Problem 20 - University Physics with Modern Physics

Question

A \(15.0-\Omega\) resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed switch. (a) At 200 ms after the switch is opened the current has decayed to 0.210 A. Calculate the inductance of the coil. (b) Calculate the time constant of the circuit. (c) How long after the switch is closed will the current reach 1.00\(\%\) of its original value?

Step-by-Step Solution

Answer

(a) The inductance is 4.32 H. (b) The time constant is 0.288 s. (c) The current reaches 1% in 1.327 s.

Step 1: Understanding the Circuit and Problem

We have a series circuit with a 15.0 Ω resistor and an inductor. When the switch is open, the current decays exponentially. We need to find the inductance of the coil, time constant, and time for current to reach 1% of its original value.

Step 2: Using the Formula for Exponential Decay

The current decay when the switch is opened follows the equation: \[ I(t) = I_0 e^{-\frac{t}{\tau}} \]where \( I(t) \) is the current at time \( t \), \( I_0 \) is the initial current, and \( \tau \) is the time constant. We know \( I(t) = 0.210 \text{ A} \) at \( t = 200 \text{ ms} \).

Step 3: Calculating the Initial Current

The initial current \( I_0 \) when the switch was closed can be determined using Ohm's Law:\[ I_0 = \frac{V}{R} = \frac{6.30 \text{ V}}{15.0 \Omega} = 0.42 \text{ A} \]

Step 4: Solving for the Time Constant (a)

By rearranging the decay equation, we solve for \( \tau \):\[ 0.210 = 0.42 e^{-\frac{0.2}{\tau}} \]\[ e^{-\frac{0.2}{\tau}} = 0.5 \]Take the natural logarithm of both sides:\[ -\frac{0.2}{\tau} = \ln(0.5) \]\[ \tau = \frac{0.2}{-\ln(0.5)} \approx 0.2 \times 1.443 = 0.288 \text{ s} \]

Step 5: Calculating the Inductance (b)

Use the relationship between the time constant \( \tau \), inductance \( L \), and resistance \( R \):\[ \tau = \frac{L}{R} \]\[ L = \tau \times R = 0.288 \text{ s} \times 15.0 \Omega = 4.32 \text{ H} \]

Step 6: Calculating the Time to Reach 1% of Initial Current (c)

To find the time it takes for the current to decay to 1% of its initial value:\[ I(t) = 0.01 \times I_0 = 0.0042 \text{ A} \]\[ 0.0042 = 0.42 e^{-\frac{t}{0.288}} \]\[ e^{-\frac{t}{0.288}} = 0.01 \]Taking the natural logarithm gives:\[ -\frac{t}{0.288} = \ln(0.01) \]\[ t = -0.288 \times \ln(0.01) \approx 0.288 \times 4.605 = 1.327 \text{ s} \]

Key Concepts

Exponential Decay

In electrical circuits, exponential decay describes how the current decreases over time when a switch is opened. This behavior is prominent in circuits that include components like resistors and inductors. When the electrical path is broken by the switch, the current doesn't simply stop immediately; instead, it reduces gradually in what we call an exponential decay.

This decay can be captured mathematically by the equation \[ I(t) = I_0 e^{-\frac{t}{\tau}} \]where:

• \( I(t) \) is the current at time \( t \)
• \( I_0 \) is the initial current
• \( \tau \) is known as the time constant
• \( e \) is the base of the natural logarithm

The decay is defined as 'exponential' because the current decreases in proportion to its current value; essentially, it rapidly declines at first and slows as it approaches zero. Understanding this concept is crucial for calculating how long it takes for currents to decrease to specific levels in practical applications.

Time Constant

The time constant, denoted as \( \tau \), is a critical factor in understanding the behavior of circuits involving resistors and inductors. It informs us about how fast or slow the exponential decay occurs.

Mathematically, \( \tau \) is described by the equation:\[ \tau = \frac{L}{R} \]where:

• \( L \) is the inductance of the coil, measured in Henries (H)
• \( R \) is the resistance, measured in Ohms (\( \Omega \))

In the context of our circuit, the time constant represents the time it takes for the current to decrease to approximately 37% of its original value – a significant point when observing exponential decay. The relationship between inductance and resistance helps determine how swiftly energy stored in an inductor is released.

A high time constant means the current decays slowly, as energy is released at a slower pace, whereas a low time constant suggests a faster decay.

Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering, vital for calculating current flow in circuits. This law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance of the conductor.

Expressed mathematically, Ohm’s Law is:\[ I = \frac{V}{R} \]where:

• \( I \) is the current measured in Amperes (A)
• \( V \) is the voltage measured in Volts (V)
• \( R \) is the resistance measured in Ohms (\( \Omega \))

In the context of the given problem, we used Ohm’s Law to determine the initial current when the switch was closed, by dividing the voltage (6.30 V) by the resistance (15.0 \( \Omega \)). Applying Ohm's Law provides a foundational step to understanding how changes in voltage or resistance will affect the current, especially when paired with elements that introduce inductance to the circuit.