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Chapter I, Introduction to Powers of 10
Text reference: Pages 6-9.
Problem: Express the resistance value of 2,500,000 Ω in engineering notation. Next,
substitute the correct metric prefix for the power of 10 used with engineering notation.
Answer: In engineering notation 2,500,000 Ω = 2.5  106 Ω.
Because the metric prefix mega (M) corresponds to 106, the value of 2,500,000 Ω can
also be expressed as 2.5 MΩ.
In summary, 2,500,000 Ω = 2.5  106 Ω = 2.5 MΩ.
Chapter 1, Electricity
Text reference: Pages 38-40.
Problem: Calculate the resistance, R, in ohms for a conductance, G, of 0.00075 S.
Answer: Resistance, R, is the reciprocal of conductance, G. Therefore:
R = 1/G
= 1/0.00075 S
= 1.333 kΩ
Chapter 2, Resistors
Text reference: Pages 59-60.
Problem: Using the four-band resistor color code, indicate the colors of the bands for a
resistor whose value is 2.4 MΩ, ±5%.
Answer:
2 → red (first band)
4 → yellow (second band)
M → is 106 so the number of zeros that need to be added after the first two digits
(2 and 4) is five, which corresponds to the color green.
(Note: 2.4 MΩ = 2.4  106 Ω = 2,400,000 Ω. Note that there are five zeros after the first
two digits, 2 and 4.)
For a tolerance of ± 5%, the fourth band is gold.
Therefore, the resistor color bands are: red, yellow, green, and gold, as read from left to
right.
Chapter 3, Ohm’s Law
Text reference: Pages 78-83.
Problem: A simple series circuit consists of a battery and a resistor. Calculate the
current, I, if the battery voltage is 12V and the resistor’s value is 2 kΩ.
Answer: Ohm’s law states that the current (I) is equal to the voltage (V) divided by the
resistance (R) or;
I = V/R
Therefore: I = 12 V/2 kΩ
= 12 V/2000 Ω
= 0.006 A or 6 x 10-3 A or 6 mA
Chapter 4, Series Circuits
Text reference: Pages 113-115.
Problem: In Figure 4-30 (page 130 in the text), determine the voltage, V2. Also, based on
the resistor voltage drops, which resistor has the highest resistance value?
Answer: Kirchhoff’s Voltage Law (KVL) states that in a series circuit, the sum of the
resistor voltage drops must add to equal the applied voltage.
Since the applied voltage (VT) is 36V, all three resistor voltage drops must add to equal
this value.
VT = 36 V
V1 = 9 V
V2 = ? V
V3 = 16 V
Since VT = V1 + V2 + V3
Then, V2 = VT − V1 − V3
= 36 V − 9 V − 16 V
= 11V
Since the current, I, is the same through each series resistor, the highest resistance value
must have the highest voltage drop. Since V3 is the highest voltage drop, R3 must have
the highest resistance value.
Chapter 5, Parallel Circuits
Text reference: Page 152.
Problem: In Figure 5-30 (page 164 of the text), solve for VA and I1.
Answer: In this circuit, R1 = 100 Ω and P1 = 2.25 W. Therefore, V1 can be calculated by
using the following equation:
V1 =
=
P1  R1
2.25W 100 
= 15 V
Since both resistors are connected in parallel with the applied voltage, VA, then;
V1 = V2 = VA = 15 V
I1 = V1/R1
= 15 V/100 Ω
= 0.15 A or 150 mA
Chapter 6, Series-Parallel Circuits
Text reference: Pages 174-175.
Problem: In Figure 6-30 (page 193 of the text), solve for RT and IT.
Answer: To calculate RT, the steps are as follows:
1. R3 and R4 are in series and are simply added. R3 + R4 = 150 Ω + 180 Ω = 330 Ω
2. R2 is in parallel with the combined resistance of R3 + R4. Using the REQ formula
for equal branch resistances; REQ = R /2 = 330 Ω/2 = 165 Ω.
3. R1 is in series with the equivalent resistance of R2, R3, and R4. Therefore, to
determine RT, add R1 to the combined equivalent resistance of R2, R3, and R4.
