
Concept explainers
List the advantages of the use of two-wire control devices.

Explanation of Solution
A two-wire control circuit can be a pressure switch, toggle switch, simple switch, limit switch, float switch, or a thermostat which breaks or makes a connection to a motor through definite on and off positions.
The advantage of two-wire control devices is that the motor is operated (or controlled) by an automatic control device without requiring the presence of a human operator. For example, if the motor is stopped by power failure while the contacts are closed, the two wire control device will not open, that remains in a closed condition. Therefore, the motor can restart when the power is restored.
Conclusion:
Thus, the advantage of the use of two-wire control devices is explained.
Want to see more full solutions like this?
Chapter 17 Solutions
Electric Motor Control
- A 110-V, three-phase, Y-connected, 8 pole, 48-slot, 6000-rpm, double-layer wound chronoun anı vonorotor boo 10 +1 urn or oilarrow_forward-7 Name some of the factors that contribute to the deterioration of organic insulators. -8 A motor is built with class H insulation. What maximum hot-spot temperature can it withstand?arrow_forwardCalculate the full-load current of a 250 hp, 230 V dc motor having an efficiency of 92 percent.arrow_forward
- Assignment #2 A 110-V, three-phase, Y-connected, 8 pole, 48-slot, 6000-rpm, double-layer wound, synchronous generator has 12 turns per coil. If one side of the coil is in slot 1, the other side is in slot 6. There are 4 parallel paths. When the generator delivers the rated load at a line voltage of 110 V, the voltage regulation is 5%. What is the flux per pole? Draw two consecutive phasegroups of one of the phase windings and connect them (a) in series and (b) in parallel showing the Start (S) and Finish (F) of both connections. (A separate drawing for each connection)arrow_forward3-4 Transmissiva Live of 120km has R= 0.2 ~2/15 X= 0.8 -2/km Y = 15H/6 5/km The line is supplies a load of 45 kV, SOMW, 0.8 lead p.f find sending voltage, Sending Current p.f. Sanding Voltage Regulation ⑨Voltage 5 Ⓒ charching coming! изу usy π cct लेarrow_forwardA (medium) single phase transmission line 100 km long has the following constants : Resistance/km = 0.25 Q; Susceptance/km = 14 × 10° siemen ; Reactance/km = 0.8 Receiving end line voltage = 66,000 V Assuming that the total capacitance of the line is localised at the receiving end alone, determine (i) the sending end current (ii) the sending end voltage (iii) regulation and (iv) supply power factor. The line is delivering 15,000 kW at 0.8 power factor Lead Draw the phasor diagram to illustrate your calculations.arrow_forward
- 1. An electromagnetic device is shown below. The coil in the left side is connected to a steady AC power source. The left coil generates a changing magnetic flux, which is = 1.5cos(120πt +л/6) T. Calculate the voltage vs generated across the right coil given the number of turns of the right coil is 5 (You only need to calculate the magnitude). Vparrow_forwardFor the closed loop system shown in figure, determine the following:arrow_forwardWhat is the open loop transfer function and feedback for thia system? Determine the type of the open loop system. Find the poles s1 and s2 of the open loop system. If the input is a step function R(s)=1/s, find the step response c(t) of the open loop.arrow_forward
- not use ai please don'tarrow_forward2. A DC generator is shown below. This DC generator is driven by a prime mover and rotating in counterclockwise direction. The armature is connected with a load resistor. (i) Using cross (x) or dot (*) to indicate the current direction of each conductor in the armature. (ii) If we want to reverse the polarity of the generated armature voltage, what can we do to? rotation S load Narrow_forward6. The figures below show the equivalent circuit of a separately excited DC generator and the approximate relationship between the flux of main field and exciting current. The field current I can be regulated by the variable resistor Ry, and the battery voltage supplying power to the exciter is 12V. The armature resistance Ro is 20, and the load is 182. For the DC generator, we aim to keep the voltage across the load (RL) constant in different speed range conditions. In the beginning, the flux is 0.12 Wb, the DC generator speed is 1000 rpm, and the generated voltage E。 is 100 V. Calculate: (1) The current flowing through the load. (2) When the speed of generator changes to 1500 rpm, how should we adjust the exciting current Ix to ensure Ę is still 100 V. (Hint: E₁ = Zno/60) (3) When the speed of generator changes to 500 rpm, how should we adjust the exciting current Ix to ensure Eo is still 100 V. (Hint: Eo = Zno/60) Rf ww (Wb) 0.17 0.15 12 V 1x F ele 1 1 2 ell Eo Ro ww 9 w RL Ix (A)arrow_forward
- Electricity for Refrigeration, Heating, and Air C...Mechanical EngineeringISBN:9781337399128Author:Russell E. SmithPublisher:Cengage Learning

