?
Ground-Fault Interrupt
18.8
SKIP
-4-
PHYSICS 212
CHAPTER 18 MULTIPLE CHOICE QUESTIONS
DIRECT-CURRENT CIRCUITS
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
C___
1. The two ends of a 3.0-Ω resistor are connected to a 9.0-V battery. What is the current through the resistor?
A. 27 A
D. 0.33 A
B. 6.3 A
E. 0.17 A
C. 3.0 A
(Basic)
B___
2. The two ends of a 3.0-Ω resistor are connected to a 9.0-V battery. What is the total power delivered by the
battery to the circuit?
A. 3.0 W
D. 0.11 W
B. 27 W
E. 0.067 W
C. 0.33 W
(Basic)
B___
3. The basic function of an electromotive force in a circuit is to do which of the following?
A. Convert electrical energy into some other form.
B. Convert some other form of energy into electrical.
C. Both choices (a) and (b) are valid.
D. None of the above choices are valid.
(Basic)
A___
4. Which voltage is not caused by a source of emf?
A. the voltage across a charged capacitor
B. the voltage across two copper-iron junctions at different temperatures
C. the voltage across the terminals of a dry cell battery
D. the voltage from an electric generator
E. None of the above choices is valid.
A___
5. Three 8.0-Ω resistors are connected in series. What is their equivalent resistance?
A. 24.0 Ω
D. 0.13 Ω
B. 8.0 Ω
E. 0.075 Ω
C. 0.38 Ω
(Basic)
C___
6. Three resistors connected in series each carry currents labeled I1, I2 and I3. Which of the following expresses
the value of the total current IT in the system made up of the three resistors in series?
A. IT = I1 + I2 + I3
D. IT = (1/I1 + 1/I2 + 1/I3)-1
B. IT = (1/I1 + 1/I2 + 1/I3)
E. IT = 3I1 = 3I2 = 3I3
C. IT = I1 = I2 = I3
A___
7. Three resistors connected in series have individual voltages labeled ∆V1, ∆V2 and ∆V3, respectively. Which of
the following expresses the value of the total voltage ∆VT taken over the three resistors together?
A. ∆VT = ∆V1 + ∆V2 + ∆V3
D. ∆VT = (1/∆V1 + 1/∆V2 + 1/∆V3)-1
B. ∆VT = (1/∆V1 + 1/∆V2 + 1/∆V3)
E. ∆VT = 3∆V1 = 3∆V2 = 3∆V3
C. ∆VT = ∆V1 = ∆V2 = ∆V3
-5-
A___
8. Three resistors with values of R1, R2 and R3, respectively, are connected in series. Which of the following
expresses the total resistance, RT, of the three resistors?
A. RT = R1 + R2 + R3
D. RT = (1/R1 + 1/R2 + 1/R3)-1
B. RT = (1/R1 + 1/R2 + 1/R3)
E. RT = 3R1 = 3R2 = 3R3
C. RT = R1 = R2 = R3
D___
9. Three resistors, with values of 2.0, 4.0 and 8.0 Ω, respectively, are connected in series. What is the overall
resistance of this combination?
A. 0.58 Ω
D. 14.0 Ω
B. 1.1 Ω
E. 19.0 Ω
C. 7.0 Ω
(Basic)
A___ 10. Three resistors, each with resistance R1, are in series in a circuit. They are replaced by one equivalent
resistor, R. Comparing this resistor to the first resistor of the initial circuit, which of the following is true?
A. The current through R equals the current through R1.
B. The voltage across R equals the voltage across R1.
C. The power given off by R equals the power given off by R1.
D. R is less than R1.
E. R is equal to R1.
C___ 11. When a light bulb is turned on, its resistance increases until it reaches operating temperature. What happens
to the current in the bulb as it is warming up?
A. It stays constant.
D. It increases at first and then decreases.
B. It increases.
E. It decreases at first and then increases.
C. It decreases.
A___ 12. Three resistors connected in parallel each carry currents labeled I1, I2 and I3. Which of the following
expresses the value of the total current IT in the combined system?
