Problem 28 - College Physics

Question

A large electromagnetic coil is connected to a \(120 \mathrm{~Hz}\) ac source. The coil has resistance \(400 \Omega,\) and at this source frequency the coil has inductive reactance \(250 \Omega\). (a) What is the inductance of the coil? (b) What must the rms voltage of the source be if the coil is to consume an average electric power of \(800 \mathrm{~W} ?\)

Step-by-Step Solution

Answer

(a) Inductance: \(0.331\,\text{H}\). (b) RMS Voltage: \(614\,\text{V}\).

Step 1: Calculate Inductance of the Coil

To find the inductance of the coil, we use the formula for inductive reactance: \( X_L = 2\pi f L \), where \( X_L \) is the inductive reactance, \( f \) is the frequency, and \( L \) is the inductance. Given \( X_L = 250 \: \Omega \) and \( f = 120 \: \text{Hz} \), we have:\[ L = \frac{X_L}{2\pi f} = \frac{250}{2\pi \times 120} \approx 0.331 \: \text{H} \]

Step 2: Calculate the Impedance of the Coil

Impedance \( Z \) in an AC circuit with resistance \( R \) and inductive reactance \( X_L \) can be found using the formula:\[ Z = \sqrt{R^2 + X_L^2} \]Substitute \( R = 400 \: \Omega \) and \( X_L = 250 \: \Omega \):\[ Z = \sqrt{400^2 + 250^2} = \sqrt{160000 + 62500} = \sqrt{222500} \approx 471.7 \: \Omega \]

Step 3: Determine RMS Voltage from Power Output

The power \( P \) in an AC circuit in terms of RMS voltage \( V \) and impedance \( Z \) is given by:\[ P = \frac{V^2}{Z} \]Rearrange to solve for \( V \):\[ V = \sqrt{P \times Z} \]Substitute \( P = 800 \: \text{W} \) and \( Z \approx 471.7 \: \Omega \):\[ V = \sqrt{800 \times 471.7} \approx \sqrt{377360} \approx 614 \: \text{V} \]

Key Concepts

Inductance

Inductance is a fundamental property of electrical circuits that describes how effectively a coil or an inductor stores energy in a magnetic field when current flows through it. This property is primarily determined by the coil's physical characteristics, including the number of turns, coil shape, and core material. Inductors resist changes in current, and this opposition is measured in henries (H).

In the context of our problem, the coil has an inductive reactance of 250 Ω at a frequency of 120 Hz. To find the inductance, we use the relationship between inductive reactance, frequency, and inductance given by the formula:

• Inductive Reactance: \( X_L = 2\pi f L \)

By rearranging this, we can solve for the inductance \( L \):

• \( L = \frac{X_L}{2\pi f} \)

Plugging in the known values lets us find that the inductance is approximately 0.331 H. This means when the current flows at this frequency, the coil behaves as if it has a certain amount of 'magnetic inertia' resisting changes in current.

AC Circuits

AC circuits are circuits powered by an alternating current (AC) source, where the current changes direction periodically. Unlike direct current (DC), which flows in a single direction, AC allows for the efficient transmission of electrical energy over long distances. This periodic change is typically sinusoidal, characterized by its frequency and amplitude.

AC circuits often use components like resistors, capacitors, and inductors, which react differently to AC compared to DC. In our scenario, the circuit is driven by a source with a frequency of 120 Hz. The behavior of the circuit under AC is analyzed using properties like impedance, a measure combining resistance and reactance, to find how much the circuit impedes the flow of electricity.

To manage these circuits properly, one needs to understand the basics of how current and voltage behave in them. Impedance (\( Z \)) in an AC circuit is crucial; it combines resistance (\( R \)) and inductive reactance (\( X_L \)) as:

• \( Z = \sqrt{R^2 + X_L^2} \)

This formula helps us calculate the overall opposition the circuit presents to the AC, giving a more comprehensive understanding than looking at resistance alone.

Inductive Reactance

Inductive reactance is an important concept when analyzing AC circuits. It represents how much an inductor resists the change in current flow when subjected to an alternating current. Unlike resistance which applies to both AC and DC, inductive reactance is specific to AC circuits and plays a vital role in determining the circuit's impedance.

The value of inductive reactance (\( X_L \)) is directly proportional to both the frequency (\( f \)) of the AC source and the inductance (\( L \)) of the coil. This relationship is expressed as:

• \( X_L = 2\pi f L \)

In our exercise, the inductive reactance is given as 250 Ω at 120 Hz, aiding in the initial calculation of the coil's inductance.

This characteristic of inductors means they are more effective at opposing higher frequencies. They tend to "block" higher frequency currents while allowing lower frequencies to pass through more easily, an important concept in the design of filters and tuners.

Electric Power Calculation

Electric power in an AC circuit is an essential aspect of understanding how much work electric current can do in the electrical system over a period of time. Power, measured in watts, indicates how much energy is consumed or produced by a device. In circuits with reactive components like inductors, we need to consider both the real and apparent parts of power due to the phase shift between current and voltage.

The average power consumed by a circuit can be determined through the RMS voltage and the impedance. The formula for power in an AC circuit is:

• \( P = \frac{V^2}{Z} \)

Where \( P \) is the average power, \( V \) is the RMS voltage, and \( Z \) is the impedance.

In our example, we want the coil to consume an average power of 800 W. By knowing the impedance calculated as approximately 471.7 Ω, we can rearrange the power formula to find the RMS voltage needed:

• \( V = \sqrt{P \times Z} \)

Substituting the given power and calculated impedance values, we determine that the RMS voltage should be approximately 614 V.