Problem 53 - College Physics

Question

A 1.50 \(\mathrm{mH}\) inductor is connected in series with a dc battery of negligible internal resistance, a 0.750 \(\mathrm{k} \Omega\) resistor, and an open switch. How long after the switch is closed will it take for (a) the current in the circuit to reach half of its maximum value, (b) the energy stored in the inductor to reach half of its maximum value? (Hint: You will have to solve an exponential equation.)

Step-by-Step Solution

Answer

(a) \( t \approx 1.39 \times 10^{-6} \) seconds; (b) \( t \approx 4.15 \times 10^{-6} \) seconds.

Step 1: Identify the given values

We are given an inductor of inductance \( L = 1.50 \, \text{mH} = 1.50 \times 10^{-3} \, \text{H} \), a resistor with resistance \( R = 0.750 \, \text{k}\Omega = 750 \, \text{\Omega} \), and a dc source in the circuit.

Step 2: Determine the time constant

For an RL circuit, the time constant \( \tau \) is given by the formula \( \tau = \frac{L}{R} \). Substitute the given values:\[ \tau = \frac{1.50 \times 10^{-3}}{750} = 2.00 \times 10^{-6} \, \text{seconds} \]

Step 3: Calculate the time for current to reach half its maximum value

The current \( I(t) \) in an RL circuit when the switch is closed is given by:\[ I(t) = I_{\text{max}} (1 - e^{-\frac{t}{\tau}}) \]Where \( I_{\text{max}} \) is the maximum current. We want \( I(t) = \frac{I_{\text{max}}}{2} \). Set up the equation:\[ \frac{I_{\text{max}}}{2} = I_{\text{max}} (1 - e^{-\frac{t}{\tau}}) \]Solve for \( t \) to give:\[ 0.5 = 1 - e^{-\frac{t}{\tau}} \]\[ e^{-\frac{t}{\tau}} = 0.5 \]Taking the natural logarithm on both sides gives:\[ -\frac{t}{\tau} = \ln(0.5) \]\[ t = -\tau \ln(0.5) \]Substitute \( \tau = 2.00 \times 10^{-6} \) into the equation:\[ t = -2.00 \times 10^{-6} \ln(0.5) \approx 1.39 \times 10^{-6} \, \text{seconds} \]

Step 4: Calculate the time for the energy in the inductor to reach half the maximum value

The energy \( U(t) \) stored in the inductor is \( U(t) = \frac{1}{2} L I(t)^2 \). We need \( U(t) = \frac{1}{2} U_{\text{max}} \).Since \( U \propto I^2 \), the current when \( U \) is half of \( U_{\text{max}} \) is \( \frac{I_{\text{max}}}{\sqrt{2}} \).We use the same equation as in Step 3:\[ \frac{I_{\text{max}}}{\sqrt{2}} = I_{\text{max}} (1 - e^{-\frac{t}{\tau}}) \]\[ \frac{1}{\sqrt{2}} = 1 - e^{-\frac{t}{\tau}} \]Solve for \( t \):\[ e^{-\frac{t}{\tau}} = 1 - \frac{1}{\sqrt{2}} \]Take natural logarithms:\[ -\frac{t}{\tau} = \ln\left(1 - \frac{1}{\sqrt{2}}\right) \]\[ t = -\tau \ln\left(1 - \frac{1}{\sqrt{2}}\right) \approx 4.15 \times 10^{-6} \, \text{seconds} \]

Key Concepts

Inductor Time Constant

The time constant in an RL circuit is a crucial parameter that helps determine how quickly current adjusts after a change, like closing a switch. An RL circuit's time constant, denoted by \( \tau \), is calculated using the formula \( \tau = \frac{L}{R} \). Here, \( L \) represents the inductance of the inductor, and \( R \) is the resistance in the circuit. The time constant is essentially the time it takes for the current to reach approximately 63.2% of its maximum value through exponential growth.

In the example provided, with an inductor of 1.50 mH and a resistor of 750 Ω, the time constant is calculated as \( 2.00 \times 10^{-6} \) seconds. This means the circuit's current will reach most of its steady state in microseconds, illustrating how RL circuits can react very quickly in many applications, whether in power electronics or signal filtering systems.

• The induced voltage across the inductor influences the rate of current change.
• Smaller time constants imply faster responses, while larger ones indicate slower adjustments.

Exponential Decay in Circuits

When studying circuits with inductors and resistors, exponential decay is a common phenomenon, especially in transient states following a sudden switch closure. In an RL circuit, the current doesn't jump instantly to its maximum but rather follows an exponential curve with time.

The formula for the current in such a circuit is \( I(t) = I_{\text{max}}(1 - e^{-\frac{t}{\tau}}) \). As time progresses, the term \( e^{-\frac{t}{\tau}} \) decays exponentially from 1 to 0, allowing the current to asymptotically approach its maximum. This is because the inductor initially resists changes in current, storing energy momentarily before gradually allowing more current to pass.

For example, when the current reaches half of its maximum value, it's a heart moment to recognize the halfway point of exponential growth leading upwards.

• Exponential functions are key to modeling these gradual changes in current.
• Analysis of these curves helps engineers predict how circuits react over time.

Energy Storage in Inductors

Inductors serve as energy storage elements in circuits, capable of storing energy in the form of magnetic fields when current passes through them. This stored energy can oppose changes in current, which is one of the reasons why inductors are used for such purposes in circuits.

The energy \( U(t) \) stored in an inductor at any time \( t \) is given by \( U(t) = \frac{1}{2} L I(t)^2 \). This relationship shows that the energy stored is proportional to the square of the current. Furthermore, when the energy reaches half of its maximum level, the current doesn't just halve; instead, it falls to \( \frac{I_{\text{max}}}{\sqrt{2}} \), providing a key point for understanding how energy and current relate differently.

In practice, this nonlinear behavior of energy storage in inductors is imperative for operations like voltage regulation, signal processing, and in the design of transformers.

• Inductors release stored energy into the circuit upon a decrease in current.
• Understanding this energy transformation is critical for effective circuit design.