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Electric Circuits. Solution Manual by James W. Nilsson, Susan Riedel

By James W. Nilsson, Susan Riedel

For sophomore point, one- or two-semester Introductory Circuit research or Circuit concept classes taught in electric or computing device Engineering Departments. brought up by way of IEEE Spectrum as an "up and coming vintage within the box of circuits,"Electric Circuits is the main usual introductory circuits textbook. This revision of either textual content and strong supplementations package deal positive factors an elevated emphasis on pupil and teacher review, a redesigned artwork application, a brand new four-color structure, and considerable new or revised difficulties all through. The 7th variation keeps the targets that experience made this article a best-seller: 1) to construct an realizing of innovations and ideas explicitly when it comes to earlier studying; 2) to stress the connection among conceptual realizing and challenge fixing methods; three) to supply scholars with a robust origin of engineering practices

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10 The interconnection is invalid because KCL is violated at the right-hand node. Summing the currents leaving, −(−5A) − 3A + 8A = 10A = 0 Note that KCL is also violated at the left-hand node. 11 Write the two KCL equations, summing the currents leaving the node: KCL, top node: 75A − 5v∆ − 25A = 0A KCL, bottom node: − 75A + 5v∆ + 25A = 0A To satisfy KCL, note that v∆ = 10 V. Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop: KVL, right loop: KVL, outer loop: − v75 − 50V + vdep − 20V = 0 20V − vdep + v∆ = 0 − v75 − 50V + v∆ = 0 2–16 CHAPTER 2.

There are many other voltage values that will satisfy the equations, too. Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. We can now calculate the power developed by the two voltage sources: pv−sources = p60 + p100 = −(60)(5) + (100)(5) = 200 W. Since the power is positive, the sources are absorbing 200 W of power, or developing −200 W of power. 7 Write the two KCL equations, summing the currents leaving the node: KCL, top node: − 30A − i8 + 10A = 0A KCL, bottom node: 30A + i8 − 10A = 0A Note that the value i8 = −20A satisfies these two equations.

We use the values of i8 , v30 and v10 stated above to calculate the power associated with each source: p30A = −(30)(−8) = 240 W p16V = −(30)(16) = −480 W CHAPTER 2. Circuit Elements 2–14 p8V = −(−20)(8) = 160 W p10V = −(10)(10) = −100 W p10A = (10)(18) = 180 W Pabs = Pdel = 580 W Power developed by the current sources: pi−sources = p30A + p10A = 240 + 180 = 420 W Since power is positive, the sources are absorbing 420 W of power, or developing −420 W of power. 8 The interconnect is valid since it does not violate Kirchhoff’s laws.

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