Problem 52 - University Physics with Modern Physics

Question

An inductor is connected to the terminals of a battery that has an emf of 12.0 \(\mathrm{V}\) and negligible internal resistance. The current is 4.86 \(\mathrm{mA}\) at 0.940 \(\mathrm{ms}\) after the connection is completed. After a long time the current is 6.45 \(\mathrm{mA} .\) What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

Step-by-Step Solution

Answer

Resistance \( R \approx 1860 \ \Omega \); Inductance \( L \approx 0.34 \ \text{H} \).

Step 1: Understand the Problem

We need to find the resistance \( R \) and the inductance \( L \) of an inductor in an RL circuit. Given values are initial EMF \( \varepsilon = 12.0 \text{ V} \), the current after a long time \( I_0 = 6.45 \text{ mA} \), and the current at \( t = 0.940 \ \text{ms} \) is \( I(t) = 4.86 \ \text{mA} \).

Step 2: Determine the Resistance

After a long time, the inductor reaches a steady state, and the current \( I_0 \) equals \( \varepsilon/R \). Rearranging gives:\[ R = \frac{\varepsilon}{I_0} = \frac{12.0}{6.45 \times 10^{-3}} \approx 1860 \ \Omega. \]

Step 3: Use the RL Circuit Current Equation

In an RL circuit, the current at any time \( t \) is given by:\[ I(t) = I_0 \left(1 - e^{-\frac{Rt}{L}}\right). \]

Step 4: Solve for the Inductance

Insert the known values into the equation from Step 3:\[ 4.86 \times 10^{-3} = 6.45 \times 10^{-3} \left(1 - e^{-\frac{1860 \times 0.940 \times 10^{-3}}{L}}\right). \]Solve for \( L \):Rearrange the equation:\[ 1 - \frac{4.86}{6.45} = e^{-\frac{1860 \times 0.940 \times 10^{-3}}{L}}. \]Then:\[ 1 - \frac{4.86}{6.45} = e^{-1.748 \times 10^{-3}/L}. \]Taking the natural logarithm:\[ -\ln\left(\frac{1 - \frac{4.86}{6.45}}{L}\right) = \frac{1.748 \times 10^{-3}}{L}. \]Solving for \( L \):\[ L = \frac{1.748 \times 10^{-3}}{-\ln\left(1 - \frac{4.86}{6.45}\right)} \approx 0.34 \ \text{H}. \]

Key Concepts

Inductor Resistance Calculation

In an RL circuit, understanding how to calculate the resistance of an inductor is essential. The resistance of the inductor becomes a pivotal part of the circuit behavior over time. When the inductor finally reaches a steady state, known as the long-term behavior, it is crucial that we address the calculation of its resistance.

At this stage, the current through the inductor becomes stable, reaching its maximum value, which is the total current flowing through the circuit. We can relate this current, often noted as \(I_0\), to the battery's electromotive force (emf) \(\varepsilon\). This relationship is given by \(I_0 = \varepsilon / R\). Here, \(R\) represents the resistance of the inductor.

To find \(R\), simply rearrange the formula to figure out the resistance:

• \(R = \varepsilon / I_0\)

Plug in the values from the problem: \(\varepsilon = 12.0\, \text{V}\) and \(I_0 = 6.45 \times 10^{-3}\, \text{A}\). From this, we calculate:

• \(R = \frac{12.0}{6.45 \times 10^{-3}} = 1860 \, \Omega\)

This calculation determines the resistance of the inductor based on the known final current through the RL circuit.

Inductance Determination

Inductance is a measure of how effectively an inductor can store energy in its magnetic field. In an RL circuit, determining the inductance \(L\) requires analyzing the circuit dynamics as current changes over time.

The current \(I(t)\) in an RL circuit at any time \(t\) can be described by the equation:

• \(I(t) = I_0 (1 - e^{-\frac{Rt}{L}})\)

Here:

• \(I_0\) is the final steady-state current,
• \(R\) is the resistance of the inductor,
• \(t\) is the time after the circuit is closed.

By inserting known values into this equation, you can solve for \(L\), the inductance of the circuit. In our example, you know the values for \(I(t) = 4.86 \times 10^{-3}\, \text{A}\), \(I_0 = 6.45 \times 10^{-3}\, \text{A}\), and \(t = 0.940 \, \text{ms}\), with the resistance \(R\) found to be \(1860 \, \Omega\).

By replacing these in the current equation, you solve for \(L\) in the expression:

• \(4.86 \times 10^{-3} = 6.45 \times 10^{-3} (1 - e^{-\frac{1860 \times 0.940 \times 10^{-3}}{L}})\)

This is simplified further to achieve:

• \(L = \frac{1.748 \times 10^{-3}}{-\ln(1 - \frac{4.86}{6.45})} \approx 0.34 \, \text{H}\)

Conquering this mathematics allows you to determine the effective inductance of the inductor in this situation.

Current Exponential Decay

A key feature of RL circuits is how current behaves over time, particularly how it changes and stabilizes when an inductor is first connected to a circuit. This behavior is often characterized by exponential decay, which provides a clear representation of how current reaches its final value.

When the switch is closed in an RL circuit, the current does not immediately jump to its final value \(I_0\). Instead, it gradually increases, following a pattern called exponential growth towards a maximum (or final) value as defined by:

• \(I(t) = I_0 (1 - e^{-\frac{Rt}{L}})\)

This formula shows how the current evolves over time, where \(e\) is the base of the natural logarithm, reflecting the asymptotic approach of \(I(t)\) towards \(I_0\).

In this decay process, the negative exponent \(-\frac{Rt}{L}\) reflects the rate at which the current grows, dictated by both the resistance \(R\) and the inductance \(L\) of the circuit.

The longer \(t\) is, the closer \(e^{-\frac{Rt}{L}}\) gets to zero, explaining why, after a prolonged period, the current essentially settles to the steady-state current \(I_0\). This model is crucial for understanding how energy is distributed and finally stabilized across the electronic components within the circuit.

Understanding this concept is vital for designing circuits where precise current stabilization and management are required.