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A \(120.0-\mathrm{V}\) motor draws a current of \(7.00 \mathrm{A}\) when running at normal speed. The resistance of the armature wire is \(0.720 \Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to \(15.0 \mathrm{A} ?\)

Short Answer

Expert verified
(a) Back EMF is 114.96 V. (b) Starting current is 166.67 A. (c) Add 7.28 Ω resistance.

Step by step solution

01

Understand the Problem

Here, we need to calculate (a) the back electromotive force (emf) when the motor runs at its normal speed, (b) the current at the motor startup before it begins rotating, and (c) the series resistance required to limit the starting current to 15.0 A.
02

Use Ohm's Law and Determine Back EMF

The total voltage provided by the source is composed of the: \[ \text{Voltage Drop across Resistance} + \text{Back EMF} \]The voltage drop across motor's armature resistance when it's running normally is:\[ V_R = I \times R = 7.00 \text{ A} \times 0.720 \Omega = 5.04 \text{ V} \]Thus, the back emf (E) is:\[ \text{Back EMF} = V - V_R = 120.0 \text{ V} - 5.04 \text{ V} = 114.96 \text{ V} \]
03

Calculate Current When the Motor Starts

When the motor is just turned on, it hasn't started generating back emf. Thus, the full voltage is dropped over the resistance:\[ V = I_{start} \times R \]Solving for the starting current (\(I_{start}\)):\[ I_{start} = \frac{V}{R} = \frac{120.0 \text{ V}}{0.720 \Omega} = 166.67 \text{ A} \]
04

Determine Required Series Resistance

To limit the starting current to 15.0 A, we use the modified equation:\[ V = I_l \times (R + R_s) \]Where \(I_l\) is the limited starting current, and \(R_s\) is the series resistance addition. Solving for \(R_s\), we have:\[ R + R_s = \frac{V}{I_l} \to R_s = \frac{120.0 \text{ V}}{15.0 \text{ A}} - 0.720 \Omega = 8.0 \Omega - 0.720 \Omega = 7.28 \Omega \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Ohm's Law is a fundamental principle used to understand electrical circuits. It explains the relationship between voltage, current, and resistance. The law is expressed with the formula:\[ V = I \times R \]where:
  • \( V \) is the voltage across the circuit in volts (V),
  • \( I \) is the current flowing through the circuit in amperes (A),
  • \( R \) is the resistance of the circuit in ohms (\( \Omega \)).
When applied correctly, Ohm's Law can help you calculate any one of these values if the other two are known. In the context of a motor, Ohm’s Law is crucial because it helps identify the voltage drop across the motor's resistance and the impact of back EMF. Knowing how to manipulate the formula allows one to work out intricate details of electric circuits efficiently.
Motor Resistance
Motor resistance is the internal opposition to the flow of electric current within the motor. This resistance is mainly due to the armature winding within the motor and is a critical factor in calculating how the motor operates in different phases, such as startup and normal operation.
The resistance influences how much current flows and how much voltage is dropped across the motor itself when voltage is applied. In our example, the motor's resistance is stated as \( 0.720 \Omega \). Understanding this allows us to determine other important parameters like the back EMF or the initial surge of current when the motor is powered on. Motor resistance directly affects energy efficiency and heat generation inside the motor.
Starting Current
Starting current, also known as inrush current, is the initial surge of current that flows into the motor when it is first turned on. This surge occurs because the motor has not begun moving yet, and thus has not started generating back EMF that would otherwise reduce the voltage across the motor.
When a motor starts, the resistance to current flow is minimal, causing a high current draw. In the exercise example, the starting current is found to be \( 166.67 \mathrm{A} \) for a motor with a resistance of \( 0.720 \Omega \) and a supply voltage of \( 120.0 \mathrm{V} \). The starting current is typically much higher than the running current, potentially causing damage or unnecessary stress to the motor. Hence, asking how to limit this starting current without impacting the motor's performance becomes essential.
Series Resistance
Series resistance involves adding a resistor in series with the motor circuit to limit the amount of starting current the motor receives. It's a crucial method because it reduces the initial current surge encountered at startup, thus protecting the motor from possible damage.
In our example, to limit the starting current to a safe \( 15.0 \mathrm{A} \), a series resistance of \( 7.28 \Omega \) is calculated. By using this additional resistance, you can ensure that the starting current does not exceed desired levels. The formula used for this calculation involves similar principles to Ohm’s Law:\[ V = I \times (R + R_s) \]where \( R_s \) is the series resistance required. This setup offers a practical solution to control the starting current, maintaining the motor's health and function.

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Most popular questions from this chapter

A piece of copper wire is formed into a single circular loop of radius \(12 \mathrm{cm} .\) A magnetic field is oriented parallel to the normal to the loop. and it increases from 0 to \(0.60 \mathrm{T}\) in a time of \(0.45 \mathrm{s}\). The wire has a resistance per unit length of \(3.3 \times 10^{-2} \Omega / \mathrm{m} .\) What is the average electrical energy dissipated in the resistance of the wire?

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A circular coil of radius 0.11 m contains a single timm and is located in a constant magnetic field of magnitude \(0.27 \mathrm{T} .\) The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to \(0.30 \mathrm{m}\) in a time of \(0.080 \mathrm{s} .\) Concepts: (i) Why is there an emf induced in the coil? (ii) Does the magnitude of the induced emf depend on whether the area is increasing or decreasing? Explain. (iii) What determines the amount of current induced in the coil? (iv) If the coil is cut so it is no longer one continuous piece, are there an induced emf and an induced current? Explain. Calculations: (a) Determine the magnitude of the emf induced in the coil. (b) The coil has a resistance of \(0.70 \Omega\). Find the magnitude of the induced current.

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