By S.A. Reza Zekavat
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For non-electrical engineering majors taking the creation to electric engineering course.
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Electrical Engineering: ideas and Applications is the results of a multi-disciplinary attempt at Michigan Technological collage to create a brand new curriculum that's beautiful, motivational, and correct to scholars via growing many application-based difficulties; and supply the optimum point of either diversity and intensity of assurance of EE themes in a curriculum package.
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By keeping the measured temperature error below a specified tolerance, the temperature of the feed to the column is kept at its optimum and costs are kept down. Net profit goes up as a result, and you get a big pat on the back—and maybe a big raise! Clearly, to do this on-the-job assignment, you must be able to apply EE skills and knowledge in a plant environment. As this scenario illustrates, you must often define the problem, select a solution strategy, and pick equipment to implement it. 1).
7, a sinusoid includes an amplitude. 2 represents the notation for the amplitude of the sinusoid. For details, refer to Chapter 6. An electric circuit consists of one or more sources. A source supplies power and energy to other elements. A source is also called an active element. All power supplied by active elements in a circuit is consumed, dissipated, or converted by other elements in the circuit. Elements that absorb energy are called passive elements. Passive and active elements in a circuit can be identified by considering the element’s voltage polarity relative to the element’s current direction.
In conclusion, by applying KCL, only four steps are needed to find an unknown current entering or leaving a node or a closed surface. The steps include the following: 1. 2. 3. 4. Determine currents entering the node (or closed surface). Determine currents leaving the node (or closed surface). Apply KCL to the nodes. Use the created set of equations to find the unknown current. 17. 4. SOLUTION • The current from only one branch enters Node 1—the one that passes through element A. Therefore, if we show the current entering Node 1 by iA, we can express this as: iA = 5A • The currents passing through elements B and C leave Node 1.