That is,
RT = 75 Ω + 165 Ω
RT = 240 Ω
With VT = 12 V, IT can be calculated using Ohm’s law:
IT = VT/ RT
RT was determined to be 240 Ω above. Therefore,
IT = 12 V/240 Ω
IT = 0.05 A or 50 x 10-3 A or 50 mA
Chapter 7, Voltage Dividers and Current Dividers
Text reference: Pages 207 and 208.
Problem: In Figure 7-19 (page 219 of the text), solve for I1.
Answer: Currents divide inversely as the branch resistances. In this circuit, use the
formula:
I1 =
=
R2
 IT
R1  R 2
100 
 120 mA
150   100 
= 48 mA
Chapter 8, Analog and Digital Multimeters
Text reference: Pages 241-245.
Problem: List two advantages of a digital multimeter (DMM) as compared to a
conventional volt-ohm-milliammeter (VOM).
Answer:
1. A DMM has a digital readout which displays the measured value directly.
2. A DMM has a high input resistance (typically 10 MΩ) when used as a
voltmeter, which generally prevents the loading effect in most circuits.
Chapter 9, Kirchhoff’s Laws
Text reference: Page 269.
Problem: What is a principal node?
Answer: A principal node is a common connection for three or more components in
a circuit where currents can combine or divide.
Chapter 10, Network Theorems
Text reference: Pages 285-288.
Problem: In Figure 10-27 (page 307 of the text), determine the Thevenin equivalent
circuit with respect to terminals A and B (mentally remove RL).
Answer: The procedure to Thevenize this circuit is as follows:
1.
2.
3.
4.
Remove RL.
Use the voltage divider rule to determine the open-circuit voltage, VAB.
VAB = VT × R2/(R1 + R2) = 20 V × 12 Ω/(4 Ω + 12 Ω) = 20 V × 0.75 = 15 V.
Short-circuit VT to determine RAB. With VT short-circuited, R1 is effectively in
parallel with R2. The “product over the sum” formula is used to calculate RAB.
5. RAB = (R1× R2)/(R1 + R2) = (4 Ω×12 Ω)/(4 Ω + 12 Ω) = 48 Ω/16 Ω = 3 Ω.
6. Therefore, VTH = 15 V and RTH = 3 Ω.
The Thevenin equivalent circuit consists of VTH and RTH in series. Terminal A is
connected to the positive side of VTH (through RTH) and terminal B is connected to
the negative side of VTH.
Chapter 11, Conductors and Insulators
Text reference: Pages 323-325.
Problem: In Figure 11-22 (page 339 of the text), answer the following questions when
the switch, S1 is closed:
a. How much voltage is across the switch?
b. How much voltage is across the lamp?
c. Will the lamp be lit?
d. What is the current in the circuit based on the specifications of the lamp?
Answer:
a.
b.
c.
d.
0V
6.3 V
Yes
150 mA
Chapter 12, Batteries
Text reference: Pages 361-364.
Problem: The output voltage of a battery drops from 6 V with no load to 5.5 V with a
load current of 300 mA. Calculate the internal resistance, ri of the battery.
Answer:
ri =
=
VNL  VFL
IL
6 V  5.5V
300 mA
= 1.67 Ω
Chapter 13, Magnetism
Text reference: Pages 388-389.
Problem: (a) What is a ferrite material? (b) What is a typical application for a ferrite?
Answer:
(a) Ferrite is the name for a nonmetallic material that has the
ferromagnetic properties of iron. Ferrites have very high permeability
like iron. However, a ferrite is a nonconducting ceramic material,
whereas iron is a conductor.
(b) An application is a ferrite core, usually adjustable, in the coils of RF
transformers. Another application is in ferrite beads. The ferrite bead
concentrates the magnetic field of the current in a wire and can serve
as a simple, economical choke, instead of a coil.
Chapter 14, Electromagnetism
Text reference: Pages 404-405.
Problem: (a.) What is meant by magnetic hysteresis? (b) What is meant by hysteresis
losses?
Answer:
(a) Hysteresis means a lagging behind. With respect to the magnetic flux in an
iron core of an electromagnet, the flux lags behind the increases and
decreases in the magnetizing force.
(b) Hysteresis loss is a loss of energy caused by a magnetizing force reversing
thousands or millions of times per second. Since a large part of the
magnetizing force must overcome the friction of the molecular dipoles, heat
is produced. The energy wasted as heat is called hysteresis loss.
