Chapter 1: Basic ConceptsEGR214 W’98 1Homework 2 Solutionspp. 59-65Problem # 3:(a) Using the current divider rule: Problem # 10KCL for the Gaussian surface: . Thus, IoutIin14Ω-------14Ω-------11Ω-------12Ω-------++-------------------------------------7A0.250.25 1 0.5++---------------------------------1A===Iin1Ω1Ω3Ω2ΩIoutI1I25A8A7A6AI15I178+++ 0=I120A–=Chapter 1: Basic ConceptsEGR214 W’98 2Problem # 11NOTE: When choosing current directions and voltage polarities, stick to the convention that current flows from positive voltage polarity to negative voltage polarity. This will allow you to compute the dissipated power using P=VI, and to find out if the device is generating (P<0) or absorbing (P>0) power.KCL @A: . Thus, KCL @C: . Thus, KCL @B: . Thus, KVL @(G-A-B-G): ⇒ KVL @(G-B-C-G): ⇒ KVL @(G-B-A-C-G): ⇒ Dissipated power: Pi = Vi IiP1 = (5 I1)(I1) = 5 (-1)2 = 5WP2 = 4V I2 = 4V (-9A) = -36WP3 = (2 I3) I3 = 2 (15)2 = 450WP4 = 1V 6A = 6W This means that the current source is absorbing power. We will see that non-ideal current and voltage sources dissipate power in the resistors that make up their internal circuitry.P5 = (-31V) 7A = -217WP6 = (-26V) 8A = -208W6A4V+-8A2Ω5Ω7AI1I2I3+-V3V1+-V5+--V4V6+-ACBG+6I17+=I11A–=78+I3=I315A=I1I28+=I29A–=V4V14V++ 0=V4V1–4–5–I14–1V===4V–V6V3++ 0=V6V–3 4+2–I34+26V–== =4V–V1–V5V3++ 0=V545I12I3–+451–()215()–+31V–== =Pdissipatedtotal∑536– 450 6 217– 208–++ 0==Chapter 1: Basic ConceptsEGR214 W’98 3Problem # 13(a) Ohm’s Law for series 12Ω + 20Ω: ⇒ Ohm’s Law for series 12Ω: KVL starting from GND: ⇒ (b) KVL starting from GND: Since ⇒ ⇒ Iin = 0.25A, a = 4 and ⇒ ⇒ ⇒ Problem # 37(a) KVL starting from GND in the direction of Iin: ⇒ V220 12+()Iin24V==Iin2432------0.75A==V112Iin12 0.75()9V== =30V–RIin20Iin12Iin++ + 0=R30 32Iin–Iin------------------------30 24–0.75------------------8Ω===30V–aV1RIin20IinV1+++ + 0=V112Iin=aV1R20+()V112-------V1++30=V130a1R20+()12--------------------++---------------------------------------360a32R++-------------------------==30V–aV1RIin20IinV1+++ + 0=a12Iin()R20+()Iin12Iin++30=RIinIin20 12 12a++()+30=R30Iin------20 12 12a++()– 120 80–40Ω===60–30Iin30 20 60–40Iin+++ + 0=Iin7070------1A==R30V1A----------30Ω==60V30ΩIin+R-+-+
View Full Document