A 120-V, series-wound dc motor draws 0.50 A from its power...

A 120-V, series-wound dc motor draws 0.50 A from its power source when operating at full speed, and it draws 2.0 A when it starts. The resistance of the armature coils is $10 \Omega$. (a) What is the resistance of the field coils? (b) What is the back emf of the motor when it is running at full speed? (c) The motor operates at a different speed and draws $1.0 \mathrm{A}$ from the source. What is the back emf in this case?

Key Concepts

Series-wound DC Motor

A series-wound DC motor is a type of motor in which the field windings are connected in series with the armature. This design means that the same current flows through both the armature and the field coils, leading to a strong interaction between the magnetic field and the armature current. It is commonly used in applications where a high starting torque is required, and its performance significantly depends on the current drawn from the power source at various operating conditions.

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering that states the relationship between voltage, current, and resistance in a circuit (V = IR). In the context of DC motor analysis, Ohm's Law is used to calculate voltage drops across various resistive components, such as the armature and field windings. This law is essential for determining unknown resistances or voltage values when the current and one of the other quantities are known.

Back Electromotive Force (EMF)

Back electromotive force (EMF) is the voltage generated by a running motor that opposes the applied voltage. It is proportional to the speed of the motor and the strength of its magnetic field. Back EMF plays a crucial role in regulating the current within the motor: at higher speeds, the back EMF increases and limits the current draw. Understanding back EMF is essential for analyzing motor performance under various load conditions and speed changes.

Armature and Field Resistances

In a DC motor, the armature resistance refers to the resistance of the coils in the armature windings, while the field resistance pertains to the resistance of the field windings. In series-wound motors, both resistances affect the overall voltage drop and performance of the motor. By analyzing the voltage drops across these components using methods like Ohm's Law, one can calculate unknown resistances or determine key operating parameters such as motor speed and back EMF.

Transcript

00:01 In the given problem, we are having a series wound dc motor, which is being operated at 120 volt, and it is drawing an initial current of 0 .50 ampere.

00:20 Then it is moving at full speed.

00:25 Full speed means there will be an induced emf also, a back emf also will be being induced at this speed.

00:34 And it draws a current i2 is equal to 2 .0 ampere when it starts at start and then the amf induced the back amf induced will be zero now the resistance of the armature coil resistance of armature has been given as 10 oom so in the first part of the problem we have to find resistance of the field coil rf we have to find for which we will use the expression for the current passing through this dc motor at the startup means when there is no back -m -f induced so at a start -up as back end of induced is zero so using holmes law i2 is equal to just we simply v by total resistance in series which is our resistance of our nature are rr plus r f so here i 2 .0 ampere potential this is 120 volt and the resistance will be 10 plus rf here rf is being used for resistance of field coil making a cross multiplication here we get 10 plus rf is equal to 120 by 2 needs to say this is 60 oh so finally we can see the value of the resistance of field coil comes out to be 60 minus 10 or we can say the value of this rf is 50 this is the answer for the first part of the question now in the second part we have to find the back emf induced in the motor when it is running at the full speed so now we will use expression again using home slope expression for current at full speed will be i1 is equal to v minus e the back in if induced divided by the net resistance again means resistance of armature foil plus resistance of field coil so here here i1 is 0 .5 ampere or this is 120 e is missing and the net resistance now will become 50 plus 10 om making across multiplication again we get 120 minus e is equal to 0 .5 into 60 or it comes out to be 30 so we can say the back amf induced is 120 minus 30 volt or this is 90 volt answer for the second part of the problem...

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