Chapter 15, Alternating Voltage and Current
Text reference: Pages 436-439.
Problem: If the sine wave in Figure 15-29 (page 464 of the text) has a peak value of
30 V, then calculate:
(a) the peak-to-peak value
(b) the rms value
(c) the average value
Answer:
(a) The peak-to-peak value is equal to twice the peak (or maximum) value of a sine
wave. Therefore, since the peak value is 30 V, 30 V × 2 = 60 V peak-to-peak.
(b) The rms (root-mean-square) value is the value of a sine wave that corresponds
to the same amount of direct current or voltage in heating power. The rms value
= 0.707 × peak value. Therefore, since the peak value is 30 V, 30 V × 0.707 =
21.21 V rms.
(c) The average value is the arithmetic average of all values in a sine wave for one
alternation. The average value = 0.637 × peak value. Therefore, since the peak
value is 30 V, 30 V × 0.637 = 19.11 V average value.
Chapter 16, Capacitance
Text reference: Page 492.
Problem: A 5-μF, 0.27-μF, and a 15-μF capacitor are in parallel. How much is the total
capacitance, CT?
Answer: Connecting capacitors in parallel is equivalent to increasing the plate area.
Therefore, the total capacitance (CT) is the sum of the individual capacitances. In the
example above,
CT = C1 + C2 + C3 + … + etc.
= 5-μF + 0.27-μF + 15-µF
= 20.27-μF
Chapter 17, Capacitive Reactance
Text reference: Pages 517-518.
Problem: In Figure 17-13 (page 527 of the text), solve for
(a) XCT
(b) I
(c) VC1, VC2, and VC3
Answer: (a) For the circuit shown in Figure 17-13, the total capacitive reactance, XCT is
equal to the sum of the series reactances. That is:
XCT = XC1 + XC2 + XC3
= 400 Ω + 800 Ω + 1.2 kΩ
= 2.4 kΩ
(b) The current, I, is equal to:
VT
= 36 V/2.4 kΩ = 0.015 A = 15 mA
XCT
(c) The voltage across each capacitive reactance is equal to IXC. Therefore,
VC1 = I × XC1
= 15 mA × 400 Ω
= 6V
VC2 = I × XC2
= 15 mA × 800 Ω
= 12V
VC3 = I × XC3
= 15 mA × 1.2 kΩ
= 18V
Chapter 18, Capacitive Circuits
Text reference: Pages 535-537.
Problem: In Figure 18-13 (page 550 of the text), solve for;
(a) the resistor voltage, VR
(b) the capacitor voltage, VC
(c) the total voltage, VT
Answer: Since the resistor and capacitor are connected in series, the current is limited by
both XC and R. The current is the same through both series connected components. Each
component has its own series voltage drop, equal to IR for the resistance and IXC for the
capacitive reactance. Therefore,
(a) VR = I × R
= 20 mA × 2 kΩ
= (20 × 10-3 A) × (2 × 103 Ω)
= 40 V
(b) VC = I × XC
= 20 mA × 1.5 kΩ
= (20 × 10-3 A) × (1.5 × 103 Ω)
= 30 V
(c)
VT = VR 2 + VC 2
VT =
40 2 + 30 2
VT =
2500
VT = 50 V
Chapter 19, Inductance
Text reference: Pages 569-575.
Problem: If RL equals 15 Ω in Figure 19-35 (page 597 of the text), solve for:
(a) the secondary voltage, VS
(b) the secondary current, IS
(c) the secondary power, Psec
Answer: The turn’s ratio is the number of turns in the primary to the number of turns in
the secondary. Therefore,
Turns ratio = NP/NS
where NP = number of turns in the primary and NS = number of turns in the secondary. In
this circuit, the turn’s ratio is 5:1 or 5/1.
(a) With unity coupling between the primary and secondary, the voltage induced in each
turn of the secondary is the same as the self-induced voltage of each turn in the primary.
Therefore, the voltage ratio is in the same proportion as the turn’s ratio. That is:
VP/VS = NP/NS
The turn’s ratio is 5:1. Therefore, VP is stepped down by a factor of 5, making VS equal
to 120/5, or 24 V.
