, SOLUTION MANUAL FOR Electronics with Discrete Components, 2nd
Edition Galvez
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,Electronics with Discrete Components
Solutions to Problems
Enrique J. Galvez
Department of Physics and Astronomy
Colgate University
2025
,2
,Chapter 1
The Basics
1. Two capacitors C1 and C2 are connected in series, with a potential V0 across
them.
(a) If the charges on the two capacitors are the same, with a value q, find the
voltages V1 and V2 across each capacitor in terms of the variables given.
Solution: V1 = q/C1 and V2 = q/C2 .
(b) Find a relation between and V0 , V1 , and V2 .
Solution: Because the are in series, V0 = V1 + V2 .
(c) Now consider replacing the two capacitors in series by one capacitor C0
and connect it to the supply with voltage V0 . If C0 draws the same charge
q from the supply, find a relation between C0 , C1 , and C2 .
Solution: From (b), V0 = q(1/C1 + 1/C2 ) = q/C0 . Therefore, 1/C0 =
1/C1 + 1/C2 .
2. Two capacitors with capacitances 3 µF and 2 µF are initially discharged. They
are connected in series, and then the two ends of the combination are connected
to a 10-V battery, as shown in Figure 1.34.
(a) If we connect a single capacitor to the same battery, what would be the
capacitance of the capacitor so that it draws the same charge from the
3
,4 CHAPTER 1. THE BASICS
battery?
Solution: Since the capacitors are in series then C = (1/3 + 1/2)−1 =
6/5 = 1.2 µF.
(b) How much charge does the battery deliver?
Solution: Because the two capacitors are in series, they have the same
charge Q. The source delivers Q. The voltages across each capacitor add
up to the source voltage V0 = 10 V. Therefore Q/C = V0 , or Q = 12 µC.
(c) Find the charges on each capacitor.
Solution: The charge on each capacitor is Q = 12 µC.
(d) Find the voltage across each capacitor.
Solution: For the top capacitor: Vtop = Q/Ctop = 4 V; and for the bottom
capacitor, Vbot = Q/Cbot = 6 V.
(e) If point B is at zero potential, what is the potential at point A between
the two capacitors?
Solution: VA = VB + 6 V = 6 V.
3. A battery with potential V0 is connected to a capacitor C. We label the charge
in the top plate as q1 and the charge in the bottom plate as q2 . Which statement
is correct?
(a) q1 = q2 = CV0
(b) q1 = −q2 = CV0
(c) q1 + q2 = CV0
(d) q1 = −q2 = 2CV0
(e) q1 = q2 = CV0 /2
Solution: Answer is (b), because the capacitor’s top plate has a charge Q, and
the bottom plate has a charge −Q. Their absolute value is equal to CV0 .
4. Suppose that we have two capacitors with capacitances 0.1 µF and 0.2 µF. We
connect them in parallel and apply a voltage of 10 V to their ends.
(a) What is the charge on each capacitor?
Solution: Because the capacitors are in parallel, the voltage across them
is the same. If C1 = 0.1 µF and C2 = 0.2 µF, then q1 = C1 V0 = 1 µC and
q2 = C2 V0 = 2 µC.
, 5
(b) What is the total amount of charge that the source has to supply to charge
the two capacitors?
Solution: The source has to supply charge to both capacitors: q = q1 +q2 =
3 µC.
(c) If we replaced both capacitors with one so that it draws the same amount
of charge from the supply, what capacitance would it have?
Solution: Ceq = (q1 + q2 )/V0 = 0.3µF.
(d) Can you generalize the previous situation for finding the equivalent capac-
itance of any two capacitors connected in parallel?
Solution: Ceq = C1 + C2 .
5. In the circuit of Figure 1.35 (in the textbook, shown below), C1 = 0.2 µF and
C2 = 0.15 µF. When a voltage of 10 V is applied to the system of capacitors,
we find that the charge of capacitor 1 is 1 µC.
(a) Find the value of C3 , and the charge on it.
