1. Chapter 1: Introduction

2. Outlines
Recognize interrelationships of electrical engineering with other fields of science and engineering.
2. List the major subfields of electrical engineering.
3. List several important reasons for studying electrical engineering.

3. 4. Define current, voltage, and power, including their units.
5. Calculate power and energy, as well as determine whether energy is supplied or absorbed by a circuit element.
6. State and apply basic circuit laws.
7. Solve for currents, voltages, and powers in simple circuits.

4. Electrical systems have two main objectives:
To gather, store, process, transport, and present information
To distribute and convert energy between various forms

5. Electrical Engineering Subdivisions
Communication systems
Computer systems
Control systems
Electromagnetics
Electronics
Photonics
Power systems
Signal processing
(Circuit design)
(Biomedical engineering)

6. Why Study Electrical Engineering?
Because it is everywhere,…

7. Headlight circuit

8. Electrical Current
Electrical current is the time rate of flow of electrical charge through a conductor or circuit element. The units are amperes (A), which are equivalent to coulombs per second (C/s).

9. Electrical Current

12. Direct Current Alternating Current
When a current is constant with time, we say that we have direct current, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.

13. .

16. Voltages
The voltage associated with a circuit element is the energy transferred per unit of charge that flows through the element. The units of voltage are volts (V), which are equivalent to joules per coulomb (J/C).

20. POWER AND ENERGY

23. KIRCHHOFF’S CURRENT LAW
The net current entering a node is zero.
Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node.

26. pp. 19
Fig. 1.18

27. Series Circuits

28. KIRCHHOFF’S VOLTAGE LAW
The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit.

33. Find voltages Find series/parallel connection

39. Resistors and Ohm’s Lawb

40. Conductance

41. Resistance Related to Physical Parameters

42. Voltage, current and power

45. Using KVL, KCL, and Ohm’s Law to Solve a Circuit

48. Problem Set
6, 14, 19, 23, 28, 33, 48, 57

49. Chapter 2: Resistive Circuits

50. Resistive Circuits
Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel.
2. Apply the voltage-division and current-division principles.
3. Solve circuits by the node-voltage technique.

51. 4. Solve circuits by the mesh-current technique.
5. Find Thévenin and Norton equivalents and apply source transformations.
6. Apply the superposition principle.
7. Draw the circuit diagram and state the principles of operation for the Wheatstone bridge.

56. Circuit Analysis using Series/Parallel Equivalents
Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.
Redraw the circuit with the equivalent resistance for the combination found in step 1.

57. 3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance.
4. Solve for the currents and voltages in the final equivalent circuit.

62. Voltage Division

64. Application of the Voltage-Division Principle

66. Current Division

69. Application of the Current-Division Principle

72. Although they are very important concepts, series/parallel equivalents and the current/voltage division principles are not sufficient to solve all circuits.

73. Node Voltage Analysis

75. Writing KCL Equations in Terms of the Node Voltages for Figure 2.16

84. End 9/27/2006

85. Circuits with Voltage Sources
We obtain dependent equations if we use all of the nodes in a network to write KCL equations.

90. Node-Voltage Analysis with a Dependent Source First, we write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source.

93. Next, we find an expression for the controlling variable ix in terms of the node voltages.

94. Substitution yields

96. Node-Voltage Analysis
1. Select a reference node and assign
variables for the unknown node
voltages. If the reference node is
chosen at one end of an
independent voltage source, one
node voltage is known at the start,
and fewer need to be computed.

97. 2. Write network equations. First, use KCL
to write current equations for nodes
and supernodes. Write as many current
equations as you can without using all
of the nodes. Then if you do not have
enough equations because of voltage
sources connected between nodes, use
KVL to write additional equations.

98. 3. If the circuit contains dependent sources, find expressions for the
controlling variables in terms of the
node voltages. Substitute into the
network equations, and obtain
equations having only the node
voltages as unknowns.

99. 5. Use the values found for the node
4. Put the equations into standard form and solve for the node voltages.
5. Use the values found for the node
voltages to calculate any other currents or voltages of interest.
4. Put the equations into standard form and solve for the node voltages.
5. Use the values found for the node voltages to calculate any other currents or
voltages of interest.

102. Mesh Current Analysis

103. Choosing the Mesh Currents
When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.
Sometimes it is said that the mesh currents are defined by “soaping the window panes.”

105. Writing Equations to Solve for Mesh Currents
If a network contains only resistances and independent voltage sources, we can write
the required equations by following each current around its mesh and applying KVL.