(b) By Ohm’s law, the amount of secondary current equals the secondary voltage divided
by the resistance in the secondary circuit. Therefore,
IS = VS/RL = 24 V/15 Ω = 1.6 A
The power dissipated by RL in the secondary is IS2 × RL or VS × IS. Therefore,
P = IS2 × RL = (1.6)2 A × 15 Ω = 2.56 A × 15 Ω = 38.4 W
or,
P = VS × IS = 24 V × 1.6 A = 38.4 W
Chapter 20, Inductive Reactance
Text reference: Pages 604-608.
Problem: List two factors that affect the inductive reactance, XL a coil. Also, how much
is the inductive reactance, XL of a coil for a steady dc current?
Answer: The two factors that affect the amount of inductive reactance, XL are:
1. The frequency (f) of the alternating current
2. The inductance (L) of the coil
The higher the frequency of the alternating current, and the greater the inductance,
the higher the inductive reactance, XL.
There is no inductive reactance, XL for a steady direct current. In this case, the only
opposition to direct current is the dc resistance of the wire used to make the coil.
Chapter 21, Inductive Circuits
Text reference: Pages 632-635.
Problem: In Figure 21-25 (page 646 of the text), change the value of XL to 8 Ω and solve
for IR, IL, IT, and ZEQ.
Answer: In this circuit, the applied voltage VA is the same across XL, R, and the
generator. Each branch has its own individual current. Therefore,
IR = VA/R = 24 V/10 Ω = 2.4 A
IL = VA/XL = 24 V/8 Ω = 3 A
IT =
IR 2 + IL2
IT =
2.4 2 + 32
IT = 14.76
IT = 3.84 A
The total impedance, ZEQ is equal to the applied voltage, VA divided by the total line
current, IT. That is:
ZEQ = VA/IT
= 24 V/3.84 A
= 6.25 Ω
Chapter 22, RC and L/R Time Constants
Text reference: Pages 656-657.
Problem: Why can arcing be a problem when coils are used in switching circuits?
Answer: A demonstration of producing a high voltage by opening an inductive
circuit is shown in Figure 22-3 (page 656 of the text).
1. With the switch, S, closed, the applied voltage, V, of 8 V is not enough to
light the 90-V neon bulb.
2. When the switch, S, is opened however, the short L/R time constant
results in a high induced voltage, VL, which lights the bulb.
3. When the switch is opened the high induced voltage is also felt across the
switch contacts and arcing occurs. Arcing can destroy the switch contacts
and under certain conditions can even cause fires or explosions.
Chapter 23, Alternating Current Circuits
Text reference: Pages 693-695.
Problem: In Figure 23-25 (page 708 of the text) change XL to 75 Ω and solve for the net
reactance, X, ZT, I, VR, VC and VL.
Answer: The net reactance, X, is the difference between XC and XL.
X = XC − XL = 125 Ω − 75 Ω = 50 Ω
ZT =
R2  X 2
= 100 2  50 2
= 111.8 Ω
I = V/ZT = 125 V/111.8 Ω = 1.12 A
VR = I × R = 1.12 A × 100 Ω = 112 V
VL = I × XL = 1.12 A × 75 Ω = 84 V
VC = I × XC = 1.12 A × 125 Ω = 140 V
Chapter 24, Complex Numbers for AC Circuits
Text reference: Page 718.
Problem: What is the definition of a complex number? In what form is the complex
number: 100 + j300 Ω?
Answer: A complex number is the combination of a real and imaginary term.
Complex numbers can be expressed in either rectangular or polar form.
The complex number 100 + j300 Ω is expressed in rectangular form where 100 Ω is
the real term and +j300 Ω is the imaginary number.
Chapter 25, Resonance
Text reference: Page 767.
Problem: What is meant by electronic tuning?
Answer: A typical application of resonant circuits is in tuning a receiver to the
carrier frequency of a desired station. Tuning is done by an air-variable capacitor.
Capacitance can be varied by adjusting the variable capacitor so that the plates are
completely meshed (highest amount of capacitance) to out of mesh (smallest amount
of capacitance). For electronic tuning, the capacitance is varied by a varactor. A
varactor is a semiconductor diode that varies in capacitance when its reverse bias
voltage is changed.