A. IT = I1 + I2 + I3
D. IT = (1/I1 + 1/I2 + 1/I3)-1
B. IT = (1/I1 + 1/I2 + 1/I3)
E. IT = 3I1 = 3I2 = 3I3
C. IT = I1 = I2 = I3
_C__ 13. Three resistors connected in parallel have the individual voltages labeled ∆V1, ∆V2 and ∆V3, respectively.
Which of the following expresses the total voltage ∆VT across the three resistors when connected in this
manner?
A. ∆VT = ∆V1 + ∆V2 + ∆V3
D. ∆VT = (1/∆V1 + 1/∆V2 + 1/∆V3)-1
B. ∆VT = (1/∆V1 + 1/∆V2 + 1/∆V3)
E. ∆VT = 3∆V1 = 3∆V2 = 3∆V3
C. ∆VT = ∆V1 = ∆V2 = ∆V3
D___ 14. Three resistors with values R1, R2 and R3, respectively, are connected in parallel. Which of the following
expresses the total resistance, RT, of the three resistors when connected in parallel?
A. RT = R1 + R2 + R3
D. RT = (1/R1 + 1/R2 + 1/R3)-1
B. RT = (1/R1 + 1/R2 + 1/R3)
E. RT = 3R1 = 3R2 = 3R3
C. RT = R1 = R2 = R3
-6-
B___ 15. Three resistors, each with resistance R1, are in parallel in a circuit. They are replaced by one equivalent
resistor, R. Compare this resistor to the first resistor of the initial circuit. Which of the following statements
is true?
A. The current through R equals the current through R1.
B. The voltage across R equals the voltage across R1.
C. The power given off by R equals the power given off by R1.
D. R is greater than R1.
E. R is less than R1.
A___ 16. If R1 < R2 < R3, and if these resistors are connected in parallel in a circuit, which one has the highest current?
A. R1
D. All have the same current.
B. R2
E. The answer depends on the internal
resistance of the battery.
C. R3
B___ 17. Household circuits are wired in ____.
A. series
B. parallel
C. both series and parallel
D. neither series nor parallel
B___ 18. In applications where electrical shocks may be more likely, such as around water in kitchens and bathrooms,
special outlets called GFI's are used. What does GFI stand for?
A. get free instantly
D. gravity-free insulator
B. ground-fault interrupter
E. guided fault isolation
C. give fast interruption
(Basic)
C___ 19. Household 120-V outlets are made to accept three-pronged plugs. One of the prongs attaches to the "live"
wire at 120 V, and another attaches to the "neutral" wire that is connected to ground. What is the round third
prong for?
A. It serves as a backup to the hot wire.
B. It lets the appliance run if the neutral wire breaks.
C. It connects the case of the appliance directly to ground for safety purposes.
D. It serves for direct current feed.
E. Nothing electrical, it is for mechanical sturdiness.
(Basic)
-7-
PHYSICS 212
1.
CHAPTER 18 ADDITIONAL QUESTIONS
DIRECT-CURRENT CIRCUITS
A battery with an emf of 12 V is connected to an external resistor whose resistance is 100 Ohms. If the current
through the external resistor is 0.11 A, what is the internal resistance of the battery? (Basic)
V = IReq
V
12V
=
I 0.11A
Req = 109.09Ω
Req =
Req = R + r
r = Req − R = 109.09Ω − 100Ω
r = 9.09Ω
2.
If a 3.0 V battery has a 0.5 Ah rating. (a) What is the average current it can maintain for 2 hours? (b) What is
the total energy the battery can supply? (Basic)
Q = 0.5 Ah = It
Q 0.5 Ah
=
t
2h
I = 0.25 A
I=
 3600 s 
Q = 0.5 Ah 
 = 1800C
 h 
U = QV = (1800C )(3.0V )
U = 5400 J
-8-
3.
