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.
Here are some important equations:
v = dw/dq
The above equation is the definition of voltage. v is voltage, w is work (energy) in joules, and q is charge in coulombs. Voltage is work done per unit charge. Separating charges requires work. for simple cases (DC), we can use V = w/Q and w = VQ.
i = dq/dt
The above equation relates current to charge. i is current in amps, q is charge in coulombs, and t is time in seconds. Current is nothing more than the flow of charge. Note that current is defined as the flow of positive charges. For simple cases (DC), we can use I = Q/t and Q = It.
v = Ri
This is Ohm's Law. v is in volts, R is resistance in ohms, and i is current in amps. It's an important equation; commit it to memory.
p = dw/dt
This is the basic power equation. p is power in watts, w is work (energy) in joules, and t is time in seconds. Don't confuse the symbol w (work) with the power unit W (watts). Power can be positive or negative. When a device absorbs power, the power is positive. When it provides power (as with a battery), the power is negative. For simple cases (DC), we can use P = w/t and w = Pt.
We can manipulate the power equation with a little math:
p = dw/dt = (dw/dt)(dq/dq) = (dw/dq)(dq/dt) = vi
This gives us the familiar power equation p = vi. Lower case is used to show that this equation is true for time-varying voltages and currents as well as DC voltages and currents.
We can further manipulate the power equation:
p = vi = (iR)i = i2R
This equation is useful when we know the current through the resistor.
Let's try manipulating the power equation again:
p = vi = v(v/R) = v2/R
This equation is especially useful when we know the voltage across a resistor. Take care when you use this equation. Make sure the voltage you use is the actual voltage across the resistor and not some voltage elsewhere in the circuit.
There are three ideal passive circuit elements: resistors, inductors, and capacitors. Resistors obey Ohm's Law, which is discussed above. We will leave the discussion of inductors and capacitors for another lesson.
An important point about passive circuit elements in general, and resistors
in particular, is that current flows into the positive end, as shown
The Ohm's law equation, v = ir, assumes that current flows into the + end as shown.
There are two ideal active circuit elements: voltage sources and current sources.
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. An ideal voltage source produces a defined voltage, regardless of the current magnitude and direction. For example, an ideal 6 V battery has 6 V across it if the current is 0 A, 100 A, or -1,000 A. Ideal voltage sources do not obey Ohm's Law. The voltage of a real voltage source will vary slightly depending on the current, because the real device has internal resistance. Unless otherwise specified, we will always assume our circuit elements are ideal. Also note that current can flow either way through a voltage. 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 is shown below.
This example is a 3 A current source. It will produce 3 A in the direction of the arrow, regardless of the magnitude and polarity of the voltage across it. For example, it could have 30 V across it or -700 V across it; that will depend on the circuit it is connected to. For example, if the 3 A current source is connected to a 50 Ω resistor, its voltage from Ohm's Law will be 3 × 50 = 150 V. Current sources do not obey Ohm's Law.
In addition to the independent sources shown above, there are also dependent
sources, those that depend on what else is happening in the circuit. An ideal
dependent voltage source is shown below.
The voltage vs may perhaps be 5V6 where V6 is a voltage elsewhere in the circuit; if V6 happens to be 3 V, then vs would be 5 × 3 = 15 V. vs doesn't have to be dependent on a voltage. For example, it could have a value of 2i7, where i7 is 2.5 A. In this instance, vs would be 2 × 2.5 = 5 V. Yes, although vs is dependent upon a current, it is still a voltage.
The device below is an ideal dependent current source.
The current is may perhaps be 3V2 where V2 is a voltage elsewhere in the circuit. If V2 is 7 V, is would be 3 × 7 = 21 A. The current does not have to be dependent on another current.
Be careful when dealing with dependent sources in circuit equations. If it's got a + sign inside, it's a voltage source, even if it's dependent on a current. If it has an arrow inside, it's a current source, even if it is dependent on a voltage.
Before going on, you should complete Tutorial 1A on the physics of electricity.
Kirchhoff's Voltage Law and Kirchhoff's Current Law are powerful tools in circuit analysis. Before we state these laws, we must understand some terminology. Consider the circuit below.
Node: A node is a point in a circuit where two or more circuit elements join. Points A, B, C, and D are nodes.
Essential Node: An essential node is a point in a circuit where three or more circuit elements join. Points B and D are essential nodes. We usually care more about the essential nodes than about the simple nodes.
Branch: A branch is a portion of a circuit that connects two nodes. AB,AD, BD, BC, CD, DAB, and BCD are branches.
Essential Branch: An essential branch is a portion of a circuit that connects two essential nodes. BD, DAB, and BCD are essential branches.