106. Using this pattern for mesh 1 of Figure 2.32(a), we have
For mesh 2, we obtain
For mesh 3, we have

107. In Figure 2.32(b)

110. Mesh Currents in Circuits Containing Current Sources
A common mistake made by beginning students is to assume that the voltages across current sources are zero. In Figure 2.35, we have:

113. Combine meshes 1 and 2 into a supermesh
Combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined.
Mesh 3:

118. Mesh-Current Analysis
1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement.

119. 2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh.

120. 3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns.

121. 4. Put the equations into standard form
4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means.
5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.

122. Thévenin Equivalent Circuits

125. Thévenin Equivalent Circuits

128. Finding the Thévenin Resistance Directly
When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit.
We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.

135. Step-by-step Thévenin/Norton-Equivalent-Circuit Analysis
1. Perform two of these:
a. Determine the open-circuit voltage Vt = voc.
b. Determine the short-circuit current In = isc.
c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals.

136. 2. Use the equation Vt = Rt In to compute the remaining value.
3. The Thévenin equivalent consists of a voltage source Vt in series with Rt .
4. The Norton equivalent consists of a current source In in parallel with Rt .

139. Source Transformations

142. Maximum Power Transfer
The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.

145. SUPERPOSITION PRINCIPLE
The superposition principle states that the total response is the sum of the responses to each of the independent sources acting individually. In equation form, this is

151. WHEATSTONE BRIDGE
The Wheatstone bridge is used by mechanical and civil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings.

152. Problem Set
1, 12, 24, 36, 39, 51, 61, 72, 75, 85, 87

153. Chapter 3 Inductance and Capacitance

154. Chapter 3 Inductance and Capacitance
1. Find the current (voltage) for a capacitance or inductance given the voltage (current) as a function
of time.
2. Compute the capacitances of parallel-plate capacitors.

155. 3. Compute the stored energies in capacitances or inductances.
4. Describe typical physical construction of capacitors and inductors and identify parasitic effects.
5. Find the voltages across mutually coupled inductances in terms of the currents.

157. CAPACITANCE

167. Capacitance of the Parallel-Plate Capacitor

173. INDUCTANCE

182. Chapter 4 Transients

183. Solve first-order RC or RL circuits.
Chapter 4 Transients
Solve first-order RC or RL circuits.
2. Understand the concepts of transient response and steady-state response.

184. 3. Relate the transient response of first-order
circuits to the time constant.
4. Solve RLC circuits in dc steady-state
conditions.
5. Solve second-order circuits.
6. Relate the step response of a second-order system to its natural frequency and damping ratio.

185. Transients
The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integrodifferential equations.

187. Discharge of a Capacitance through a Resistance

190. The time interval τ = RC is called the time constant of the circuit.

192. (Refer to p.151 and p.152.)

193. DC STEADY STATE
The steps in determining the forced response for RLC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.

196. RL CIRCUITS
The steps involved in solving simple circuits containing dc sources, resistances, and one energy-storage element (inductance or capacitance) are:

197. 1. Apply Kirchhoff’s current and voltage
laws to write the circuit equation. 2. If the equation contains integrals,
differentiate each term in the equation
to produce a pure differential equation. 3. Assume a solution of the form K1 +
K2est.

198. 4. Substitute the solution into the
differential equation to determine the
values of K1 and s . (Alternatively, we
can determine K1 by solving the circuit
in steady state as discussed in Section
4.2.) 5. Use the initial conditions to determine
the value of K Write the final solution.

201. RL Transient Analysis
Time constant is

207. RC AND RL CIRCUITS WITH GENERAL SOURCES
The general solution consists of two parts.

208. The particular solution (also called the forced response) is any expression that satisfies the equation.
In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution.

209. The homogeneous equation is obtained by setting the forcing function to zero.
The complementary solution (also called the natural response) is obtained by solving the homogeneous equation.

210. Step-by-Step Solution
Circuits containing a resistance, a source,
and an inductance (or a capacitance)
1. Write the circuit equation and reduce it to a first-order differential equation.

211. 4. Use initial conditions to find the value of K.
2. Find a particular solution. The details of this step depend on the form of the forcing function. We illustrate several types of forcing functions in examples, exercises, and problems.
3. Obtain the complete solution by adding the particular solution to the complementary solution given by Equation 4.44, which contains the arbitrary constant K.
4. Use initial conditions to find the value of K.

212. (Refer to equation (4.48))

219. SECOND-ORDER CIRCUITS

220. Undamped resonant frequency
Damping coefficient
Undamped resonant frequency

222. Then the complementary solution is
1. Overdamped case (ζ > 1). If ζ > 1 (or equivalently, if α > ω0), the roots of the characteristic equation are real and distinct.
Then the complementary solution is
In this case, we say that the circuit is
overdamped.

