Problem 78 - Fundamentals of Physics

Question

An electric motor connected to a \(120 \mathrm{~V}, 60.0 \mathrm{~Hz}\) ac outlet does mechanical work at the rate of \(0.100 \mathrm{hp}(1 \mathrm{hp}=746 \mathrm{~W})\). (a) If the motor draws an rms current of \(0.650 \mathrm{~A},\) what is its effective resistance, relative to power transfer? (b) Is this the same as the resistance of the motor's coils, as measured with an ohmmeter with the motor disconnected from the outlet?

Step-by-Step Solution

Answer

(a) Effective resistance is 114.3 Ω. (b) No, it includes reactance, unlike coil resistance.

Step 1: Understand the Given Values

The problem provides the following values: - Voltage ( V_{ ext{rms}} ) = 120 V - Frequency = 60.0 Hz - Mechanical Power ( P_{ ext{mech}} ) = 0.100 hp (1 hp = 746 W) - RMS Current ( I_{ ext{rms}} ) = 0.650 A We need to calculate the effective resistance using these values.

Step 2: Convert Horsepower to Watts

We know that 1 horsepower is 746 Watts. Therefore, the mechanical power in watts is given by:\[P_{ ext{mech}} = 0.100 \times 746 = 74.6 \text{ W}\]

Step 3: Calculate the Power Factor, P_f

The power factor is given by the ratio of mechanical power to apparent power (the product of voltage and current):\[P_f = \frac{P_{ ext{mech}}}{V_{rms} \times I_{rms}} = \frac{74.6}{120 \times 0.650}\]Calculate this to get the power factor.

Step 4: Solve for Effective Resistance

Using the power factor calculated, the effective resistance can be found using:\[R_{ ext{eff}} = \frac{V_{rms}}{I_{rms}} \times P_f\]Substitute the values:\[R_{ ext{eff}} = \frac{120}{0.650} \times P_f\]Calculate this to find the effective resistance.

Step 5: Compare Effective Resistance with Coil Resistance

When the motor is running, its effective resistance reflects power transfer including losses (inductive reactance, etc.). The coil resistance measured by an ohmmeter is the DC resistance, which does not include reactance or other effects in AC operations.

Key Concepts

Effective Resistance

Effective resistance in an AC circuit, like that of an electric motor, represents how much the motor opposes the current but taking into account the nature of AC currents, which include phase differences between voltage and current. It is different from regular DC resistance, which measures pure opposition to current flow without phasing considerations.

To find the effective resistance (\( R_{\text{eff}} \)), we use the formula:\[R_{\text{eff}} = \frac{V_{rms}}{I_{rms}} \times P_f\]Here, \( V_{rms} \) is the root mean square voltage, \( I_{rms} \) is the root mean square current, and \( P_f \) is the power factor, a measure of how effectively current is being converted into useful work.

Remember, the effective resistance includes not only the resistance of the coils but also additional factors like inductance, which becomes significant in AC circuits.

Power Factor

The power factor (\( P_f \)) is a key indicator of how effectively the motor uses electrical power to perform work. It relates to the phase difference between voltage and current in an AC circuit.

A power factor of 1 means the motor is perfectly efficient in converting energy from electrical to mechanical form, while a lower value indicates inefficiencies, often due to reactance in the circuit.

We calculate the power factor using:\[P_f = \frac{P_{\text{mech}}}{V_{rms} \times I_{rms}}\]Where \( P_{\text{mech}} \) is the mechanical power output in watts, \( V_{\text{rms}} \) is the root mean square voltage, and \( I_{rms} \) is the root mean square current.

This formula shows us how much of the electrical power is being effectively turned into mechanical work, critical for assessing the motor's performance.

Mechanical Power

Mechanical power (\( P_{\text{mech}} \)) is the amount of work your electric motor performs per unit of time. For electric motors, mechanical power is often given in horsepower, which needs conversion to watts for electrical calculations.

One horsepower (\( ext{hp} \)) equals 746 watts. In this exercise, we converted 0.100 hp to get 74.6 watts as the motor's mechanical output.

Understanding mechanical power in terms of watts makes it easier to integrate with electrical calculations, especially when determining things like the power factor or effective resistance.

RMS Current

RMS, or root mean square current (\( I_{\text{rms}} \)), is used in AC (Alternating Current) circuits to express the effective value of an alternating current. It enables us to equate an AC current to a DC current in terms of the energy it can deliver.

Think of \( I_{\text{rms}} \) as a measure that allows direct comparison to a hypothetically equivalent DC circuit. It's calculated by taking the square root of the average of the squares of all values in an AC current waveform over one cycle.

In our exercise, an \( I_{\text{rms}} \) of 0.650 A was used to determine other parameters of the motor’s function, emphasizing how important this value is for accurate electrical analysis of AC systems.