Chap_2_Solut_Dorf

Chap_2_Solut_Dorf - Chapter 2 Circuit Elements Exercises Ex...

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Chapter 2 Circuit Elements Exercises Ex. 2.2-1 () 12 1 2 11 superposition is satisfied homogeneity is satisfied Therefore the element is linear. mi i m i m i ma i am i += + =⇒ Ex. 2.2-2 ( ) ( ) 1 2 1 2 superposition is not satisfied Therefore the element is not linear. m i i b mi mi b mi b mi b ++ = + + + + + Ex. 2.4-1 ( ) 2 2 10 1 W 100 v P R == = Ex. 2.4-2 22 2 (10 cos ) 10 cos W 10 vt Pt R = Ex. 2.7-1 1.2 A, 24 V 4 ( 1.2) 4.8 A cd d iv i = −= =− = i d and v d adhere to the passive convention so (24) ( 4.8) 115.2 W dd Pv i = = is the power received by the dependent source

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Ex. 2.8-1 45 , 2 mA, 20 k p IR θ ° = == Ω 45 (20 k ) 2.5 k 360 360 aa R p = ⇒= Ω 33 (2 10 )(2.5 10 ) 5 V m v × = Ex. 2.8-2 A 10 V, 280 A, 1 for AD590 K K (280 A) 1 280 K A vi k i ik T T k μ = ° ° ⎛⎞ ° ⎜⎟ =⇒ = = = ⎝⎠ Ex. 2.9-1 At t = 4 s both switches are open, so i = 0 A. Ex. 2.9.2 At t = 4 s the switch is in the up position, so = (2 mA)(3 k ) = 6V R = Ω . At t = 6 s the switch is in the down position, so v = 0 V.
Problems Section 2-2 Engineering and Linear Models P2.2-1 The element is not linear. For example, doubling the current from 2 A to 4 A does not double the voltage. Hence, the property of homogeneity is not satisfied. P2.2-2 (a) The data points do indeed lie on a straight line. The slope of the line is 0.12 V/A and the line passes through the origin so the equation of the line is 0.12 vi = . The element is indeed linear. (b) When i = 40 mA, v = (0.12 V/A) × (40 mA) = (0.12 V/A) × (0.04 A) = 4.8 mV (c) When v = 4 V, 4 33.3 A 0.12 i == P2.2-3 (a) The data points do indeed lie on a straight line. The slope of the line is 256.5 V/A and the line passes through the origin so the equation of the line is 256.5 = . The element is indeed linear. (b) When i = 4 mA, v = (256.5 V/A) × (4 mA) = (256.5 V/A) × (0.004 A) = 1.026 V (c) When v = 12 V, 12 0.04678 256.5 i A = 46.78 mA. P2.2-4 Let i = 1 A , then v = 3 i + 5 = 8 V. Next 2 i = 2A but 16 = 2 v 3(2 i ) + 5 = 11. . Hence, the property of homogeneity is not satisfied. The element is not linear. P2.2-5 (a) 0.4 3.2 V 10 40 8 vvv v =+= = 0.08 A 40 v i (b) 2 2 0.4 0.8 0 10 2 5 vv v v =+ + = Using the quadratic formula 0.2 1.8 0.8, 1.0 V 2 v ± When v = 0.8 V then 2 0.8 0.32 A 2 i . When v = -1.0 V then () 2 1 0.5 A 2 i .

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(c) 2 2 0.4 0.8 0.8 0 10 2 5 vv v v =++ + + = Using the quadratic formula 0.2 0.04 3.2 2 v −± = So there is no real solution to the equation.
Section 2-4

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Chap_2_Solut_Dorf - Chapter 2 Circuit Elements Exercises Ex...

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