HW2Solution

HW2Solution - EE202 Circuit Analysis 1 Homework – Chapter...

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Unformatted text preview: EE202 Circuit Analysis 1 Homework – Chapter 2 Solution 2.3 The interconnection in Fig. P2.3 is not valid. 4A and 5A current sources in the rightmost branch do not supply the same current in the same direction. 2.9 = 80 + = −60 − = 80 + 100 = 180 = −60 − 100 = −160 = −60 × 12 = −720 = 160 × 12 = 1920 = −60 × =− =− =− × × 4 = −180 × 4 = −720 = 80 × 4 = 320 = −100 × 2 × 4 = −800 ×2 = 80 × = 720 + 720 + 800 = 2240 2.13 = no-load voltage of battery = internal resistance of battery = resistance of wire between battery and switch = resistance of wire between switch and lamp A = resistance of lamp A = resistance of lamp B = resistance of wire between lamp A and lamp B = resistance of frame between battery and lamp A = resistance of frame between lamp A and lamp B = switch 2.14 a) = Δ 420 − 100 = = 20 Ω Δ 16 − 0 = 0, When = 100 . Thus, the ideal current source is 100 20 = 5 . b) 5+ = ⇒ × 20 = − × 5 Ω =1 = −4 × 5 = 80 = × 5 = −4 2.17 a) b) = Δ 75 − 30 = = 200 Ω Δ 0.225 − 0 = 0, = 75 . Thus, the voltage source is 75 . When c) = d) 75 = 125 200 + 400 = 75 = 375 200 e) From table, the actual short-circuit current is 500 mA. f) The plot of voltage versus current constructed in part (a) is not linear. Since the proposed circuit model is a linear model, it cannot be used to predict the nonlinear behavior exhibited by the plotted data. 2.18 a, b) × 20 = × 80 50 = + ×4+ c) = d) Ω × 20 ⇒ =2 = 0.5 × 80 = 0.5 × 80 = 40 + × 4 = 2.5 × 4 = 25 = = = Ω Ω × 20 = 2 × 80 = 0.5 + × 20 = 80 × 80 = 20 = 50 × 2.5 = 125 e) = 50 × 2.19 a) × 80 = 4= + b) = c) =− Ω Ω Ω × 30 + 90 ⇒ = 2.4 = 1.6 × 80 = 2.4 × 80 = 192 × 4 = −192 × 4 = −768 × 80 = 2.4 × 30 = 1.6 × 90 = 1.6 = 768 = 460.8 + 76.8 + 230.4 = 768 × 80 = 460.8 × 30 = 76.8 × 90 = 230.4 = = = Σ Σ 2.23 = = 240 = + = 240 = = = =5 . = + ⇒ + = = = − × 18 = 5 × 18 = 90 . + 60 + 90 ⇒ =5−2=3 = 90 . . = = = + = = =2 . =5 . + = + + = × 10 = 3 × 10 = 30 . = =1 . = 30 + 60 + 90 = 180 . + 180 ⇒ = 60 . =3+1=4 60 = 15 Ω 4 2.29 a) = 0 because no current can exist in a single conductor connecting two parts of a circuit. b, c) = = 10 . = × 5 = 10 × 50 × 5 = 50 ⇒ = −60 = −240 × 2000 = × 500 + = −6 × 10 = −6 × 10 2.36 a) = = 250 400 + 50 + 200 250 400 + 50 + 200 = b) = 2.39 × 10 × = = 5 = 1414.23 2.39 × 10 × 59.17 4 250 400 + 50 + 200 × 400 = 59.17 × 200 = 29.59 × 50 = 7.40 = 2.39 × 10 = × 5 = 7071.13 2.39 × 10 × 29.59 10 = 5 = 70422.54 2.39 × 10 × 7.40 25 2.39 × 10 × c) All values are much greater than a few minutes. ...
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This document was uploaded on 06/11/2010.

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