EBK MATLAB: AN INTRODUCTION WITH APPLIC
5th Edition
ISBN: 8220102007642
Author: GILAT
Publisher: YUZU
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Chapter 10, Problem 3P
The ladder of a fire truck can be elevated (increase of angle
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10.34 Determine the power readings of the two wattmetersshown in the circuit of Fig. P10.34 given that ZY = (15− j5) W
10.29 A 208-V (rms) balanced three-phase source supports twoloads connected in parallel. Each load is itself a balanced threephaseload. Determine the line current, given that load 1 is 12 kVAat pf 1 = 0.7 leading and load 2 is 18 kVA at pf 2 = 0.9 lagging.
10.31 A 240-V (rms), 60-Hz Y-source is connected to a balancedthree-phase Y-load by four wires, one of which is the neutral wire.If the load is 400 kVA at pf old = 0.6 lagging, what size capacitorsshould be added to change the power factor to pf new = 0.95lagging?
Chapter 10 Solutions
EBK MATLAB: AN INTRODUCTION WITH APPLIC
Ch. 10 - Prob. 1PCh. 10 - Prob. 2PCh. 10 - The ladder of a fire truck can be elevated...Ch. 10 - Prob. 4PCh. 10 - Prob. 5PCh. 10 - a 3-D mesh plot of the function z=cos2Re0.2R where...Ch. 10 - Prob. 7PCh. 10 - Prob. 8PCh. 10 - Prob. 9PCh. 10 - Molecules of a gas in a container are moving...
Ch. 10 - Plank’s distribution law gives the black-body...Ch. 10 - Prob. 12PCh. 10 - An RLC circuit with an alternating voltage source...Ch. 10 - 14. A defect in a crystal lattice where a row of...Ch. 10 - 15. The current I flowing through a semiconductor...Ch. 10 - The equation for the streamlines for uniform flow...Ch. 10 - 18. The deflection w of a clamped circular...Ch. 10 - The Verhulst model, given in the following...Ch. 10 - The geometry of a ship hull (Wigleyhull) can be...Ch. 10 - Prob. 20PCh. 10 - 22. A ball thrown up falls back to the floor and...
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- Cable A Cable A is a coaxial cable of constant cross section. The metal regions are shaded in grey and are made of copper. The solid central wire has radius a = 5mm, the outer tube inner radius b = 20mm and thickness t = 5mm. The dielectric spacer is Teflon, of relative permittivity &r = 2.1 and breakdown strength 350kV/cm. A potential difference of 1kV is applied across the conductors, with centre conductor positive and outer conductor earthed. Before undertaking any COMSOL simulations we'll first perform some theoretical analysis of Cable A based on the EN2076 lectures, to make sense of the simulations. Calculate the radial electric field of cable A at radial positions r b. Also calculate the maximum operating voltage of cable A, assuming a safety margin of ×2, and indicate where on the cable's cross section dielectric breakdown is most likely to occur.arrow_forward: For the gravity concrete dam shown in the figure, the following data are available: The factor of safety against sliding (F.S sliding)=1.2 Unit weight of concrete (Yconc)=24 KN/m³ - Neglect( Wave pressure, silt pressure, ice force and earth quake force) μ=0.65, (Ywater) = 9.81 KN/m³ Find factor of safety against overturning (F.S overturning) 6m3 80m Smarrow_forwardI need help checking if its correct -E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a) R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b) R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c) (R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d) Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be: E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a) -R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b) -R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c) R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d) These two equations will be implemented to the matrix formula I = inv(A) * b R11 R12 (R1 + R3 + R4) -R4 -R4 R4 + R2arrow_forward
- 10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ Varrow_forwardApply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) Warrow_forward10.8 In the network of Fig. P10.8, Za = Zb = Zc = (25+ j5) W.Determine the line currents.arrow_forward
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- B. For the oscillator circuit shown in frequency, feedback and open-loop gains. +10 V name the circuit, derive and find the oscillation P.Av +9 V -000 4₁ 5 mH w R₁ C₂ HH 1 με w 100 pF R₂ T R CA www. 470 pF w ww www 1000 pF HH 1μF C₁ HH 1μF Ra ww HI 4₁ 000 1.5 mH H 4 AF 000 10 mHarrow_forwardI want to check if the current that I have from using the mesh analysis is correct? I1 = 0.214mA I2 = -0.429mAarrow_forwardI want to find the current by using mesh analysis pleasearrow_forward
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