Solution: The voltage across capacitor 1 is V1 = q1 /C1 = 5 V, so the
voltage across capacitors 2 and 3 must also be 5 V (to add to the supply
voltage). The charge on capacitor 1 must be split onto capacitors 2 and 3
because they are in parallel (i.e., the wire that connects the bottom plate
of 1 to the top plates of 2 and 3 is initially neutral, so q2 +q3 −q1 = 0). The
charge on capacitor 2 is q2 = (5V)(0.15µF) = 0.75 µC. Thus q3 = 0.25 µC.
The capacitance of 3 is then C3 = (0.25µC)/(5V) = 0.05 µF.
(b) What is the equivalent capacitance of the arrangement?
Solution: The capacitance of 2 and 3 in parallel is the sum of the two:
C23 = C2 + C3 = 0.2 µF. They are in series with capacitor 1, so Ceq =
(1/C1 + 1/C23 )−1 = 0.1 µF. Check: The source must deliver the charge
Q = Ceq V0 = 1 µC, which is also equal to the charge on capacitor 1.
6. We apply 1.2 V to the 0.5 µF capacitor of Figure 1.36 (in the textbook, shown
next). We then disconnect the capacitor from the power supply. Subsequently,
,6 CHAPTER 1. THE BASICS
we connect an uncharged 1-µF capacitor in parallel with the other capacitor.
Find the charge on the 1-µF capacitor. (Note that the voltage across it is not
1.2 V.)
Solution: When we connect the supply to the first capacitor, it gets a charge
q1 = (0.5µF)(1.2V) = 0.6 µC. We then disconnect it from the supply and
connect it to the second capacitor. The charges rearrange so that the two
capacitors have the same voltage across them. We can also think that the two
capacitors now act as a single capacitor of capacitance C ′ = C1 + C2 = 1.5 µF.
The charge on it is the initial one, so the new voltage across the capacitors is
V ′ = q1 /C ′ = 0.4 V. The charges on each capacitor are then q1′ = C1 V ′ = 0.2 µC
and q2 = C2 V ′ = 0.4 µC. Check: The two charges indeed add to the original
charge.
7. The potential of point A in Figure 1.37 (in the textbook, shown below) is 0 V.
Find the potential of point B.
Solution: We have a single loop. The sum of voltages must add to zero. The
total supply voltage is 20V − 4V = 16 V. The voltage drops across the resistors,
their value times the current, must be equal to this voltage. Since the resistors
are in series in the loop, then their combined resistance is 8 kΩ. The current is
then I = (16V)/(8kΩ) = 2 mA. The potential of B is the potential on A plus the
drops on the 5-kΩ and 2-kΩ resistors, (5kΩ)(2mA) = 10 V and (2kΩ)(2mA) = 4
V, respectively. Therefore VB = 14 V.
8. The colors of the bands of the resistors of Figure 1.7 (in the textbook, shown
, 7
next) are: (a) Orange, orange, red, white; (b) Brown, black, red, gold; (c)
Brown, red, blue, orange, green.
(a) Determine their value and tolerance.
Solution: (a) 33 × 102 Ω = 3.3kΩ, 20%; (b) 10 × 102 Ω = 1kΩ, 5%; and (c)
126 × 103 Ω = 126kΩ, 0.5%.
(b) A measurement of the actual resistance of (b) gave 992 Ω. Should we
return it as defective? Explain.
Solution: No, because the difference between the expected and measured
is 0.8%, which is within the 5% tolerance of the resistor.
9. Calculate the equivalent resistance of the network in Figure 1.38 (in the text-
book, shown below).
Solution: In the upper-right side of the circuit is a wire oriented diagonally.
Notice that it meets with another wire to short a resistor. The potential of that
wire is the same. Let us call this potential, VC . Between the wire at potential
A and the one at C there is only one resistor. Between C and B there are three
resistors in parallel. Therefore we can reduce the arrangement as two resistors
in series: R between A and C, and R/3 between C and B. The equivalent
resistance between A and B is 4R/3.
, 8 CHAPTER 1. THE BASICS
10. If point A is at zero potential in Figure 1.39 (in the textbook, shown next),
what is the potential at points B and C?
