DELMAR'S STANDARD TEXT OF ELECTRICITY
6th Edition
ISBN: 9780357323380
Author: Herman
Publisher: CENGAGE C
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 6, Problem 4PA
To determine
The ohmic value and minimum power rating of each resistor.
Expert Solution & Answer
Trending nowThis is a popular solution!
Students have asked these similar questions
3.) The time delay fuse using the strict standard rating for 2Ø, 230 V, 7.5 hp SCIM motor is
A. 25 A B. 30 A C. 35 A D. Any of the above
is the DC motor in figure 1 turning in the same direction as that in figure 3 ?
a- yes
b- no
A direct current motor is connected to a 250-volt dc supply. The armature has a current draw of 165 amps when operating on a full load. You have been assigned the task of connecting two resistors in the armature circuit to provide speed control for the motor. When both resistors are connected in the circuit, the armature is to be limited to 50% of the full-load current draw. When only one resistor is connected in the circuit, the armaturecurrent is to be limited to 85% of full-load current. Determine the ohmic value and minimum power rating of each resistor.
Chapter 6 Solutions
DELMAR'S STANDARD TEXT OF ELECTRICITY
Ch. 6 - A series circuit has individual resistor values of...Ch. 6 - A series circuit contains four resistors. The...Ch. 6 - A series circuit contains five resistors. The...Ch. 6 - A circuit has three resistors connected in series....Ch. 6 - A circuit has four resistors connected in series....Ch. 6 - Define a series circuit.Ch. 6 - State the three rules for series circuits.Ch. 6 - A series circuit has resistance values of...Ch. 6 - If a voltage of 24 V is applied to the circuit in...Ch. 6 - Referring to the circuit described in Questions 8...
Ch. 6 - A series circuit contains the following values of...Ch. 6 - Prob. 1PACh. 6 - Three wire-wound resistors have the following...Ch. 6 - You are an electrician working in an industrial...Ch. 6 - Prob. 4PACh. 6 - Using the three rules for series circuits and Ohms...Ch. 6 - Use the general voltage divider formula to...Ch. 6 - Find all missing values for the circuit shown in...Ch. 6 - Find all missing values for the circuit shown in...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Complete the wiring diagrams as per Worksheets 5-A and 5-B then solve for "A" through "G" for the DC motor on Worksheets 5-A and 5-B when it is running at capacity and its specifications are as follows:Rated input = 125 volts, 16 ampsArmature Resistance = 0.3 ohms(Ω)Shunt Field Resistance = 50 ohms (Ω)Series Field Resistance = 0.6 ohm (Ω)Rheostat Resistance = 12.5 ohm (Ω)A) Voltage drop of armature (IR-arm)a) 4.05 V b) 120.95 V c) 125 V d) 129.5 VB) Voltage generated by motor (E-cemf)a) 4.05 V b) 120.95 V c) 125 V d) 129.5 VC) power consumed by shynt field(p-shunt)a) 54.68 W, b) 312.5 W, c) 1632.83 W, d) 2000 WD) power consumend by armature resistance( P-arm)a) 54.68 W , b) 312.5 W, c) 1632.83 W , d) 3000 Warrow_forwardComplete the wiring diagrams as per Worksheets 5-A and 5-B then solve for "A" through "G" for the DC motor on Worksheets 5-A and 5-B when it is running at capacity and its specifications are as follows:Rated input = 125 volts, 16 ampsArmature Resistance = 0.3 ohms(Ω)Shunt Field Resistance = 50 ohms (Ω)Series Field Resistance = 0.6 ohm (Ω)Rheostat Resistance = 12.5 ohm (Ω)A) Shunt field current (I-shunt)a) 2.5 A b) 13.5A c) 16 A d) 18.5 AB) Armature current (I-arm)a) 2.5 A b) 13.5A c) 16 A d) 18.5 AC) Voltage drop of shunt field (I-shunt)a) 4.05 V b) 120.95 V c) 125 V d) 129.5 VD) Voltage drop of armature (IR-arm)arrow_forwardassume that a series motor is taking the same current with the conditions that 440v motor takes a 25 a and runs at 800 rpm the armature resistance including the resistance of the series winding is 0.77ohms calculate the speed will it run if a 1.88 resistance is connected in series with the supply voltagearrow_forward
- Explain how does the current will be reduced during the starting of the motor.arrow_forwardA series motor runs at 800rpm when the voltage is 400v and the current is 25 ampere the armature resistance is 0.4ohm and the series field resistance is 0.2ohm determine the resistance to be connected in series to reduce the speed to 600rpm with same currentarrow_forwardA 400V, 60HZ, 6-pole wye connected , 30 induction motor draws 75KW with a line current of 100A. The core loss is 2KW, Friction and Windage loss is 1.2KW. The stator resistance between two terminals is 0.352. What is the efficiency if the motor runs at a slip of 2.5%?arrow_forward
- A DC series motor of resistance 1 ohm across terminals runs at 1000 rpm at 250 V taking a current of 20 A. When an additional resistance of 6 ohm is inserted in series and taking the same current, the new speed would be ?arrow_forwardA 220 V shunt motor driving its normal load draws an armature current of 50 A from a 220 V dc source. The armature resistance of this motor including brushes is 0.25 ohms. How much armature current (minimum) Will this motor draw from a 200 V DC source when driving the same load with the field adjusted to maintain the same speed.arrow_forwardEA5 kW, 20O V dc shunt motor has armature resistance of 1 ohm and shunt field resistance of 100 ohm. At no load, the motor draws 6 A from 200 V supply and runs at 1000 rpm. What is the total copper loss of the machine? (a) 400 W (c) 36 W (b) 16 W (d) 416 Warrow_forward
- Complete the wiring diagrams as per Worksheets 7-A and 7-B then solve for "A" through "K" for the DC motor on Worksheets 7-A and 7-B when it is running at capacity and its specifications are as follows:Rated input = 125 volts, 16 ampsArmature Resistance = 0.3 ohms(Ω)Shunt Field Resistance = 50 ohms (Ω)Series Field Resistance 0.6 ohms (Ω)Rheostat Resistance = 12.5 ohms (Ω)A)power consumed by series field(P-series)a)56.58W b)153.6W c)265.42W d)1523.WB)power consumed by shunt field(P-shunt)a)56.58W b)153.6W c)265.42W d)1523.56WC)power consumed by armature resistance(P-arm)a)56.58W b)153.6W c)265.42W d)1523.56WD)percent efficiency of the machine( % eff) if rotational loses are 850 Wa)30% b)64% c)86% d)99%arrow_forwardPlease answer part B 1. This is a question about a wind turbine generator a) The blades of a 20 kW, 400 V machine rotate at 2 revs per second at rated wind speed ) Calculate the speed of the generator if the gear box ratio is 1:10 i Calculate the current generated at rated power ) This four pole machine has a lap wound armature consisting of 16 coils each with a resistance of 0.3 ohms. Calculate the resistance of the armature. iv) Draw the circuit diagram for the generator with this armature winding resistance and the shunt resistance is 50 ohms V) Calculate the winding losses and hence the efficiency of the wind turbine at rated wind speed. b) The actual wind speed is lower than rated, so that the blades rotate at half the speed. The output voltage is now 190 V i) What is the induced emf at rated speed? ii) What is the emf at actual speed iii) What is the actual output current? iv) What is the actual power generated?arrow_forwardWith aid of a diagrams explain how the starter of a DC motor worksarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning