The author of these lessons is Dr. Thomas G. Cleaver, Professor Emeritus of Electrical Engineering, The University of Louisville. The author wishes to acknowledge the contributions of Dr. Joseph D. Cole and Dr. William H. Pierce.

Below is a summary of the Code of Professional Practice and Conduct for professional engineers and land surveyors in Kentucky [Kentucky Engineer, Vol. 37, No. 8, Sept. 2001].

The engineer ... shall conduct his practice in order to protect the public health, safety, and welfare.

A licensee shall issue all professional communications and work products in an objective and truthful manner. A licensee shall be objective and truthful in all professional reports, statements or testimony and shall include all material facts.

A licensee shall avoid conflicts of interest.

A licensee shall not knowingly associate with any person engaging in fraudulent, illegal or dishonest activities. A licensee shall not aid or abet the illegal practice of engineering.

A licensee shall perform his services only in the area of his competence.

The professional engineer ... shall avoid conduct likely to discredit or reflect unfavorably upon the dignity or honor of his profession.

Professional ethics and personal ethics are really no different. When you
graduate as engineers, you will be expected to practice your profession with
honor and integrity. This means that you must tell the truth, never take
credit for the work of another, treat others with respect, and give your
employer good value for your salary. As a student, your role is very similar.
You tell the truth; you do not copy homework or cheat on exams; you treat your
classmates and your instructors with respect; you work hard at your classes.
For your instructor, there is **nothing** more important than your growth as
people of strong character.

You are expected to know the International System (SI) of units (the metric system), because units in electricity are based on this system. For some practice with SI units, click on this game.

Note that current is defined as the flow of **positive** charges.

v=Ri is Ohm's Law. It's an important equation; commit it to memory.

Mega (10^{6}), kilo (10^{3}), milli (10^{-3}), micro
(10^{-6}), nano(10^{-9}), and pico(10^{-12}) are the
most frequently used prefixes. Memorize them.

The basic power equation is p = iv. Lower case is used to show that this
equation is true for time-varying voltages and currents. If you know the
current through a resistor, use p = i^{2}R. If you know the voltage
across a resistor, use p = v^{2}/R.

Two voltage sources are shown below.

The one on the left is the standard representation for a DC voltage source. The
one on the right is, technically, the symbol for a battery. For our purposes,
we can consider both symbols to represent general, ideal, DC voltage
sources.

Be sure you understand the definitions of branch, node, and loop. Elements in a circuit, such as resistors, voltage sources, current sources, etc., are connected by nodes. A node is a point of connection between two or more elements. No elements are enclosed inside it. Essential nodes connect 3 or more elements. No elements are enclosed inside it. A supernode can have elements inside it but only specific ones. Supernodes are formed by enclosing a voltage source between two essential nodes and any resistors that share both those essential nodes with the voltage source.

Note the subscripting convention. v_{ab} is positive if point a is
more positive than b.

Lower case letters (v, i) indicate time-dependent quantities.

Upper case letters (V, I) indicate time-independent quantities, such as DC values.

A voltage source has the property that its voltage is independent of the current through it. Thus, a "6 Volt battery" will have 6 volts across its terminals whether it is connected to a 1 Ω resistor, or to a 1 million Ω resistor. It should be understood that what we are talking about here is an "ideal" voltage source. The voltage of a real voltage source will depend somewhat on the current through it. Unless otherwise specified, we will always be using ideal elements. Also note that current can flow either way through a battery. It supplies energy when current flows out of the positive end; it absorbs energy ("charging" the battery) when current flows into the positive end.

An ideal current source has the property that its current is independent of the voltage across it. A "5 amp source" will supply 5 amps to a short circuit with zero voltage or 5 amps to a 1 million Ω resistor causing a 5 million volt drop. The voltage across a current source (both magnitude and direction) is always unknown until the rest of the circuit is defined.

Although Kirchhoff's Laws will work for any direction chosen, for consistency, we will always go around loops in the clockwise direction, and we will count voltage drops as positive.

Click here to see a demonstration of how voltage, resistance, and current are related.

Watch your signs when doing Kirchhoff's Laws problems. This is the principal source of student errors.

Nodes that are connected by a wire (a short circuit) can (and should) be treated as a single node.

Before going on, you should complete Tutorial 1A on **the
physics of electricity**.

*This section was written by William H. Pierce.*

A voltage generator imposes its stated voltage as the **difference** in
voltage between its two terminals. In many cases, that does not mean that the
voltage to ground is the stated value of the generator. In the circuit below,
for example, V_{a} is not 2 V, nor is V_{b} 2 V or -2 V. What
is true is that

V_{a} - V_{b} = 2 V.

