Electric Motor Control
10th Edition
ISBN: 9781133702818
Author: Herman
Publisher: CENGAGE L
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Textbook Question
Chapter 4, Problem 3SQ
What do the following abbreviations stand for?
- a. SPST
- b. SPDT
- c. DPST
- d. DPDT
- e. NO
- f. NC
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HANDWRITTEN SOLUTION NO
DO NOT USE AI NEED HANDWRITTEN SOLUTION
Each branch of a three-phase star-connected load
consists of a coil of resistance 4.2 Ω and reactance
5.6 Ω. The load is supplied at a line voltage of 400 V,
50 Hz. The total active power supplied to the load
is measured by the two-wattmeter method. Draw a
circuit diagram of the wattmeter connections and
calculate their separate readings. Derive any formula
used in your calculations.
ANS:
13.1 kW, 1.71 kW
Chapter 4 Solutions
Electric Motor Control
Ch. 4 - Identify the following symbols.Ch. 4 - Electrical symbols usually conform to which...Ch. 4 - What do the following abbreviations stand for? a....Ch. 4 - Single-pole and double-pole switch symbols are...Ch. 4 - The symbol shown is a a. polarized capacitor. b....Ch. 4 - The symbol shown is a a. normally closed float...Ch. 4 - The symbol shown is a(n) a. iron core transformer....Ch. 4 - The symbol shown is a a. normally open pressure...Ch. 4 - The symbol shown is a a. double-acting push...Ch. 4 - If you were installing the circuit in Figure 412,...
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- Three non-reactive loads are connected in delta across a three-phase, three-wire, 400 V supply in the following way: (i) 10 kW across R and Y lines; (ii) 6 kW across Y and B lines; (iii) 4 kW across B and R lines. Draw a phasor diagram showing the three line voltages and the load currents and determine: (a) the current in the B line and its phase relationship to the line voltage VBR; (b) the reading of a wattmeter whose current coils are connected in the B line and whose voltage circuit is connected across the B and R lines. The phase rotation is R–Y–B. Where would a second wattmeter be connected for the two-wattmeter method and what would be its reading? ANS: 21.8 A, 36°35′ lagging; 7 kW; 13 kWarrow_forwardNEED HANDWRITTEN SOLUTION DO NOT USE AI OR CHATGPTarrow_forwardA factory has the following load with power factor of 0.85 lagging in each phase. Between the red and yellow phases 40 A, between the yellow and blue phases 50 A, and between the blue and red phases 60 A. If the supply is 415 V, three-phase, calculate the line currents. Draw a phasor diagram for phase and line quantities. Ensure to draw all necessary diagrams ANS: IR = 87.178<-68.380 A; IY = 78.102<-178.120 A; IB = 95.394<61.210 A.arrow_forward
- Answer question D only using by hand first darw cylinder then calculate show me starrow_forwardThe phase currents in a delta-connected three-phase load are as follows: between the red and yellow lines, 30 A at p.f. 0.707 leading; between the yellow and blue lines, 20 A at unity p.f.; between the blue and red lines, 25 A at p.f. 0.866 lagging. Calculate the line currents and draw the complete phasor diagram. ANS: 21.6 A in R, 49.6 A in Y, 43.5 A in Barrow_forward. Two wattmeters connected to measure the input to a balanced three-phase circuit indicate 2500 W and 500 W respectively. Find the power factor of the circuit: (a) when both readings are positive; (b) when the latter reading is obtained after reversing the con nections to the current-coil of one instrument. Draw the phasor and connection diagrams. ANS: 0.655, 0.359arrow_forward
- Explain the advantage of connecting the low-voltage winding of distribution transformers in star. A factory has the following load with power factor of 0.9 lagging in each phase. Red phase 40 A, yellow phase 50 A and blue phase 60 A. If the supply is 400 V, three phase, four-wire, calculate the current in the neutral and the total active power. Draw a phasor diagram for phase and line quantities. Assume that, relative to the current in the red phase, the current in the yellow phase lags by 120° and that in the blue phase leads by 120°. ANS: 17.3 A, 31.2 kWarrow_forwardA three-phase, 400 V system has the following load connected in delta: between the red and yellow lines, a non-reactive resistor of 100 Ω; between the yellow and blue lines, a coil having a reactance of 60 Ω and negligible resistance; between the blue and red lines, a loss-free capacitor having a reactance of 130 Ω. Calculate: (a) the phase currents; (b) the line currents. Assume the phase sequence to be R–Y, Y–B and B–R. Also, draw the complete phasor diagram. ANS: 4.00 A, 6.67 A, 3.08 A, 6.85 A, 10.33 A, 5.8 Aarrow_forwardWith the aid of a circuit diagram, show that two wattmeters can be connected to read the total power in a three-phase, three-wire system. Two wattmeters connected to read the total power in a three-phase system supplying a balanced load read 10.5 kW and −2.5 kW respectively. Calculate the total active power. Drawing suitable phasor diagrams, explain the significance of: (a) equal wattmeter readings; (b) a zero reading on one wattmeter. ANS: 8 kWarrow_forward
- A factory has the following load with power factor of 0.9 lagging in each phase. Red phase 40 A, yellow phase 50 A and blue phase 60 A. If the supply is 400 V, three-phase, four-wire, calculate the current in the neutral and the total active power. Draw a phasor diagram for phase and line quantities. Assume that, relative to the current in the red phase, the current in the yellow phase lags by 120° and that in the blue phase leads by 120°.arrow_forwardFundimentals of Energy Systems Q8arrow_forwardTwo wattmeters are used to measure power in a three phase, three-wire network. Show by means of connection and complexor (phasor) diagrams that the sum of the wattmeter readings will measure the total active power. Two such wattmeters read 120 W and 50 W when connected to measure the active power taken by a balanced three-phase load. Find the power factor of the load. If one wattmeter tends to read in the reverse direction, explain what changes may have occurred in the circuit ANS: 0.815arrow_forward
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