Question

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

R = 2.48 Ω, L = 2.33 H.

Step-by-step solution

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.

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.

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{ \( \Omega \)} \).

Verify Initial Condition Equation

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

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.

Calculate the Inductance \( L \)

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

Summary of Results

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

Key concepts

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.