A 240-V, 60-Hz, two-pole universal motor operates at a speed of 12,000 rev / min on full load an

A 240-V, 60-Hz, two-pole universal motor operates at a speed...

Key Concepts

Universal Motor

A universal motor is a type of electric motor that can operate on either AC or DC supply because its field winding is connected in series with the armature. Its design allows for high starting torque and high speed, making it common in household appliances and power tools. While having a relatively simple construction, the performance of a universal motor is sensitive to load variations and its speed can vary significantly with changes in the applied voltage and load conditions.

Back Electromotive Force (Back EMF)

Back EMF is the voltage generated by a running motor that opposes the applied voltage. It results from the motor’s rotational motion in the magnetic field, and its magnitude is directly related to the speed of the motor and the strength of the magnetic field. Back EMF is essential in determining the net voltage across the motor’s internal resistive and reactive components, which in turn governs the current and overall performance of the motor.

Electrical Impedance in AC Circuits

In AC circuits, impedance represents the total opposition to the flow of alternating current and includes both resistive (real) and reactive (imaginary) components. Understanding impedance is crucial when analyzing motors because it affects how voltage divides among the motor’s windings and components, influencing the current flow, phase shift, and overall power consumption. This concept is fundamental to calculating voltage drops and determining the effective voltage available for driving the motor.

Mechanical Power Development

Mechanical power development in a motor refers to the conversion of electrical energy into mechanical energy, which is the useful work output of the motor. This power is calculated by accounting for losses within the motor, such as resistive (copper) losses and other internal dissipation, and represents the actual power available for performing mechanical tasks. It is an important metric for assessing the performance and capability of the motor in practical applications.

Motor Efficiency

Motor efficiency is a measure of how effectively a motor converts electrical input power into useful mechanical output power. It is defined as the ratio of the mechanical power delivered to the load to the electrical power consumed from the source. High efficiency indicates that a greater proportion of the input energy is converted into useful work, while low efficiency signifies substantial energy losses, typically in the form of heat and other dissipative effects.

Power Factor

The power factor in an AC circuit is the cosine of the phase angle between the voltage and the current waveforms. It quantifies how effectively the current is being converted into useful work and indicates the degree to which the circuit's resistive component dominates over the reactive component. In motor applications, a high power factor means that the motor draws current more efficiently relative to the actual work performed, while a low power factor suggests a higher presence of reactive power and less efficient utilization of electrical energy.

Transcript

00:01 In the given problem, we are having a 120 volt motor, means the voltage supplied to it is 120 volt.

00:13 It is a series wound motor in which the field resistance is 80 -oom and the armature coil is having a resistance of 10 -oom.

00:26 So the net resistance of the motor will become 80 -on plus 10 -oom.

00:32 Means to say this becomes 90 om now when it is rotating at the full speed a back emf induced in it has been given as 75 volt in the first part of the problem we have to find the initial current drawn by the motor here the word initial means there is no emf is induced in it so simply using om's law the expression for current comes out to be just v by r this v is 120 volt and r is the net resistance which is 90 oom so current drawn by the coil comes out to be 4 by 3 amp this is the answer for the first part of the problem now in the second part when the motor is operating at full speed so definitely now the net voltage at the full speed will become a potential applied minus back emf induced means it will be 120 minus 75 volt and we have to find the current drawn by the battery now so in part b the expression for current now will become this v minus e divided by the net resistance so here this is 120 minus 75 divided by 90 om or we can say it becomes 45 volt by 90 om so finally this current now comes out to be 0 .5 ampere this is the answer for second part of the problem in the third part of the problem we have to find the power output of the source and we know the source voltage is 120 volt and the current now is just half amp so the power output of the source will be p is equal to v into i for v this is for i this is half and here and here this is volt into ampere so this power will come out to be 60 watt the answer for third part of the problem in the fourth part of the problem we have to find the power output of the motor and the power output of the motor will be simply power output of the motor of the motor will be found using the emf induced in it means e into i, e was 75 volt into half again ampere...

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