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.

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.