Three resistors are connected in series. Their values are 2 Ohms, 4 Ohms, and 5 Ohms. What is the potential
drop across the 4 Ohms resistor when the group is connected to a 12 V battery?
Req = R1 + R2 + R3
Req = 2Ω + 4Ω + 5Ω
Req = 11Ω
V = IReq
I=
V
12V
=
Req 11Ω
I = 1.091A
V2 = IR2 = (1.091A)(4Ω)
V2 = 4.36V
4.
Three resistors are connected in parallel. Their values are 2 Ohms, 4 Ohms, and 5 Ohms. What is the potential
drop across the 4 Ohms resistor when the group is connected to a 12 V battery?
V2 = 12V
-9-
5.
A capacitor of 5 µf is connected in parallel with a 1.25 MΩ resistor and an 18 V battery. After charging the
capacitor a switch is opened disconnecting the battery. (A) How long does it take for the capacitor to loose 80%
of its initial charge? (B) What is the power loss through the resistor at this time? (C) How much energy remains
in the capacitor at this time?
A
.2Q0 = Q0 e
.2 = e
t
− RC
t
− RC
t
RC
t = − RC ln(.2) = −(1.25M Ω)(5µ f )ln(.2)
ln(.2) = −
t = 10.06s
B
I0 =
V0
18V
=
= 1.44 × 10−5 A
6
R 1.25 × 10 Ω
I = I0e
t
− RC
−5
= (1.44 × 10 A)e


10.06 s
−

 (1.25 M Ω )(5 µ f ) 
I = 2.88 × 10−6 A
P = I 2 R = (2.88 × 10−6 ) 2 (1.25 × 106 Ω)
P = 1.04 × 10−5 W
C
VC = V0 e
t
− RC
= (18V )e


10.06 s
−

 (1.25 M Ω )(5 µ f ) 
VC = 3.6V
U = 12 CVC2 = 12 (5µ f )(3.6V ) 2
U = 9.00µ J
-10-
PHYSICS 212 CHAPTER 18 PROBLEMS
DIRECT-CURRENT CIRCUITS
PRACTICE PROBLEMS:
HOMEWORK PROBLEMS:
1.
2, 6, 7, 9, 11, 34, 41
1, 3, 5, 30, 31, 36, 39, 40, 45, 60
A battery having an emf of 9.00 V delivers 117 mA when connected to a 72.0-S load. Determine the
internal resistance of the battery.
From ∆V = I ( R + r ) , the internal resistance is
r=
3.
∆V
9.00 V
−R=
− 72.0 Ω = 4.92 Ω
I
0.117 A
A lightbulb marked “75 W [at] 120 V” is screwed into a socket at one end of a long extension cord in
which each of the two conductors has a resistance of 0.800 S. The other end of the extension cord is
plugged into a 120-V outlet. Draw a circuit diagram, and find the actual power of the bulb in the circuit
described.
For the bulb in use as intended,
Rbulb =
( ∆V )
P
2
=
(120 V )
2
75.0 W
= 192 Ω
Now, presuming the bulb resistance is unchanged,
the current in the circuit shown is
I=
∆V
120 V
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
P = I 2 Rbulb = ( 0.620 A ) (192 Ω ) = 73.8 W
2
-11-
5.
a) Find the equivalent resistance between points a and b in Figure P18.5. (b) Calculate the current in
each resistor if a potential difference of 34.0 V is applied between points a and b.
Figure P18.5
(a)
The equivalent resistance of the two parallel resistors is
−1
1 
 1
Rp = 
+
 = 4.12 Ω
 7.00 Ω 10.0 Ω 
Thus,
Rab = R4 + R p + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
Also,
( ∆V ) p = I ab R p = (1.99 A )( 4.12 Ω ) = 8.18 V
and I10 =
Rab
( ∆V ) p
R7
( ∆V ) p
R10
=
=
=
34.0 V
= 1.99 A , so
17.1 Ω
I ab =
Then, I 7 =
30.
( ∆V )ab
b)
8.18 V
= 1.17 A
7.00 Ω
8.18 V
= 0.818 A
10.0 Ω
Show that J = RC has units of time.
