By James W. Nilsson, Susan Riedel
For sophomore point, one- or two-semester Introductory Circuit research or Circuit conception classes taught in electric or desktop Engineering Departments. pointed out by means of IEEE Spectrum as an "up and coming vintage within the box of circuits,"Electric Circuits is the main ordinary introductory circuits textbook. This revision of either textual content and powerful supplementations package deal beneficial properties an elevated emphasis on pupil and teacher review, a redesigned paintings application, a brand new four-color layout, and plentiful new or revised difficulties all through. The 7th variation keeps the pursuits that experience made this article a best-seller: 1) to construct an realizing of strategies and ideas explicitly when it comes to past studying; 2) to stress the connection among conceptual figuring out and challenge fixing methods; three) to supply scholars with a powerful beginning of engineering practices
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Extra resources for Electric Circuits. Solution Manual
Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. 8 [a] 5 20 = 100/25 = 4 Ω 9 18 = 162/27 = 6 Ω Rab = 5 + 12 + 3 = 20 Ω 5 20 + 9 18 + 10 = 20 Ω 20 30 = 600/50 = 12 Ω 3–12 CHAPTER 3. 082 = 128 µ W 50 The equivalent resistance to the right of the 10 Ω resistor is (6 + 5 (8 + 12)) = 6 + 5 20 = 6 + 4 = 10 Ω. We can use current division to see that the current then splits equally between the two 10 Ω branches.
Thus the value of the voltage source must be 24 V. The model, valid for 0 ≤ is ≤ 24 A, is shown below: R= 2–26 CHAPTER 2. Circuit Elements [c] The circuit is shown below: Write a KVL equation in the clockwise direction, starting below the voltage source. 25 Ω [d] The circuit is shown below: Write a KVL equation in the clockwise direction, starting below the voltage source. 25 Ω [e] The short circuit current can be found in the table of values (or from the plot) as the value of the current is when the voltage vs = 0.
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.