By James H. Bentley

The box of electric engineering is especially innovative-new items and new principles are continu best friend being built. but most of these suggestions are according to the basic ideas of electric engineering: Ohm's legislations, Kirchhoff's legislation, suggestions keep an eye on, waveforms, capacitance, resistance, inductance, electrical energy, magnetism, present, voltage, energy, power. it truly is those simple basics that are verified for within the specialist Engineering exam (PE Exam). this article presents an geared up evaluation of the elemental electric engineering basics. it truly is an outgrowth of an electric engineering refresher direction taught through the writer to applicants getting ready for the pro Engineering Examination-a path which has enabled ratings of electric engineers in Minnesota and Wisconsin to effectively go the PE examination. the fabric is consultant of the kind of questions showing within the PE assessments ready via the nationwide Council of Engineering Examiners (NCEE) during the last twelve years. every one challenge within the textual content has been conscientiously chosen to demonstrate a selected notion. integrated with each one challenge is at the least one resolution. even if the ideas were rigorously checked, either by way of the writer and via scholars, there is adjustments of interpretation. additionally, from time to time convinced assumptions may have to be made sooner than challenge answer, and because those to person, the ultimate resolution can also vary. The assumptions will range from person writer has tried to maintain the necessities for assumptions and interpretation to a mini mum.

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**Extra resources for A Programmed Review for Electrical Engineering**

**Sample text**

C. WI = Wtot C I V 2 = ! 25 X 10-3 joule Cz V2 = ! 75 X 10-3 joule d. Energy lost in the resistor is the difference between the initial total capacitor energy and the final total capacitor energy. 875 X 10-3 joule If your solutions are correct, go on to Problem 2-9. If your solutions are not correct, review pp. 2, 3, 6, 7,10,15 and 16. PROBLEM 2-9. POWER, ENERGY, AND CHARGE In the steady-state circuit below, calculate: a. b. c. d. e. voltage across the capacitor; stored charge; energy; current delivered by the battery; power delivered by the battery.

L o. ±L E 120~ 0 . 91 0 = 2448 watts If your answer is correct, go on to Problem 2-12. If your answer is not correct, review pp. 11 and 12. COMPENSATING CIRCUITS The problems in this section apply to a variety of electrical engineering specialities including control theory and transient analysis. Use is made of partial fractions (see Chapter 5) and the Laplace transform. PROBLEM 2-12. SIMPLE LAG CIRCUIT Derive the equation for output voltage across the capacitor of the simple lag circuit shown below after the switch is closed.

PROBLEM 2·11. L o. ±L E 120~ 0 . 91 0 = 2448 watts If your answer is correct, go on to Problem 2-12. If your answer is not correct, review pp. 11 and 12. COMPENSATING CIRCUITS The problems in this section apply to a variety of electrical engineering specialities including control theory and transient analysis. Use is made of partial fractions (see Chapter 5) and the Laplace transform. PROBLEM 2-12. SIMPLE LAG CIRCUIT Derive the equation for output voltage across the capacitor of the simple lag circuit shown below after the switch is closed.