Solution: We can calculate the current flowing through the loop by adding the
voltage sources, 16 V, and dividing that by the equivalent resistance of three
resistors in series, or 8 kΩ. We get I = 2 mA. The potential of the positive
side of the battery above A is 8 V. The potential of point B is one resistor drop
below: VB = 8V − (1kΩ)(2mA) = 6 V. Point C is another resistor drop below
B: VC = VB − (5kΩ)(2mA) = −4 V. Check: The potential of the negative side
of the battery below A is −8 V. The potential of point C is one resistor drop
above this: VC = −8V + (2kΩ)(2mA) = −4 V.
11. Many of the circuits that you will use in the lab will be powered by +12 V
power supplies. What is the smallest value 1/8-W resistor that we can apply
the full voltage of the power supply without burning it?
Solution: Given are: V0 = 12 V, and Pmax = 0.125 W. We know that P = V 2 /R,
so Rmin = V 2 /Pmax = 1152Ω.
12. Be aware of the dangers of electricity. Human skin can exhibit large variations
in electrical resistance. Although dry skin may have a resistance of 100 kΩ, wet
and tender skin may have resistances as low as 1 kΩ. If electrocution is caused
by currents above 50 mA, what applied voltages would cause electrocution for
(a) dry and (b) wet skin?
Solution: Lethal current is Ileth = 50 mA. (a) For dry skin Vleth = (50mA)(100kΩ) =
5000 V. (b) For wet skin Vleth = (50mA)(1kΩ) = 50 V.
13. Using only the concepts of equivalent resistance and voltage divider, calculate
the voltage between points A and B of Figure 1.40 (in the textbook, shown
below). Hint: First find the voltage drop across the 8 k resistor, but do not
ignore the resistor ladder to the right of it.
Edition Galvez
Important Notes
The file organized chapter by chapter.
A sample of selected pages has been provided for preview.
All available appendices and Excel files (if included in the original resources) are
provided.
We continuously update our files to ensure you receive the latest and most accurate
editions.
New editions are added regularly – stay connected for updates!
⚠️Note on Answer Keys: If the answer key is not included within the chapter
questions, you will find the complete answers and solutions at the end of each
chapter.
✅ Why Buy From Us?
📚 Complete & organized chapter-by-chapter – no missing content, no guessing.
⚡ Instant digital delivery – get your file the moment you pay, no waiting.
📅 Always up to date – we track new editions so you always get the latest version.
💬 Friendly support – real humans ready to help, anytime you need us.
🔒 Safe & secure – thousands of satisfied students trust us every semester.
🛡️Our Guarantees
💰 Money-Back Guarantee: Not satisfied? We offer a full refund – no questions asked.
🔄 Wrong File? No Problem: Contact us and we will replace it immediately with the
correct version, free of charge.
⏰ 24/7 Support: We are always here – reach out anytime and expect a fast response.
,Electronics with Discrete Components
Solutions to Problems
Enrique J. Galvez
Department of Physics and Astronomy
Colgate University
2025
,2
,Chapter 1
The Basics
1. Two capacitors C1 and C2 are connected in series, with a potential V0 across
them.
(a) If the charges on the two capacitors are the same, with a value q, find the
voltages V1 and V2 across each capacitor in terms of the variables given.
Solution: V1 = q/C1 and V2 = q/C2 .
(b) Find a relation between and V0 , V1 , and V2 .
Solution: Because the are in series, V0 = V1 + V2 .
(c) Now consider replacing the two capacitors in series by one capacitor C0
and connect it to the supply with voltage V0 . If C0 draws the same charge
q from the supply, find a relation between C0 , C1 , and C2 .
Solution: From (b), V0 = q(1/C1 + 1/C2 ) = q/C0 . Therefore, 1/C0 =
1/C1 + 1/C2 .
2. Two capacitors with capacitances 3 µF and 2 µF are initially discharged. They
are connected in series, and then the two ends of the combination are connected
to a 10-V battery, as shown in Figure 1.34.
(a) If we connect a single capacitor to the same battery, what would be the
capacitance of the capacitor so that it draws the same charge from the
3
,4 CHAPTER 1. THE BASICS
battery?
Solution: Since the capacitors are in series then C = (1/3 + 1/2)−1 =
6/5 = 1.2 µF.
(b) How much charge does the battery deliver?