223. In this case, we say that the circuit is
2. Critically damped case (ζ = 1). If ζ = 1 (or equivalently, if α = ω0 ), the roots are real and equal. Then the complementary solution is
In this case, we say that the circuit is
critically damped.

224. 3. Underdamped case (ζ < 1)
3. Underdamped case (ζ < 1). Finally, if ζ < 1 (or equivalently, if α < ω0), the roots are complex. (By the term complex, we mean that the roots involve the square root of –1.) In other words, the roots are of the form
in which j is the square root of -1 and the natural frequency is given by

225. For complex roots, the complementary solution is of the form
In electrical engineering, we use j rather than i to stand for square root of -1, because we use i for current.
For complex roots, the complementary solution is of the form
In this case, we say that the circuit is underdamped.

228. (See text book for solution, particularly on the i.c.)

236. (Dual of the series circuit)

237. Problem Set
4, 7, 15, 21, 30, 35, 37, 45, 48

238. Chapter 5 Steady-State Sinusoidal Analysis

239. Chapter 5 Steady-State Sinusoidal Analysis
1. Identify the frequency, angular frequency, peak value, rms value, and phase of a sinusoidal signal.
2. Solve steady-state ac circuits using phasors and complex impedances.

240. 3. Compute power for steady-state ac circuits.
4. Find Thévenin and Norton equivalent
circuits.
5. Determine load impedances for
maximum power transfer.
6. Solve balanced three-phase circuits.

242. SINUSOIDAL CURRENTS AND VOLTAGES
Vm is the peak value
ω is the angular frequency in radians per second
θ is the phase angle
T is the period

243. Frequency
Angular frequency

244. Root-Mean-Square Values

245. RMS Value of a Sinusoid
The rms value for a sinusoid is the peak value divided by the square root of two. This is not true for other periodic waveforms such as square waves or triangular waves.

247. Phasor Definition

248. Adding Sinusoids Using Phasors
Step 1: Determine the phasor for each term.
Step 2: Add the phasors using complex arithmetic.
Step 3: Convert the sum to polar form.
Step 4: Write the result as a time function.

249. Using Phasors to Add Sinusoids

252. Sinusoids can be visualized as the real-axis projection of vectors rotating in the complex plane. The phasor for a sinusoid is a snapshot of the corresponding rotating vector at t = 0.

253. Phase Relationships
To determine phase relationships from a phasor diagram, consider the phasors to rotate counterclockwise. Then when standing at a fixed point, if V1 arrives first followed by V2 after a rotation of θ , we say that V1 leads V2 by θ . Alternatively, we could say that V2 lags V1 by θ . (Usually, we take θ as the smaller angle between the two phasors.)

254. To determine phase relationships between sinusoids from their plots versus time, find the shortest time interval tp between positive peaks of the two waveforms. Then, the phase angle is
θ = (tp/T ) × 360°. If the peak of v1(t) occurs first, we say that v1(t) leads v2(t) or that v2(t) lags v1(t).

258. COMPLEX IMPEDANCES

263. Kirchhoff’s Laws in Phasor Form
We can apply KVL directly to phasors. The sum of the phasor voltages equals zero for any closed path.
The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.

264. Circuit Analysis Using Phasors and Impedances
1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)

265. 2. Replace inductances by their complex impedances ZL = jωL. Replace
capacitances by their complex impedances
ZC = 1/(jωC). Resistances have impedances
equal to their resistances.
3. Analyze the circuit using any of the techniques
studied earlier in Chapter 2, performing the
calculations with complex arithmetic.

277. AC Power Calculations

287. THÉVENIN EQUIVALENT CIRCUITS

288. The Thévenin voltage is equal to the open-circuit phasor voltage of the original circuit.
We can find the Thévenin impedance by zeroing the independent sources and determining the impedance looking into the circuit terminals.

289. The Thévenin impedance equals the open-circuit voltage divided by the short-circuit current.

294. Maximum Average Power Transfer
If the load can take on any complex value, maximum power transfer is attained for a load impedance equal to the complex conjugate of the Thévenin impedance.
If the load is required to be a pure resistance, maximum power transfer is attained for a load resistance equal to the magnitude of the Thévenin impedance.

297. BALANCED THREE-PHASE CIRCUITS
Much of the power used by business and industry is supplied by three-phase distribution systems. Plant engineers need to be familiar with three-phase power.

299. Phase Sequence
Three-phase sources can have either a positive or negative phase sequence.
The direction of rotation of certain three-phase motors can be reversed by changing the phase sequence.

301. Wye–Wye Connection
Three-phase sources and loads can be connected either in a wye configuration or in a delta configuration.
The key to understanding the various three-phase
configurations is a careful examination of the wye–wye circuit.