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A series-wound DC motor draws a current of 0.50 amps from its 120-volt power source when it operates at full speed.
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The resistance of the armature coils of the motor is 10 ohms.
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What is the resistance of the motorβs field coils?
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What is the back emf of the motor when it is operating at full speed?
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The motor operates at a different speed at which it draws a current of 1.0 amps from its power source.
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What is the back emf of the motor when it operates at this speed?
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In this statement, weβre told that when the motor is running at full speed, itβs current is 0.50 amperes, what weβll call πΌ sub π.
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Weβre also told that itβs powered by a potential difference of 120 volts, what weβll call π.
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The resistance of the coils in the armature is 10 ohms, a value weβll call π
sub π.
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And in part one, we want to solve for the resistance of the field coils, what weβll call π
sub π.
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In part two, we want to solve for the back emf of the motor when itβs running at full speed.
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Weβll call that π sub π.
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In part three, weβre told that the motor drops back from full speed.
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And at this new lower speed, has a current running through it of 1.0 amps.
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Weβll call that current πΌ.
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At this new lower motor speed, we want to once more solve for the back emf.
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Weβll call that, simply, π.
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We can start on our solution by drawing a cross section diagram of this motor.
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In the middle of our motor, we have the armature.
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The armature is what spins while the rest of the motor stays still.
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Itβs called the stator.
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The armature is equipped with four coils of wire wound in series, weβre told, in this DC motor.
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The resistance of each one of these coils is π
sub π, given as 10 ohms.
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The rotation of the armature, which is typically connected to a shaft which lets the motor do useful work, is due to its interaction with the field created by the stator.
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The stator itself has four coils also wound in series through which it passes current.
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These coils are called field coils.
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And in part one, we want to solve for the resistance of each of the four coils.
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To figure out π
sub π, we can use Ohmβs law that the potential difference across a component or series of components is equal to the current running through those components multiplied by the resistance π
.
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In our series-wound DC motor, running at full speed, the potential difference, 120 volts, is equal to the current at full speed, 0.50 amps, multiplied by the sum of the total resistance of our armature and our field coils.
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There are four each for both the armature and the field coils.
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So when we rearrange and solve for π
sub π, itβs equal to π over four πΌ sub π minus π
sub π.
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Weβre given π, πΌ sub π, and π
sub π in the problem statement.
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So weβre now ready to plug in and solve for π
sub π.
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When we compute this term, we find that π
sub π is equal to 50 ohms.
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Thatβs the resistance of each one of the four field coils.
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Next, with the motor still running at full speed, we wanna calculate the back emf generated, π sub π.
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We can recall that anytime a current πΌ is generated in a circuit, that current induces an emf that opposes the current.
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Almost as though an electrical circuit works to maintain its inertia.
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This back emf π can be fit into ohmβs law as a second voltage term.
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This one in opposition to the one causing the current πΌ.
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So if we want to solve for the back emf of the motor when itβs running at full speed, we can write that the current πΌ at full speed is equal to π minus π sub π over π
sub π plus π
sub π, the total resistance of an armature and field coil.
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When we rearrange this equation to solve for π sub π, we find itβs equal to π minus πΌ sub π times the quantity π
sub π plus π
sub π.
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We know each of these four values and can plug in now to solve for π sub π.
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When we enter these values on our calculator, we find that π sub π is 90 volts.
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Thatβs the back emf generated when the motor is running at full speed.
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Finally, we want to solve for π which weβve called the back emf when the motor is running at less than full speed.
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To calculate this, we can reuse our equation for π sub π.
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Weβll simply need to change out the current from πΌ sub π to πΌ, which then means weβre solving for π.
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Entering these values in with our new value of 1.0 amps for current, when we calculate the back emf π, we find that itβs 60 volts.
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Thatβs the back emf generated when the motor is running at less than full speed.