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A coil has a resistance of \(48.0 \Omega\). At a frequency of \(80.0 \mathrm{~Hz}\) the voltage across the coil leads the current in it by \(53^{\circ}\). Determine the inductance of the coil.

Short Answer

Expert verified
The inductance of the coil is approximately 0.127 H.

Step by step solution

01

Calculate Inductive Reactance

The phase angle \( \theta \) between the voltage and the current is given as \(53^{\circ}\). Using the relation \( \tan \theta = \frac{X_L}{R} \), where \( X_L \) is the inductive reactance and \( R = 48.0 \Omega \) is the resistance, we calculate \( X_L \). First, find \( \tan 53^{\circ} \).
02

Solve for Inductive Reactance

Calculate \( X_L = R \cdot \tan 53^{\circ} \). Substitute \( R = 48.0 \Omega \) and \( \tan 53^{\circ} \) (approximately 1.327) to find \( X_L \). Thus, \( X_L = 48.0 \Omega \times 1.327 \approx 63.696 \Omega \).
03

Determine Inductance

Inductive reactance \( X_L \) relates to inductance \( L \) by the formula \( X_L = 2\pi f L \), where \( f = 80.0 \mathrm{~Hz} \). Solve for \( L \): \( L = \frac{X_L}{2\pi f} \).
04

Calculate Inductance

Substitute \( X_L = 63.696 \Omega \) and \( f = 80.0 \mathrm{~Hz} \) into \( L = \frac{63.696 \Omega}{2\pi \times 80.0 \mathrm{~Hz}} \). Compute this to find \( L \approx 0.1267 \mathrm{~H} \).

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Key Concepts

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

Inductive Reactance
Inductive reactance is a crucial concept in understanding how coils, like inductors, behave in alternating current (AC) circuits. It is the opposition presented by the inductor to the change in current passing through it. The formula to calculate inductive reactance \( X_L \) is given by:
  • \( X_L = 2\pi f L \)
where \( f \) represents the frequency of the AC current and \( L \) is the inductance of the coil.
In the original exercise, our task was to find this reactance at a frequency of 80 Hz. Using this formula, we started by substituting appropriate values to find \( X_L \). The resistance \( R \) was given as 48.0 \( \Omega \). It is important to note that reactance shares dimensional units with resistance, \( \Omega \), emphasizing its role as a type of resistance in AC circuits. Understanding inductive reactance is key to predicting how the coil will react under varying frequencies.
Phase Angle
The phase angle in AC circuits is an indicator of the phase difference between the voltage across a component and the current through it. Manipulating these phase differences helps us analyze and design efficient AC systems. When voltage leads or lags behind the current, this difference is expressed as an angle, typically measured in degrees.
In the presented problem, we're given a phase angle of 53°. This means the voltage across the coil leads the current by this angle. The formula involving the phase angle is:
  • \( \tan \theta = \frac{X_L}{R} \)
Knowing the phase angle allows us to solve for the inductive reactance, linking it directly to values given for resistance. Thus, understanding the phase relationship plays a critical role in the correct design and function of AC circuits. Phase angles affect everything from power quality to stability in systems.
Inductance Calculation
Calculating inductance, denoted as \( L \), helps us quantify how well a coil can store energy in its magnetic field when current flows through it. The relationship between inductance and inductive reactance is captured by the equation:
  • \( L = \frac{X_L}{2\pi f} \)
This equation highlights how both frequency and reactance contribute to determining inductance.
In our exercise, we've determined \( X_L \) first to facilitate the inductance calculation. By substituting \( X_L = 63.696 \Omega \) and \( f = 80.0 \mathrm{Hz} \) into the formula, we calculated the inductance to be approximately 0.1267 H (henries). This calculation is essential for designing components that are intended to function optimally within specific frequencies and system requirements.
AC Circuit Analysis
AC Circuit Analysis involves studying circuits that operate with alternating current. This kind of analysis is crucial for understanding the behavior of circuit components like resistors, capacitors, and inductors when they are driven by AC sources.
Incorporating elements such as inductors introduces unique challenges due to their frequency-dependent properties, such as inductive reactance. Analyzing this involves using key concepts such as Ohm's and Kirchhoff's laws adapted for AC circuits. For example, considering AC source frequency and phase angles becomes imperative for accurate circuit modeling.
  • Calculations such as those performed in our exercise play a fundamental role in determining the characteristics and performance of such circuits.
  • Factors like phase angle and inductance inform circuit design decisions, ensuring efficiency and reliability.
The ultimate goal of AC Circuit Analysis is to predict how the system will behave under various operating conditions, guiding towards optimal performance strategies.

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Most popular questions from this chapter

An \(L-R-C\) series circuit is connected to an ac source of constant voltage amplitude \(V\) and variable angular frequency \(\omega\). (a) Show that the current amplitude, as a function of \(\omega\), is $$ I=\frac{V}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}} $$ (b) Show that the average power dissipated in the resistor is $$ P=\frac{V^{2} R / 2}{R^{2}+(\omega L-1 / \omega C)^{2}} $$ (c) Show that \(I\) and \(P\) are both maximum when \(\omega=1 / \sqrt{L C}\), the resonance frequency of the circuit.

An LCR series circuit with \(100 \Omega\) resistance is connected to an \(A C\) source of \(200 V\) and angular frequency 300 radians/sec. When only the capacitance is removed The current lags behind the voltage by \(60^{\circ} .\) When only the inductance is removed the current leads the voltage \(60^{\circ}\). Calculate the current and the power dissipated in the LCR circuit.

A Step-Down Transformer. A transformer connected to a \(120-\mathrm{V}(\mathrm{rms})\) ac line is to supply \(12.0 \mathrm{~V}\) (rms) to a portable electronic device. The load resistance in the secondary is \(5.00 \Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply?

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{~V}, R=80.0 \Omega\), and the reactance of the capacitor is \(480 \Omega .\) The voltage amplitude across the capacitor is \(360 \mathrm{~V}\). (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

In an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance \(X_{C}\) of the capacitor is \(200 \Omega\) and the voltage amplitude across the capacitor is \(600 \mathrm{~V}\). The circuit has \(R=300 \Omega\). What is the voltage amplitude of the source?

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