Solution: Because the two capacitors are in series, they have the same
charge Q. The source delivers Q. The voltages across each capacitor add
up to the source voltage V0 = 10 V. Therefore Q/C = V0 , or Q = 12 µC.
(c) Find the charges on each capacitor.
Solution: The charge on each capacitor is Q = 12 µC.
(d) Find the voltage across each capacitor.
Solution: For the top capacitor: Vtop = Q/Ctop = 4 V; and for the bottom
capacitor, Vbot = Q/Cbot = 6 V.
(e) If point B is at zero potential, what is the potential at point A between
the two capacitors?
Solution: VA = VB + 6 V = 6 V.
3. A battery with potential V0 is connected to a capacitor C. We label the charge
in the top plate as q1 and the charge in the bottom plate as q2 . Which statement
is correct?
(a) q1 = q2 = CV0
(b) q1 = −q2 = CV0
(c) q1 + q2 = CV0
(d) q1 = −q2 = 2CV0
(e) q1 = q2 = CV0 /2
Solution: Answer is (b), because the capacitor’s top plate has a charge Q, and
the bottom plate has a charge −Q. Their absolute value is equal to CV0 .
4. Suppose that we have two capacitors with capacitances 0.1 µF and 0.2 µF. We
connect them in parallel and apply a voltage of 10 V to their ends.
(a) What is the charge on each capacitor?
Solution: Because the capacitors are in parallel, the voltage across them
is the same. If C1 = 0.1 µF and C2 = 0.2 µF, then q1 = C1 V0 = 1 µC and
q2 = C2 V0 = 2 µC.
, 5
(b) What is the total amount of charge that the source has to supply to charge
the two capacitors?
Solution: The source has to supply charge to both capacitors: q = q1 +q2 =
3 µC.
(c) If we replaced both capacitors with one so that it draws the same amount
of charge from the supply, what capacitance would it have?
Solution: Ceq = (q1 + q2 )/V0 = 0.3µF.
(d) Can you generalize the previous situation for finding the equivalent capac-
itance of any two capacitors connected in parallel?
Solution: Ceq = C1 + C2 .
5. In the circuit of Figure 1.35 (in the textbook, shown below), C1 = 0.2 µF and
C2 = 0.15 µF. When a voltage of 10 V is applied to the system of capacitors,
we find that the charge of capacitor 1 is 1 µC.
(a) Find the value of C3 , and the charge on it.
Solution: The voltage across capacitor 1 is V1 = q1 /C1 = 5 V, so the
voltage across capacitors 2 and 3 must also be 5 V (to add to the supply
voltage). The charge on capacitor 1 must be split onto capacitors 2 and 3
because they are in parallel (i.e., the wire that connects the bottom plate
of 1 to the top plates of 2 and 3 is initially neutral, so q2 +q3 −q1 = 0). The
charge on capacitor 2 is q2 = (5V)(0.15µF) = 0.75 µC. Thus q3 = 0.25 µC.
The capacitance of 3 is then C3 = (0.25µC)/(5V) = 0.05 µF.
(b) What is the equivalent capacitance of the arrangement?
Solution: The capacitance of 2 and 3 in parallel is the sum of the two:
C23 = C2 + C3 = 0.2 µF. They are in series with capacitor 1, so Ceq =
(1/C1 + 1/C23 )−1 = 0.1 µF. Check: The source must deliver the charge
Q = Ceq V0 = 1 µC, which is also equal to the charge on capacitor 1.
6. We apply 1.2 V to the 0.5 µF capacitor of Figure 1.36 (in the textbook, shown
next). We then disconnect the capacitor from the power supply. Subsequently,
,6 CHAPTER 1. THE BASICS
we connect an uncharged 1-µF capacitor in parallel with the other capacitor.
Find the charge on the 1-µF capacitor. (Note that the voltage across it is not
1.2 V.)
Solution: When we connect the supply to the first capacitor, it gets a charge
q1 = (0.5µF)(1.2V) = 0.6 µC. We then disconnect it from the supply and
connect it to the second capacitor. The charges rearrange so that the two
capacitors have the same voltage across them. We can also think that the two
capacitors now act as a single capacitor of capacitance C ′ = C1 + C2 = 1.5 µF.