Loop: A loop is a closed path. ABDA, BCDB, and ABCDA are loops.
Mesh: A mesh is a loop with no loops inside it. ABDA and BCDB are meshes. We usually choose to write Kirchhoff's Voltage Law equations for meshes rather than loops.
There is a subscripting convention for naming voltages. VAB is the voltage across the 100 Ω resistor. It is also the voltage of point A with respect to point B. It is positive if point A is more positive than point B. It is negative if point B is more positive than point A. For example if a 2 A current were flowing from A to B, VAB would be 200 V. If a 4 A current were flowing from B to A, VAB would be -400 V.
Lower case letters (v, i) indicate time-dependent quantities.
Upper case letters (V, I) indicate time-independent quantities, such as DC values.
Kirchhoff's Current Law is frequently abbreviated as KCL. It states that the algebraic sum of the currents at any node is zero. This law is derived from conservation of charge in physics: Charge cannot be created or destroyed. For the purposes of KCL, it means that no charge is allowed to accumulate at a point. What goes in must come out. Let's clarify the statement of KCL mathematically:
Σnode iout = 0
What we mean is that we'll count the currents going out of the node (i.e. leaving the node) as positive and those entering as negative. This is just a convention, but we will stick to it.
Σnode iout = 1 - 7 + 2 + I = 0
I = -1 + 7 - 2 = 4 A
The current entering the node (7 A) was taken as negative. The others were leaving the node, so they were positive.
To determine the number of KCL equations you need to solve a circuit, count the essential nodes and subtract one. For example, if there are 5 essential nodes, you will need 4 KCL equations.
Kirchhoff's Voltage Law is frequently abbreviated as KVL. It states that the algebraic sum of the voltages around any loop is zero. This law is derived from conservation of energy in physics: Energy gained by going around a loop must eventually be lost when you return to the starting point. Voltage, you will remember, is energy per unit charge (dw/dq). Let's clarify the statement of KCL mathematically:
Σloop vdrop = 0
What we mean is that we'll count the voltage drops as positive and the voltage rises as negative. This is just a convention, but we will stick to it.
A portion of a circuit is shown below.
The task is to find the the unknown voltage Vgap. The 10 Ω resistors will not figure into the solution. They are there only to show that there are other circuit elements connected at the nodes. The voltages across the 20 Ω resistor the 2 A current source are known (perhaps my measurement). We will apply KVL by going around the loop clockwise starting and ending at point A:
Σloop vdrop = 40 + 25 - 6 + Vgap - 18 = 0
Vgap = -40 - 25 + 6 + 18 = -41 V
Going clockwise, starting from point A, the first circuit element encountered is the 40 V source. It is entered in the KVL as positive, because voltage drops (from + to -) are taken as positive. Some people remember this rule by noting that the + sign is the first thing you come to. Next comes the 20 Ω resistor; its voltage is 25 V, and it is also positive because it is a voltage drop. The 6 V battery is next, and its voltage is entered as negative because it is a voltage rise. The gap is next. There will be no current, but certainly there may be a voltage across the gap. There can be a voltage across anything: resistors, inductors, batteries, bananas, gaps ... anything. After the gap we come to the 2 A current source with a voltage rise (therefore negative) of 18 V. This completes the loop from point A to point A, and we set the KVL equal to zero. We then solve for the unknown Vgap.
Although Kirchhoff's Voltage Law 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.
Here is a demo of how voltage, resistance, and current are related: http://micro.magnet.fsu.edu/electromag/java/ohmslaw/.
To determine the number of KVL equations you need to solve a circuit, count the meshes. For example, if there are 5 meshes, you will need 5 KVL equations.
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.
KCL can be extended beyond just junction points in circuits. It will work for any portion of a circuit. Consider the circuit below.
The supernode is circled. KCL will apply to this supernode. That means that the sum of the currents out of the supernode is zero. Mathematically, that is:
Σsupernode Iout = -3 + I = 0
This solves to:
I = 3 A
We don't need to know the values of the resistors, the voltage of the battery, or the current of the other current source to solve for I.
We could have circled the current source and resistor on the left as a supernode instead. In this case, the KCL equation would be:
Σsupernode Iout = 3 - I = 0
This also solves to:
I = 3 A
Any part of a circuit can be circled as a supernode. The circuit elements inside the supernode will not contribute to the resulting KCL equation. The only currents to use in the KCL will be those that interface with the supernode.
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 = 20 Ω, then I
I = V/R = (Va - Vb)/R = (500 - 600)/20 = -5 A.
Yes, that's minus 5.
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.
At this time you should complete Tutorial 1 on basic circuit laws.