The time constant is τ = R C . Considering units, we find

 Volts   Coulombs   Coulombs  
Coulombs
R C → ( Ohms )( Farads ) = 
 = Second

=
=
 Amperes   Volts   Amperes   Coulombs Second 
or τ = R C has units of time.
-12-
31.
39.
Consider a series RC circuit for which C = 6.0 :F, R = 2.0 × 106 S, and , = 20 V. Find (a) the time
constant of the circuit and (b) the maximum charge on the capacitor after a switch in the circuit is closed.
(a)
τ = R C = ( 2.0 × 106 Ω )( 6.0 × 10 −6 F ) = 12 s
(b)
Qmax = C ε = 6.0 × 10 −6 F ( 20 V ) = 1.2 × 10 − 4 C
(
)
A heating element in a stove is designed to dissipate 3 000 W when connected to 240 V. (a) Assuming
that the resistance is constant, calculate the current in the heating element if it is connected to 120 V. (b)
Calculate the power it dissipates at that voltage.
From P = ( ∆V )
R=
2
( ∆V )
P
R , the resistance of the element is
2
=
( 240 V )
2
3 000 W
= 19.2 Ω
When the element is connected to a 120-V source, we find that
(a)
(b)
40.
∆V 120 V
=
= 6.25 A , and
R 19.2 Ω
P = ( ∆V ) I = (120 V )( 6.25 A ) = 750 W
I=
Your toaster oven and coffeemaker each dissipate 1 200 W of power. Can you operate them together if
the 120-V line that feeds them has a circuit breaker rated at 15 A? Explain.
The maximum power available from this line is
Pmax = ( ∆V ) I max = (120 V )(15 A ) = 1 800 W
Thus, the combined power requirements (2 400 W) exceeds the available power, and you
cannot operate the two appliances together
-13-
45.
Find the equivalent resistance between points a and b in Figure P18.45.
Figure P18.45
The resistive network between a an b reduces, in the stages shown below, to an equivalent resistance of
Req = 7.5 Ω .
60.
The circuit in Figure P18.60 contains two resistors, R1 = 2.0 kS
and R2 = 3.0 kS, and two capacitors, C1 = 2.0 :F and C2 = 3.0
:F, connected to a battery with emf , = 120 V. If there are no
charges on the capacitors before switch S is closed, determine the
charges q1 and q2 on capacitors C1 and C2, respectively, as
functions of time, after the switch is closed. [Hint: First
reconstruct the circuit so that it becomes a simple RC circuit
containing a single resistor and single capacitor in series,
connected to the battery, and then determine the total charge q
stored in the circuit.]
The total resistance in the circuit is
−1
−1
 1
 1
1 
1 
R= +
+
 =
 = 1.2 kΩ
 2.0 kΩ 3.0 kΩ 
 R1 R2 
and the total capacitance is C = C1 + C 2 = 2.0 µ F+3.0 µ F=5.0 µ F
Thus, Qmax = C ε = ( 5.0 µ F )(120 V ) = 600 µ C
-14-
Figure P18.60
and
τ = RC = (1.2 × 103 Ω )( 5.0 × 10−6 F ) = 6.0 × 10−3 s =
6.0 s
1 000
The total stored charge at any time t is then
(
Q = Q1 + Q2 = Qmax 1 − e − t τ
) or Q
1
(
+ Q2 = ( 600 µ C ) 1 − e −1 000 t
6.0 s
)
(1)
Since the capacitors are in parallel with each other, the same potential difference exists across both at
any time.
Therefore,
( ∆V )C =
C 
Q1 Q2
, or Q2 =  2  Q1 = 1.5 Q1
=
C1 C2
 C1 
Solving equations (1) and (2) simultaneously gives
Q1 =
and
( 240 µ C ) (1 − e −1 000 t 6.0 s )
(
Q2 = ( 360 µ C ) 1 − e −1 000 t 6.0 s
)
-15-
(2)