DELMAR'S STANDARD TEXT OF ELECTRICITY
DELMAR'S STANDARD TEXT OF ELECTRICITY
6th Edition
ISBN: 9780357323380
Author: Herman
Publisher: CENGAGE C
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Chapter 9, Problem 10RQ
To determine

Value of Voltage drop E3

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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.
: 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 Sm
I 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 + R2
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