Quiz 1 with solutions (click here)
Summary of Passive Sign Convention and Chapters 1-4 of Text click here (PowerPoint file)
Quiz 2 with solutions (click here). Use superposition or Thevenin equivalent circuit transformations to solve Problem 1, Kirchoff's laws to solve Problem 2, the direct method to solve Problem 3, and Thevenin equivalent circuits to solve Problem 4.
Selected homework problems from the text (see below)
Quiz 1 covered through Chapter 3 with selected additional topics from the lectures. The homework problems here address the entire course. You should look at them and work any one of them that you feel will help you review or fill in areas that you need for the Final Examination. Solutions are in the text for problems cited here unless otherwise noted. Focus on problems that you consider the best use of your study time.
Chapter 1 problems begin on page 16.
Systems of Units
1.3-1, 1.3-2, and 1.3-3
1.5-1, 1.5-2, 1.7-1 (no solutions posted)
Chapter 2 problems begin on page 44.
2.2-1, 2.2-2 (no solutions posted)
Resistors (note that voltage sources are represented by "+" and "-" signs in circles, while current sources are represented by arrows inside circles)
2.4-1, 2.4-2, 2.4-3
Voltmeter and ammeter approximations and errors
2.6-4 (no solution posted)
Chapter 3 problems begin on page 86.
Kirchoff's current law (page 56): The algebraic sum of the currents into a node at any instant is zero.
Kirchoff's voltage law (page 57): The algebraic sum of the voltage drops around any loop in a circuit is zero.
See examples beginning on page 57.
Note the description of resistors in series as a voltage divider in Section 3.3, pages 61-63 and examples beginning on page 63.
Note the description of resistors in parallel as a current divider in Section 3.4, pages 66-68, with examples beginning on page 68.
3.2-2, 3.2-3 (no solutions posted)
3.3-1 is an excellent example of a simple voltage divider (no solution posted).
3.3-4 shows how two voltage dividers can be compared. This is an early presentation of the principles of a Wheatstone Bridge.
3.4-1 is an excellent example of a simple current divider (no solution posted).
Section 3.6 shows how to analyze a complex circuit by partitioning the circuit into blocks and replacing each block with a simple equivalent circuit.
3.6-1 and 3.6-2 show the principles of circuit analysis by the methods of Section 3.6.
Chapter 4 problems begin on page 141.
Section 4.7 introduces dependent sources, denoted by diamonds instead of the circles that are used for constant sources. Note again that "+" and "-" signs are used inside the diamonds for controlled voltage sources while arrows are used inside the diamonds for controlled current sources.
Section 4.8 beginning on page 130 compares and discusses the node voltage method, in which node voltages are the principal unknowns in the equations for solving a circuit, and the current mesh method, in which currents are the principal unknowns in solving a circuit. Node voltage equations can be converted to mesh equations by the use of Ohm's law and vice versa. The introductory paragraphs are an excellent overview of the subject.
Section 4.9 introduces the use of Matlab to solve circuits.
4.2-1, 4.2-4 (no solutions posted)
4.3-7 (no solution posted) is a general problem using a loaded bridge.
4.4-7 is an interesting problem that can be solved by direct methods by noting that the left side of the circuit can be partially solved by simple equations, and that this solution can be used to solve the rest of the circuit by simple equations. Write the node or mesh equations and see if you can identify the equations that allow this. Compare your approach to an intuitive approach based on simply examining the circuit and looking for simple local relationships. This problem has been solved in Mathcad and presented in class (PDF, Mathcad).
Chapter 5 problems begin on page 186.
Source transformations, superposition, Thevenin and Norton equivalent circuits are presented in this chapter. These methods provide methods to solve complex circuits quickly.
5.2-1 shows the power of the use of equivalent circuits. Partial solution is posted as part of the problem statement.
5.2-2 is another compelling example showing how equivalent circuits can be used to reduce a complex problem to a sequence of simple problems.
5.3-1 is a simple example of superposition.
5.3-2 was used in Quiz 2.
5.4-1 is a simple use of source transformations to solve a circuit quickly.
5.4-2 is more of a tutorial than a problem.
5.5-1 shows the difference between a Thevenin equivalent circuit (a voltage source in series with a resistance) and a Norton equivalent circuit (a current source in parallel with a resistance). The question and answer are an outside-the-box trick question based on physical assumptions about the lab equipment and measurements.
Chapter 6 problems begin on page 236.
6.3-2 is a little more difficult. Ans:
6.3-4 is an example of a voltage-to-current converter (no solution posted). Ans:
6.3-6 is an example of a non-inverting amplifier.
6.3-7 is an example of two inverting amplifiers connected in series. Ans:
6.3-9 is an example of a voltage follower.