Problem 25 - University Physics with Modern Physics

Question

A 35.0 -V battery with negligible internal resistance, a \(50.0-\Omega\) resistor, and a \(1.25 \mathrm{mH}\) inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed. (a) How long after closing the switch will the current through the inductor reach one-half of its maximum value? (b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

Step-by-Step Solution

Answer

(a) 1.73 x 10^-5 s (b) 3.76 x 10^-5 s

Step 1: Identify the Circuit Parameters

Identify and list the given values from the problem: - Voltage of the battery, \( V = 35.0 \text{ V} \)- Resistance, \( R = 50.0 \Omega \)- Inductance, \( L = 1.25 \text{ mH} = 1.25 \times 10^{-3} \text{ H} \)- The switch is closed at \( t = 0 \).

Step 2: Expressions for Current in an LR Circuit

For a series \( RL \) circuit, the current at any time \( t \) after the switch is closed is given by:\[ i(t) = I_{max} (1 - e^{-t/\tau}) \]where \( I_{max} = \frac{V}{R} \) is the maximum current and \( \tau = \frac{L}{R} \) is the time constant of the circuit.

Step 3: Calculate the Maximum Current

Calculate \( I_{max} \) using the formula:\[ I_{max} = \frac{V}{R} = \frac{35.0 \text{ V}}{50.0 \Omega} = 0.7 \text{ A} \]

Step 4: Calculate the Time Constant

Calculate the time constant \( \tau \):\[ \tau = \frac{L}{R} = \frac{1.25 \times 10^{-3} \text{ H}}{50.0 \Omega} = 2.5 \times 10^{-5} \text{ s} \]

Step 5: Solve for Time to Reach Half Maximum Current

We need to find \( t \) when \( i(t) = 0.5 I_{max} \):\[ 0.5 I_{max} = I_{max} (1 - e^{-t/\tau}) \]Solve for \( e^{-t/\tau} \):\[ 0.5 = 1 - e^{-t/\tau} \Rightarrow e^{-t/\tau} = 0.5 \]Take the natural log of both sides:\[ -\frac{t}{\tau} = \ln(0.5) \]Solve for \( t \):\[ t = -\tau \ln(0.5) = 2.5 \times 10^{-5} \times \ln(2) = 1.73 \times 10^{-5} \text{ s} \]

Step 6: Relationship between Energy and Current in Inductor

The energy \( U \) stored in an inductor is given by:\[ U = \frac{1}{2} L i^2 \]When current \( i = 0.5 I_{max} \), energy becomes:\[ U = \frac{1}{2} L (0.5 I_{max})^2 = \frac{1}{4} \left( \frac{1}{2} L I_{max}^2 \right) \]The energy is one-quarter of its maximum when the current is half of its maximum.

Step 7: Calculate Time for Energy to Reach Half Maximum

Set the energy equation for half maximum energy:\[ 0.5 \cdot \left( \frac{1}{2} L I_{max}^2 \right) = \frac{1}{2} L i^2 \]\[ I^2 = \frac{1}{2} I_{max}^2 \]This corresponds to \( i = \frac{I_{max}}{\sqrt{2}} \).Use the current expression:\[ \frac{I_{max}}{\sqrt{2}} = I_{max} (1 - e^{-t/\tau}) \]\[ \frac{1}{\sqrt{2}} = 1 - e^{-t/\tau} \]Solve for \( t \):\[ e^{-t/\tau} = 1 - \frac{1}{\sqrt{2}} \]\[ t = -\tau \ln \left( 1 - \frac{1}{\sqrt{2}} \right) \approx 3.76 \times 10^{-5} \text{ s} \]

Key Concepts

Time Constant

The time constant, denoted as \( \tau \), is a fundamental concept in RL circuits. It represents the time taken for the current to reach approximately 63.2% of its maximum value in a circuit, or discharging to about 36.8% of its initial value. This constant is calculated using the formula \( \tau = \frac{L}{R} \), where \( L \) is the inductance measured in henrys (H) and \( R \) is the resistance measured in ohms (Ω).

In the given exercise, the time constant is \( 2.5 \times 10^{-5} \) seconds or 25 microseconds. This means that after one time constant, the current in the circuit will have risen to 63.2% of its maximum steady state value. Time constant helps in analyzing how quickly a given RL circuit responds to changes in voltage or current, which is crucial for applications that require precise control over timing.

Understanding the time constant can also provide insights into the rate at which energy is dissipated or stored in the circuit, demonstrating its importance in both theoretical calculations and practical implementations.

Inductor Energy

An inductor stores energy in a magnetic field when electrical current passes through it. The energy stored in an inductor can be calculated using the formula: \( U = \frac{1}{2} L i^2 \), where \( U \) is energy in joules, \( L \) is inductance in henrys, and \( i \) is current in amperes. The energy depends quadratically on the current, meaning that small changes in current can dramatically affect the energy stored.

In the scenario where the current reaches half of its maximum, the energy stored is only a quarter of the maximum possible energy. This is because the energy is proportional to the square of the current, revealing an important aspect of energy storage in inductors. For example, if \( I_{max} \) is the maximum current and \( i = 0.5 I_{max} \), then the energy \( U \) is \( \frac{1}{4} \times \frac{1}{2} L I_{max}^2 \).

Understanding how energy is stored and released in an inductor is key for designing circuits that involve inductive loads, making this concept critically important for engineers and technologists working in fields such as power electronics and signal processing.

Maximum Current

For a DC-powered RL circuit, when the switch closes, the circuit initially resists changes in current. Over time, current gradually increases until it reaches a steady state known as the maximum current. This is determined by Ohm's Law, expressed in the formula: \( I_{max} = \frac{V}{R} \), where \( V \) is voltage, and \( R \) is resistance.

In the given exercise, the maximum current calculated is 0.7 A, determined by the source voltage of 35.0 V divided by a 50.0 Ω resistor. The maximum current is crucial because it sets the upper limit of current flow in the circuit, and designers must ensure that components can handle this level.

Knowing the maximum current allows engineers to appropriately size and protect circuit elements. This understanding mitigates risks of overheating and ensures that inductors operate within their specifications, preserving their durability and performance.

Exponential Growth in Circuits

In RL circuits, current does not reach its maximum value instantaneously, but rather grows in an exponential manner. This behavior can be described by the equation \( i(t) = I_{max} (1 - e^{-t/\tau}) \), which shows how current increases over time once the switch in a circuit is closed. Initially, the rate of increase is rapid, but it slows down as it approaches \( I_{max} \).

The exponential growth reflects the nature of inductors, which oppose changes in current. It illustrates how circuits stabilize over time. The time constant \( \tau \) again plays a crucial role, as it determines the speed of this process. For circuits requiring rapid response times, small inductance or high resistance values are chosen to decrease \( \tau \).

Understanding this exponential behavior helps in predicting the response of RL circuits more accurately, which is critical in applications such as signal processing, where precise timing is essential. This knowledge allows engineers to design circuits that meet specific performance criteria, especially in dynamic environments.