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 (106), kilo (103), 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 = i2R. If you know the voltage across a resistor, use p = v2/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. vab 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, Va is not 2 V, nor is Vb 2 V or -2 V. What
is true is that
Va - Vb = 2 V.
We can solve that equation to get
Va = Vb + 2 V.
Of course, if the minus terminal were at the top, it would be
Va = Vb - 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 = (Va - Vb)/R.
Notice that Va, the first term, is associated with where the current comes into the resistor, and Vb, the second term, is associated with where the current comes out.
If Va = 500 V, Vb = 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 Va, 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
Va = 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.
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 V1, V2, V3, and V4 in the following equations:
6V1 - 12 V2 + 8V3 = 11
4V1 - 9V2 + 15V3 - 7V4 = -19
2V2 - 3V3 + V4 = 5
-10V1 - 13V2 + 14V3 + 6V4 = 0
Describe the tools you use and how to use them to solve this problem.