referring to figure 15.1 the following data are given for the two voltage and the three impedance 418ELECTRICAL CIRCUITS-ALTERNATING CURRENTthree impedances: E; = 120/30°, E: = 120/-30°, 21 = 2 ohms (resistance)Z; = 2/90° ohms (inductive reactance), Z = 10/-53.1°. CRlculate the vof the load current It and the voltage across the load, using the superpositiontheorem.5. Referring to Fig. 15-2a, interchange the sources of emf E, and E, makingE, = 120/-30° and E, 120/30°, and calculate the values of IL, EL, and P,8. The following particulars are given for the T-network of Fig. 15.4: E = 120/02Z. = 40 ohms (resistance), Z, = (15.6 + j19.2) ohms (resistance-inductive resc-tance), Z. = -j30 ohms (capscitive reactance), ZL = -j40 ohms (capacitiveresctance). Using Thévenin's theorem calculate the current through and thevoltage across the load impedance ZL.7. Using Thévenin's theorem calculate the load current and voltage, given the follow-ing dets: E = 120/0°, Z. = Z, = j9, Z. = 12, Zz = (3.68 - j8.76). (Refer toFig. 15:4.)8. Using the dats of Prob. 7, calculate the short-circuit current I, as illustratedin Fig. 15-5, for use in Eq. (145) (page 402); after evaluating this check the loadcurrent as determined by Thevenin's theorem in Prob. 7.9. The following information is given. in connection with Fig. 15-4: E = 120,0°,Z. = 12/-90°, Z, = 12/90°, Z. = 12, ZL = 8.5/45°. Calculate the load er-rent Is and the load voltage EL, using Thévenin's theorem.20. Solve Prob. 9 using Norton's theorem.11. Referring to Prob. 9, assume the same values for E, Z,, Z, and Z. If three losdimpedances Zı = S.5/45°, ZL: = 8.5/-45°, and Zz: = 8.5/0° are connected tothe output terminsls, calculate the values of I Lı, ILa, and IL:.A8. The current I = 15/-30° when E = 120/10° in Fig. 15.8a. Whst will be thecurrent I in Fig. 15-85 if E = 110/40°?18. In Fig. 15.8a the current I = 4.33 - j2.5 when E 110/0°. For what voltagein Fig. 15.8b will the current I be 6/30°?A. When switch S: is open and switch S, is closed in Fig. 15-9, the currents areI = (8 + j5) and I: (4 -j3). Calculate the current I when switch S: isopen and (a) switch S: is closed to the right, and (b) switch S; is closed to the left.16. A 48-volt battery has an internal resistance of 0.22 ohm and-is connected to avariable-resistance load through a line resistance of 1.28 ohms. For what valueof load resistance will the load power be a maximum, snd what will be the loidcurrent, load power, and power loss under this condition?16. A 228-volt constant-potential generator delivers load to a variable impedancewhose values ofR and X are readily adjustable. Assuming a "looking-bsek"impedance of (6 + j4.5) ohms calculate, for maximum load power, (a) the losdimpedance, (b) the current, (c) the load power and power factor.17. If the load in Prob. 16 is a variable resistor calculate, for maximum load power,the load resistance and power.18. If the load in Prob. 16 contains a constant inductive reactance of 3.5 ohms and avariable resistance, calculate, for maximum load power, (a) the load resistanceand impedance, (b) the load current and power.19. If the load in Prob. 16 contains a constant resistance of 4.8 ohms and a variavieinductive reactance, calculate, for maximum load power, (e) the load reaetanceand impedance, (b) the lond current and power.20. Referring to Fig. 15-13a, transfora the upper delta into an equivalent stardetermine the total current I as in Example 13. them so that they correspond to properly labeled minus (-) andsimultaneously to yield the three currents.eas-tilled tubes, are unilateral, wherea resistors, induetors, anda done for d-e eircuits. However, when there are two or moreurrent and voltage equations are set up in essentially the same wayeinclade silicon, selenium, and copper-oxide unita as well as vacuume of it will be necessary to take the added preeaution tos14)aymbols on the eireuit diagrams; this is essential for the pur-near.305All types of reetifiers, andand iron-coretors,eed-Thisas inlars,rebraich(+)(+)the(-)(-)prop-aWs,ages,mple,ed tosteadstate-5, andLition,artheroe impedancesa pair of line impedances. Note particularly the polaritythroughaeines on the generators since they relate to the angles attached tonolar form of the voltages. Specifically E/a refers to a voltage riseh the generator from a to b, i.e., from minus to plus; also E/8s to a voltage rise through the generator from e to d, i.e., fromnus to plus. If, for example, E1/aetangular form will be (86.6 + j50) volts. Conversely,-(E/a) repre-ts & voltage through the generator from b to a, i.e., from plus to minus.that-(100/30°) volts is, in the rectangular form, (-86.6-j50) volta.Belerring to Fig. 15-1, assume that it is desired to determine the phasorrents I1, I, and I, the indicated arrow directions of which are cbosenanly. Since IL is the geometric sum of I, and I, two voltage equa-e may be written as follows:100/30° volts, its correspondingoff toced asa of ance volt-sentedilateralis pro-not foructors,as-filledE/a- I,Z1- ILZ 0E/B- (IL-1)Z, - IZ=0lved

Answered: current I in Fig. 15-85 if E = 110/40°?…

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...