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An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection is completed. After a long time, the current is 6.45 mA. What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

Short Answer

Expert verified
R = 2.48 Ω, L = 2.33 H.

Step by step solution

01

Understand the Problem Scenario

The problem involves an inductor connected to a battery with an emf of 16.0 V. Initially, the current is 4.86 mA at 0.940 ms. After a long time, the current stabilizes at 6.45 mA. We need to find the resistance \( R \) and the inductance \( L \) of the inductor.
02

Identify Key Equations

For an RL circuit, the current \( I(t) \) as a function of time \( t \) is given by the equation \( I(t) = I_0 (1 - e^{-t/\tau}) \), where \( I_0 = \frac{\text{emf}}{R} \) is the final steady-state current, and \( \tau = \frac{L}{R} \) is the time constant.
03

Calculate the Resistance \( R \)

Given that the current after a long time is 6.45 mA, we use \( I_0 = \frac{\text{emf}}{R} \). Rearranging gives \( R = \frac{\text{emf}}{I_0} = \frac{16.0 \text{ V}}{6.45 \times 10^{-3} \text{ A}} \approx 2.48 \times 10^3 \text{ \( \Omega \)} \).
04

Verify Initial Condition Equation

We use the initial condition \( I(0.940 \text{ ms}) = 4.86 \text{ mA} \) in the current equation: \( 4.86 \times 10^{-3} = 6.45 \times 10^{-3} (1 - e^{-0.940 \times 10^{-3} / \tau}) \).
05

Solve for Time Constant \( \tau \)

Rearranging and solving the equation gives \( e^{-0.940 \times 10^{-3} / \tau} = 1 - \frac{4.86}{6.45} \). This simplifies to \( \tau = \frac{0.940 \times 10^{-3}}{- \ln(1 - \frac{4.86}{6.45})} \approx 0.940 \times 10^{-3} \text{ s} \), as exponentials can be approximate at small times.
06

Calculate the Inductance \( L \)

With \( \tau = \frac{L}{R} \) and \( \tau \approx 0.940 \times 10^{-3} \text{ s} \), we solve for \( L \) using \( L = \tau \times R = 0.940 \times 10^{-3} \times 2.48 \times 10^3 \approx 2.33 \text{ H} \).
07

Summary of Results

The resistance \( R \) of the inductor is approximately \( 2.48 \text{ \( \Omega \)} \), and the inductance \( L \) is approximately \( 2.33 \text{ H} \).

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Key Concepts

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

Inductor resistance
Understanding the resistance in an RL circuit involving an inductor is crucial. The inductor resistance, often denoted by \( R \), refers to the opposition to the flow of electric current within the inductive component. This resistance causes energy loss in the form of heat. In practice,
  • resistance is caused by the inherent imperfections and material properties of the inductor coil.
  • It is always considered when calculating the maximum current in an RL circuit.
Knowing the final steady-state current, we can calculate the inductor's resistance using Ohm's Law, which states \( R = \frac{\text{emf}}{I_0} \), where emf is the electromotive force and \( I_0 \) is the final current. This calculation is fundamental in electronics design and aids in predicting how the circuit will perform under stable conditions.
When dealing with circuits over time, resistive heating can lead to inefficiencies, affecting total circuit behavior.
Inductance calculation
The inductance of an inductor, symbolized as \( L \), is its ability to store energy in a magnetic field when electrical current flows through it. It is a crucial parameter in an RL circuit as it affects how the circuit resists changes in current.
  • Inductance is measured in Henries (H).
  • A larger inductance means a greater capacity to oppose changes in current.
Using the induction equation \( I(t) = I_0 (1 - e^{-t/\tau}) \), where \( \tau \) is the time constant, we derive \( L \) from \( \tau = \frac{L}{R} \). Solving for \( L \), we find \( L = \tau \times R \).
In practical applications, calculating inductance is essential for designing filters, transformers, and tuning circuits.
The precise calculation ensures the components behave correctly in dynamic scenarios, such as during power surges or load changes.
Time constant in circuits
The time constant, denoted as \( \tau \), is a key concept in analyzing RL circuits. It characterizes the time it takes for the current to change significantly in response to a voltage change.
  • Mathematically, it is represented as \( \tau = \frac{L}{R} \).
  • A shorter time constant means the circuit reacts quickly to changes in voltage.
  • Conversely, a longer time constant means the circuit responds more slowly.
The time constant represents the period required for the current to reach approximately 63.2% of its total change from initial to final values.
In our example, solving for \( \tau \) involves setting the exponential factor such that it accounts for the observed current at a specific time, giving insights into how fast the circuit stabilizes.
Understanding the time constant helps in controlling the transient behavior of circuits during initial startup or sudden power increases, a critical aspect in both the analysis and design of time-sensitive electronic systems.

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Most popular questions from this chapter

A solenoid 25.0 cm long and with a cross-sectional area of 0.500 cm\(^2\) contains 400 turns of wire and carries a current of 80.0 A. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic field if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

A 7.00-\(\mu\)F capacitor is initially charged to a potential of 16.0 V. It is then connected in series with a 3.75-mH inductor. (a) What is the total energy stored in this circuit? (b) What is the maximum current in the inductor? What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

In an \(L\)-\(C\) circuit, \(L = 85.0\) mH and \(C = 3.20 \, \mu \mathrm{F}\). During the oscillations the maximum current in the inductor is 0.850 mA. (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 mA?

A capacitor with capacitance \(6.00 \times 10^{-5}\) F is charged by connecting it to a 12.0-V battery. The capacitor is disconnected from the battery and connected across an inductor with \(L = 1.50\) H. (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation)? (b) What is the initial charge on the capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 s after the connection to the inductor is made? Interpret the sign of your answer. (e) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

A 10.0-cm-long solenoid of diameter 0.400 cm is wound uniformly with 800 turns. A second coil with 50 turns is wound around the solenoid at its center. What is the mutual inductance of the combination of the two coils?

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