Chapter 26, Filters
Text reference: Pages 794-803.
Problem: Suppose that a low-pass filter has a cutoff frequency of 1 kHz. If the input
voltage for a signal at this frequency is 250 mV, how much is the output voltage, Vout?
Answer: The ability of any filter to reduce the amplitude of undesired frequencies is
called the attenuation of the filter. The frequency at which the attenuation reduces the
output voltage to 70.7% is the cutoff frequency. In this filter circuit, the output voltage at
the cutoff frequency is equal to:
Vout = 250 mV × 0.707 = 177 mV
Chapter 27, Diodes and Diode Applications
Text reference: Pages 858-863.
Problem: In Figure 27-36 (page 869 of the text), change RL to 220 Ω and solve for the
following:
(a) IS
(b) IL
(c) IZ
Answer: The voltage dropped across the series resistor, RS, is VIN − VZ.
(a) Therefore, the current IS, through the series resistor is calculated as:
IS =
=
VIN  VZ
RS
12 V  8.2 V
50 
= 76 mA
(b) The current through the load resistor, RL is calculated as:
IL = VZ/RL = 8.2 V/220 Ω = 37.27 mA
(c) Because the zener is in parallel with RL, the series current, IS equals IZ + IL. By
rearranging the formula
IZ = IS − IL = 76 mA − 37.27 mA = 38.73 mA
Chapter 28, Bipolar Junction Transistors
Text reference: Pages 885-887.
Problem: When using an analog VOM for testing the BE and CB junctions of a silicon
transistor, what should the meter read for both polarities of the meter leads if the diode is:
(a) good?
(b) shorted?
(c) open?
Answer:
(a) A low reading for one polarity (forward-biased) and a high reading for the
other polarity (reverse-biased). The reverse to forward resistance ratio
should be equal to or greater than 1000:1.
(b) A low resistance reading across the junction for both polarities of the meter
leads.
(c) A high resistance reading across the junction for both polarities of the meter
leads.
Chapter 29, Transistor Amplifiers
Text reference: Pages 921-931.
Problem: What is the main application of an emitter follower?
Answer: The emitter follower, more properly known as a common-collector
amplifier, provides both current gain and power gain. The output signal is in phase
with the input signal. Moreover, an important characteristic of an emitter follower
is that it has a high input impedance and a low output impedance, which makes it
ideal for impedance matching applications.
Chapter 30, Field Effect Transistors
Text reference: Page 973-975.
Problem: List two reasons why E-MOSFETs are typically used in computers.
Answer:
1. E-MOSFETs take up very little space on a chip (an integrated circuit).
2. E-MOSFETs consume extremely little power.
Chapter 31, Power Amplifiers
Text reference: Page 1007-1012.
Problem: What is the main application of a class C amplifier?
Answer: Because of their high distortion, class C amplifiers cannot be used in audio
circuitry where full reproduction of the input signal is required. Class C amplifiers,
however, can be used as tuned rf amplifiers where undesired harmonic frequencies
can be filtered out, passing only the fundamental frequency to the load. In some
cases, it might be desirable to tune the LC tank circuit of a class C amplifier to a
harmonic (multiple) of the input frequency. This allows the class C amplifier to
function as a frequency multiplier.
Chapter 32, Thyristors
Text reference: Pages 1029-1032.
Problem: In Figure 32-15 (page 1035 of the text), what voltage does the emitter voltage,
VE, need to reach to make the UJT conduct?
Answer: The UJT conducts once the emitter voltage, VE, exceeds ηVBB + VD. Therefore,
VE = ηVBB + VD
= (0.7 × 15 V) + 0.7
= 10.5 + 0.7
= 11.2 V
Chapter 33, Operational Amplifiers
Text reference: Page 1075-1077.
Problem: In the circuit shown in Figure 33-43 (page 1087 of the text), what value of Vin
causes the output to be at:
(a) +Vsat
(b) −Vsat
Answer: In this circuit, the inverting input is grounded while the input signal is applied to
the noninverting input. The comparator compares Vin to the zero volt reference on the
inverting input. Therefore,
(a) When Vin is positive, Vout is driven to +Vsat.
(b) When Vin is negative, Vout is driven to −Vsat.