HW2Solution

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

## This document was uploaded on 06/11/2010.

Ask a homework question - tutors are online