EBK ELECTRICAL WIRING RESIDENTIAL
19th Edition
ISBN: 9781337516549
Author: Simmons
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 20.1, Problem 5R
a.
To determine
Find whether the given statement is true or false.
b.
To determine
Find whether the given statement is true or false.
c.
To determine
Find whether the given statement is true or false.
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Section 15-3 reversing motors using magnetic starters.Section 15-3Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
First 1. Wire the motor to operate in forward and reverse at 115 VAC.
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Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
First 1. Wire the motor to operate in forward and reverse at 115 VAC.
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- B:A 20 MVA transformer which may be called upon to operate at 30% overload, feeds 11 KV busbars through a circuit breaker: other circuit breakers supply outgoing feeders. The transformer circuit breaker is equipped with 1000/5 A CTS and the feeder circuit breakers with 400/5 A CTS and all sets of CTs feed induction type over current relays. The relays on the feeder circuits breakers have a 125% plug setting, and 0.3 time setting. If 3 ph fault current of 5000 A flows from the transformer to one of the feeders, find the operating time of the feeder relay, the minimum plug setting of the transformer relay and its time setting assuming a discrimination time margin of 0.5 sec. Relays having the following characteristics for TMS=1 PSM T in sec. 2 3.6 5 10 15 20 10 6 3.9 2.8 2.2 2.1arrow_forward10.34 Determine the power readings of the two wattmetersshown in the circuit of Fig. P10.34 given that ZY = (15− j5) Warrow_forward10.29 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.arrow_forward
- 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?arrow_forwardCable 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_forward
- 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 + R2arrow_forward10.2 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_forward
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