ECE35 Homework 2 Solutions for Chapter 3 Problems, Assignments of Electrical and Electronics Engineering

Download ECE35 Homework 2 Solutions for Chapter 3 Problems and more Assignments Electrical and Electronics Engineering in PDF only on Docsity! ECE35 Win’10 Homework 2 Solutions Assignment: Finish reading chapter 3, and do at least the following problems: Problems: 3.2-8, 3.2-9, 3.2-10, 3.3-4, 3.3-7, 3.3-8, 3.4-4, 3.4-9, 3.4-14, 3.5-1, 3.6-1, 3.6-8, 3.6-15, 3.6-28, Design Problem 3-3 P3.2-8 P3.2-9 P3.2-10 KCL at node b: KCL at node a: P3.3-4 P3.3-7 All of the elements are connected in series. Replace the series voltage sources with a single equivalent voltage having voltage 12 + 12 – 18 = 6 V. Replace the series 15 Ω, 5 Ω and 20 Ω resistors by a single equivalent resistance of 15 + 5 + 20 = 40 Ω. By voltage division P3.3-8 Use voltage division to get Then The power supplied by the dependent source is given by P3.6-15 So the circuit is equivalent to Then P3.6-28 Replace the ammeter by the equivalent short circuit and label the current measured by the meter as im. The 10-Ω resistor at the right of the circuit is in parallel with the short circuit that replaced the ammeter so it’s voltage is zero as shown. Ohm’s law indicates that the current in that 10-Ω resistor is also zero. So im = 8va Look at the upper part of the circuit. We need to know the voltage va, so first find the current through that path; call it ia. Current divider: ia = 3A * { 3 Ω / (3 Ω +6 Ω) } = 1A Then the voltage across the 4 Ω resistor is va = 4Ω *1A = 4V, And im = 8v = 8 (A/V) * 4V = 32A DP3-3 V,, = 200 mV 10 10 = OV,, =—— (120) (0.2 *"T04R ° 1o+R ) 2) letv=16 = 240 = R=5Q 10+R 16 “. P =— = 25.6W 10