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Student solutions manual to accompany Engineering circuit by William Hart Hayt

By William Hart Hayt

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All Rights Reserved. CHAPTER THREE SOLUTIONS 16. 5 V. (b) By KVL, - VG + VGS + 2000ID = 0 Therefore, VGS = VG – 2(2) = -1 V. Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved. CHAPTER THREE SOLUTIONS 17. Applying KVL around this series circuit, -120 + 30ix + 40ix + 20ix + vx + 20 + 10ix = 0 where vx is defined across the unknown element X, with the “+” reference on top. Simplifying, we find that 100ix + vx = 100 To solve further we require specific information about the element X and its properties.

1], we find i1 = 1 A. (c) Applying KVL, -20 + 10i1 + 90 + 40i1 - 15 i1 = 0 Solving, i1 = - 2A. Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved. CHAPTER THREE SOLUTIONS 19. 8v3 v3 = -50i1 So that we may write Eq. 75 A. 5 V, no further information is required to determine its value. 6 W of power. Therefore, none of the conditions specified in (a) to (d) can be met by this circuit. Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc.

All Rights Reserved. CHAPTER THREE SOLUTIONS 13. 62 kW = 180 W = 360 W = 675 W = 405 W and it is a simple matter to check that these values indeed sum to zero as they should. Engineering Circuit Analysis, 6th Edition Copyright ©2002 McGraw-Hill, Inc. All Rights Reserved. CHAPTER THREE SOLUTIONS 14. Refer to the labeled diagram below. 5 = 24 W. 5 = 20 – 6 = 14 V therefore p14 = 142/ 14 = 14 W. 5 – 14/14 = 3 A Therefore v2 = 2(3) = 6 V and p2 = 62/2 = 18 W. 5 W. 5 = - IS, thefore IS = -1 A. 5 W. A quick check assures us that these power quantities sum to zero.

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