HW5solutionsWIN10

# HW5solutionsWIN10 - Use the negative resistance converter,...

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ECE35 Winter 2010 Homework 5 Solutions Chapter 6: 6.3-1, 6.3-7, 6.4-4, 6.4-5, 6.4-6, 6.4-11, 6.4-17, 6.5-6; Begin Chapter 7: 7.2-13, 7.3-1 (no plot) P6.3-1 P6.3-7

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P6.4-4 Ohm’s law: KVL: P6.4-5
P6.4-6 KCL at node b: KCL at node a: This is correct. So . Nope; it’s +15/13V. And the numeric answers in the book are wrong, too. P6.4-11 Node equations: and so with the given values:

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P6.4-17 Represent this circuit by node equations. and So We require Try then e.g. P6.5-6

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Unformatted text preview: Use the negative resistance converter, entry ( h ) in Figure 6.6-1. P7.2-13 We’ll write and solve a node equation. Label the node voltages as shown. Apply KCL at node a to get so Then P7.3-1 Given The capacitor voltage is given by For In particular, For In particular, For Now the power and energy are calculated as ___________________ Actually, it’s 0.04(t-2) 3 for 2<t<6 and _______________ Actually, it’s 0.64(t-4) 2 for 6<t Note: graphs were not required....
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## This note was uploaded on 02/16/2010 for the course ECE ECE 35 taught by Professor Josephford during the Winter '10 term at UCSD.

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HW5solutionsWIN10 - Use the negative resistance converter,...

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