Chapter 1: Introduction

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.

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.

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

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

Why Study Electrical Engineering? Because it is everywhere,…

Headlight circuit

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).

Electrical Current

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.

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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).

POWER AND ENERGY

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.

pp. 19 Fig. 1.18

Series Circuits

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

Find voltages Find series/parallel connection

Resistors and Ohm’s Law b

Conductance

Resistance Related to Physical Parameters

Voltage, current and power

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

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

Chapter 2: Resistive Circuits

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.

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.

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.

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.

Voltage Division

Application of the Voltage-Division Principle

Current Division

Application of the Current-Division Principle

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

Node Voltage Analysis

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

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

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.

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

Substitution yields

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.

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.

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.

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.

Mesh Current Analysis

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.”

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.

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

In Figure 2.32(b)

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:

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:

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.

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.

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.

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.

Thévenin Equivalent Circuits

Thévenin Equivalent Circuits

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.

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.

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 .

Source Transformations

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

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

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.

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

Chapter 3 Inductance and Capacitance

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.

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.

CAPACITANCE

Capacitance of the Parallel-Plate Capacitor

INDUCTANCE

Chapter 4 Transients

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.

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.

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.

Discharge of a Capacitance through a Resistance

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

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

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.

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

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.

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 K2. 6. Write the final solution.

RL Transient Analysis Time constant is

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

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.

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.

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.

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.

(Refer to equation (4.48))

SECOND-ORDER CIRCUITS

Undamped resonant frequency Damping coefficient Undamped resonant frequency

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.

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.

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

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.

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

(Dual of the series circuit)

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

Chapter 5 Steady-State Sinusoidal Analysis

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.

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.

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

Frequency Angular frequency

Root-Mean-Square Values

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.

Phasor Definition

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.

Using Phasors to Add Sinusoids

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.

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.)

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).

COMPLEX IMPEDANCES

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.

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.)

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.

AC Power Calculations

THÉVENIN EQUIVALENT CIRCUITS

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.

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

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.

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.

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.

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.