The charge on it is the initial one, so the new voltage across the capacitors is
V ′ = q1 /C ′ = 0.4 V. The charges on each capacitor are then q1′ = C1 V ′ = 0.2 µC
and q2 = C2 V ′ = 0.4 µC. Check: The two charges indeed add to the original
charge.
7. The potential of point A in Figure 1.37 (in the textbook, shown below) is 0 V.
Find the potential of point B.
Solution: We have a single loop. The sum of voltages must add to zero. The
total supply voltage is 20V − 4V = 16 V. The voltage drops across the resistors,
their value times the current, must be equal to this voltage. Since the resistors
are in series in the loop, then their combined resistance is 8 kΩ. The current is
then I = (16V)/(8kΩ) = 2 mA. The potential of B is the potential on A plus the
drops on the 5-kΩ and 2-kΩ resistors, (5kΩ)(2mA) = 10 V and (2kΩ)(2mA) = 4
V, respectively. Therefore VB = 14 V.
8. The colors of the bands of the resistors of Figure 1.7 (in the textbook, shown
, 7
next) are: (a) Orange, orange, red, white; (b) Brown, black, red, gold; (c)
Brown, red, blue, orange, green.
(a) Determine their value and tolerance.
Solution: (a) 33 × 102 Ω = 3.3kΩ, 20%; (b) 10 × 102 Ω = 1kΩ, 5%; and (c)
126 × 103 Ω = 126kΩ, 0.5%.
(b) A measurement of the actual resistance of (b) gave 992 Ω. Should we
return it as defective? Explain.
Solution: No, because the difference between the expected and measured
is 0.8%, which is within the 5% tolerance of the resistor.
9. Calculate the equivalent resistance of the network in Figure 1.38 (in the text-
book, shown below).
Solution: In the upper-right side of the circuit is a wire oriented diagonally.
Notice that it meets with another wire to short a resistor. The potential of that
wire is the same. Let us call this potential, VC . Between the wire at potential
A and the one at C there is only one resistor. Between C and B there are three
resistors in parallel. Therefore we can reduce the arrangement as two resistors
in series: R between A and C, and R/3 between C and B. The equivalent
resistance between A and B is 4R/3.
, 8 CHAPTER 1. THE BASICS
10. If point A is at zero potential in Figure 1.39 (in the textbook, shown next),
what is the potential at points B and C?
Solution: We can calculate the current flowing through the loop by adding the
voltage sources, 16 V, and dividing that by the equivalent resistance of three
resistors in series, or 8 kΩ. We get I = 2 mA. The potential of the positive
side of the battery above A is 8 V. The potential of point B is one resistor drop
below: VB = 8V − (1kΩ)(2mA) = 6 V. Point C is another resistor drop below
B: VC = VB − (5kΩ)(2mA) = −4 V. Check: The potential of the negative side
of the battery below A is −8 V. The potential of point C is one resistor drop
above this: VC = −8V + (2kΩ)(2mA) = −4 V.
11. Many of the circuits that you will use in the lab will be powered by +12 V
power supplies. What is the smallest value 1/8-W resistor that we can apply
the full voltage of the power supply without burning it?
Solution: Given are: V0 = 12 V, and Pmax = 0.125 W. We know that P = V 2 /R,
so Rmin = V 2 /Pmax = 1152Ω.
12. Be aware of the dangers of electricity. Human skin can exhibit large variations
in electrical resistance. Although dry skin may have a resistance of 100 kΩ, wet
and tender skin may have resistances as low as 1 kΩ. If electrocution is caused
by currents above 50 mA, what applied voltages would cause electrocution for
(a) dry and (b) wet skin?
Solution: Lethal current is Ileth = 50 mA. (a) For dry skin Vleth = (50mA)(100kΩ) =
5000 V. (b) For wet skin Vleth = (50mA)(1kΩ) = 50 V.
13. Using only the concepts of equivalent resistance and voltage divider, calculate
the voltage between points A and B of Figure 1.40 (in the textbook, shown
below). Hint: First find the voltage drop across the 8 k resistor, but do not
ignore the resistor ladder to the right of it.