We can solve that equation to get

V_{a} = V_{b} + 2 V.

Of course, if the minus terminal were at the top, it would be

V_{a} = V_{b} - 2 V.

In Ohm's Law, I = V/R, the voltage V must be the **difference** in
voltage of the two terminals of the resistor. The current arrow points into the
positive end of a resistor, and comes out the negative end. This also
determines which terms in the Ohm's Law equation we take as positive and
negative. In this example,

I = (V_{a} - V_{b})/R.

Notice that V_{a}, the first term, is associated with where the
current comes into the resistor, and V_{b}, the second term, is
associated with where the current comes out.

If V_{a} = 500 V, V_{b} = 600 V, and R = 10 Ω, then I
will be -10 A. Yes, that's **minus** 10.

Remember that the laws for a single component only apply to the difference of the voltages on its terminals.

A current is defined by its location in the circuit, and an arrow indicating the direction in which it is defined. A current arrow hitting a resistor creates a voltage difference that is IR, with the positive sign of this difference defined on the side the arrow first hits. It will be helpful to actually write a + sign on the side the arrow first hits, in order to avoid mistakes. This arrow and + sign technique should also be used for inductors and capacitors.

In finding the voltage of a point when only differences are known, begin at
ground (0 volts) and take a sum of the signed numbers using the sign on the
**far** side of the difference. (The far side is the one closer to the point
and farther from ground). Consider the circuit below.

To find V_{a}, begin at ground, add (+3 V) for the battery on the
left, add (+2 × 4 V) for the contribution from the 4 Ω resistor, add (-4
× 5 V) for the contribution from the 5 Ω resistor, and add (-7 V) for the
contribution from the battery on the right. This gives

V_{a} = 3 + 8 - 20 - 7 = -16 V.

Before going on to the homework, you should complete Tutorial 1 on **basic
circuit laws**.

**Important:** Sketch each circuit. Label all circuits with appropriate
currents and voltages.

For the first three problems, there are no "right" or "wrong" answers. You will be given credit as long as your answers are thoughtful and sincere.

- You meet with three of your classmates at the library to work together on
the calculus homework. Do you think this helps you learn the material
faster or better? How strong is the temptation to copy a solution, rather
than to learn how to solve the problem? Where does one draw the line
between copying and getting help?

- During an exam, you see a student pass a note to another student. What
do you do?

- Someone a year ahead of you in engineering offers to give you all his old
homeworks and labs. You know your teacher frequently uses problems and
labs from previous terms. What do you do?

- How much energy does it take to move an electron from the negative
terminal of a 1.5 V battery to the positive terminal? Be careful of your signs.

- E = 36 V, R1 = 8 Ω, R2 = 4 Ω. Find the current through R1
and the power consumed by R2. Do not solve by combining resistors in
series. Use Kirchhoff's Laws. The answers are integers. Hint: Define
voltages across the two resistors. Write one KVL equation and two Ohm's Law
equations.

- E = 500 V. All the resistors shown are identical. Each one consumes 400
W. Give the value of any one resistor. Do not solve by combining resistors
in series. Use Kirchhoff's Laws. The answer is an integer. Hint: Define
voltages across the resistors. Write a KVL equation.

- E = 72 V, R1 = 60 Ω, R2 = 40 Ω. Find I. Do not solve by
combining resistors in parallel. Use Kirchhoff's Laws. The answer is an
integer. Hint: Define currents in the two resistors. Write two KVL
equations, one KCL equation, and two Ohm's Law equations.

- E = 200 V, R1 = 8 Ω, R2 = 30 Ω, R3 = 20 Ω. Do not
combine resistors in series and parallel. Using Kirchhoff's Laws, find the
current in each resistor. The answers are integers. Hint: Define voltages
and currents. Write one KCL equation, two KVL equations, and three Ohm's
Law equations.

- E1 = 12 V, E2 = 20 V, R1 = 8 Ω, R2 = 3 Ω, R3 = 15 Ω.
Find the voltage at point A with respect to ground. The answer is an
integer.

**Bonus ** (no partial credit). It is wise to become adept at
solving simultaneous equations using modern tools. Advanced calculators have
this capability, as do computer algebra systems, such as Maple. Use either a
calculator with equation-solving capability or a computer algebra program to
find V_{1}, V_{2}, V_{3}, and V_{4} in the
following equations:

6V_{1} - 12 V_{2} + 8V_{3} = 11

4V_{1} - 9V_{2} + 15V_{3} - 7V_{4} = -19

2V_{2} - 3V_{3} + V_{4} = 5

-10V_{1} - 13V_{2} + 14V_{3} + 6V_{4} = 0

Describe the tools you use and how to use them to solve this problem.