Problem le Capactive impedance WC JXL= jwL Vs. bcos (100t+35°) =6235°V > lasin (1004 + 35') = 12 cos(100++35°-90°) = 12 2-65° V W = 100 rad/sec jx300j100x300x 0.001 =j30 jxnoo =j100 x 400 x 0.001 = √40 jx200 = 3100 x 200+ 0.001 jxe= = jao 100x Sox 10 = -200 50~ 400 m H + лик VSI 300MH 200H aroar 3702 j40 jao Given circut www 70- 3j30 jaoo + Vsa node Vi V₁-V V₁ j40 J30 + + VI-6235' So j = O V₁ (- -)+ V = 435° U, (0.02-0.0583) + jo.025 v.0.0983+ 83+0.06883 - 1 Node Vo Vo-V₁ Vo + j40 200 + Vo-V₁ j40 V.-124-650 70+ jao =0 Vo Vo 122-650 + + -j200 70+ jao V₁ j40 +Volot j + 10-;20 122-650 200 7027 20° 72.801215.94 0.0250, + V. (0.0132-10.02377) = 0.1641-80.94° jooasv, +vo l0.0132 - j0,02377) = 0.026 - j0.163-2 Vo7.820-3.032 = 7.642-23.36 Vo=7647 cos (100+-23.36°

Problem le Capactive impedance WC JXL= jwL Vs. bcos (100t+35°) =6235°V > lasin (1004 + 35') = 12 cos(100++35°-90°) = 12 2-65° V W = 100 rad/sec jx300j100x300x 0.001 =j30 jxnoo =j100 x 400 x 0.001 = √40 jx200 = 3100 x 200+ 0.001 jxe= = jao 100x Sox 10 = -200 50~ 400 m H + лик VSI 300MH 200H aroar 3702 j40 jao Given circut www 70- 3j30 jaoo + Vsa node Vi V₁-V V₁ j40 J30 + + VI-6235' So j = O V₁ (- -)+ V = 435° U, (0.02-0.0583) + jo.025 v.0.0983+ 83+0.06883 - 1 Node Vo Vo-V₁ Vo + j40 200 + Vo-V₁ j40 V.-124-650 70+ jao =0 Vo Vo 122-650 + + -j200 70+ jao V₁ j40 +Volot j + 10-;20 122-650 200 7027 20° 72.801215.94 0.0250, + V. (0.0132-10.02377) = 0.1641-80.94° jooasv, +vo l0.0132 - j0,02377) = 0.026 - j0.163-2 Vo7.820-3.032 = 7.642-23.36 Vo=7647 cos (100+-23.36°

Introductory Circuit Analysis (13th Edition)

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...

See similar textbooks

Similar questions

A certain impedance is given by 1012° N. Does the current lead or lag the voltage, and by how many degrees? Fill in the blanks. The current the voltage by
Draw a seven-segment display, show the number highlighted in figure 1 (d3 -d0 are the input, a -g are the outputs). III. Simulation Wavetorm Editor - Cers/Ahmad Fainuz/Desanop/DLD Lab Practice/bedseg - beaeg - bcaseg 2018o03121936sim.vwwt Ee ep Ea w smutation search altera.com Haster Time Bar: ops Pointer: s29 us Start End iterval 529 us o ps 0 ps २ 2.56 us S12 us 75 us 10.24 us 12 s 15.36 us 172 us value at Name Ops do Figure 1
2-A resultant current is made of two components: a 10 A D.C. component and a sinusoidal component of maximum value 14.14. The average value of the resultant current is amperes (a) 0 (b) 24.14 (c) 10 (d) 4.14 and r.m.s. value is - amperes (e) 10 (f) 14.14 (g) 24.14 (h) 100
Determine the rms current in the figure for Vrms = 16 V and C = 74µF V f=10 kHz
From step1, how did you get that 50+(80*j60/80+j60) = 78.8+j38.4? also in step 2, how did you get that 5/(0.0125-0.0167) = 143.63 + j191.89?
Shown in the figure below is an "RC" circuit drive by an AC power source. The AC power source has an RMS voltage of Vps (RMS) = 11.58 Volts and is running at a frequency of f = 1.326e+04 Hz. resistor has a resistance of R = 3750 2 and the capacitor has an capacitance of C = 3.17e-09 Farads. Vps R ww Write the FORMULA for the total impedance of the circuit Ztot = Write the FORMULA for the phase of the total impedance of the circuit tot = Determine the numerical value of Ztot = Determine the numerical value of Pztot= Determine the current through the circuit: • I(PEAK) = • I(RMS) = Determine the voltage across the resistor: Amps Amps S2 C degrees
I PERFORm THE FOLLOWING OPERATIONS. (EXPRESS THE ANSWERS IN RECTANGULAR PORM, & Stow THE COMPLET SOLUTIONS). a) (4+j3) C6-je). (34j3) -(2+j3) c) (6 /20°.) (201-46")sjs) 2 30° 2. 860 140 Q LO) +J00 tj400) 3tj9
acras C relative Soura? An LRC senes circuit with R:14052 =D27mH, & C = \.9 MF is powered by an ac Voltage source of peak uoltage V9:250V frequency fa48O H&. A) Determine the peak cucrent that flaws in this circuit! B)Determine the phase angle of the Source voltage relative to the current? IDetermine the peak noHage acosi R DIDetermine the phase angle afthe valtage acress R relative to the sourse Peak তिाक F Phase ang le of the uaHage accast Lelative te the source oodage. across L? G Prak untage C. H) Phase anye of the voltage
QUESTION 5 A resistor, a capacitor, an inductor, and a mystery component are connected in parallel. The resistor has a resistance of 300N, the capacitor has a reactance of 2500, and the inductor has a reactance of 6000. The total impedance of the circuit is 59.52-j119.640. What is the mystery component? O there is no mystery component O resistor O inductor O capacitor
The sum of the following two e.m.fs will be e1 = 10 sin wt and e2 = 10 cos wt O 10 O 20 sin wt O 14.14 cos wt O 14.14 sin (wt + TT/4)
(b) Figure Q2 shows a power system network. The generators at Bus 1 and Bus 2 are represented by their equivalent current sources with the reactance in per unit on a 100MVA base. The lines are represented by n model where the series and shunt reactances are expressed in per unit on a 100MVA base as well. However, the load located at Bus 3 and Bus 4 are expressed in MW and MVar. Reconstruct the network impedance shown in the figure into the network admittance by assuming the voltage magnitude at Bus 3 and Bus 4 are 1.0 per unit, respectively. (i) (11) Calculate the bus admittance matrix of the system.
8. Desipn a phanse-lead empensatos s ame k that close-loop poles are at se-tjau5." chane Fhe rerf compansater ot $5-2. Sketeh the unid-s lap sespania of He campesalad aud mucompençaled rysdoms. s(0,55+1) Sheteh the