Problem 19 - University Physics with Modern Physics

Question

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible intermal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is \(0.500 \mathrm{A} ;(\mathrm{c})\) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

Step-by-Step Solution

Answer

(a) 2.40 A/s; (b) 1.60 A/s; (c) 0.432 A; (d) 0.75 A.

Step 1: Understanding the Circuit

The circuit consists of an inductor with an inductance of \( L = 2.50 \ \mathrm{H} \) and a resistance of \( R = 8.00 \ \Omega \), connected to a battery providing an emf \( \varepsilon = 6.00 \ \mathrm{V} \). The circuit behaves as an RL circuit. For RL circuits, the transient response of current is characterized by an exponential function depending on the time constant, \( \tau = \frac{L}{R} \).

Step 2: Calculate the Initial Rate of Increase of Current

Initially, when the switch is closed, the current starts from zero and increases at a certain rate. The formula to find the initial rate of increase of the current is derived from \( \varepsilon - iR = L \frac{di}{dt} \). At \( t = 0 \), \( i = 0 \), so \( \frac{di}{dt} = \frac{\varepsilon}{L} = \frac{6.00}{2.50} = 2.40 \ \mathrm{A/s} \).

Step 3: Calculate the Rate of Increase of Current at 0.500 A

At an instant when the current is \( i = 0.500 \ \mathrm{A} \), use the relation \( \varepsilon - iR = L \frac{di}{dt} \), therefore \( \frac{di}{dt} = \frac{\varepsilon - iR}{L} = \frac{6.00 - (0.500)(8.00)}{2.50} = 1.60 \ \mathrm{A/s} \).

Step 4: Calculate the Current after 0.250 s

Use the formula for the current in an RL circuit: \( i(t) = \frac{\varepsilon}{R} (1 - e^{-t/\tau}) \), where \( \tau = \frac{L}{R} = 0.3125 \ \mathrm{s} \). Plug in \( t = 0.250 \ \mathrm{s} \), \[ i(0.250) = \frac{6.00}{8.00} (1 - e^{-0.250/0.3125}) \approx 0.432 \ \mathrm{A} \].

Step 5: Determine the Final Steady-State Current

At steady state, the current reaches its maximum value as the exponential term goes to zero. The current is given by \( i = \frac{\varepsilon}{R} = \frac{6.00}{8.00} = 0.75 \ \mathrm{A} \).

Key Concepts

Transient Response

In an RL circuit, when a change is initiated (like closing a switch), the system doesn’t reach its final conditions instantaneously. Instead, it goes through a **transient response** phase. This phase is crucial because it describes how the circuit transitions from one state to another.

The transient response is characterized by how the current changes over time, usually following an exponential curve. Immediately after a switch is closed, the current starts from zero. It gradually increases as it approaches a steady-state value. This period, where the current is adjusting to reach its final value, is known as the transient state of the circuit.

Understanding transient response is vital because it impacts how quickly the circuit can react to changes, which is important for designing circuits that need to respond to signals swiftly.

Time Constant

The **time constant** (\[\tau = \frac{L}{R}\]) plays a key role in determining the speed of the transient response in an RL circuit. The time constant is essentially a measure of how quickly the current reaches its steady-state value after a change occurs.

A larger time constant means the current takes longer to reach its steady state. Conversely, a smaller time constant indicates a quicker response. For instance, in our example, with an inductance of 2.50 H and a resistance of 8.00 Ω, the time constant is 0.3125 seconds. This tells us the timescale over which the current changes significantly.

In practical terms, in about one time constant, the current will have reached about 63% of its final steady-state value. This concept helps in designing circuits to ensure they meet performance requirements for speed and efficiency.

Inductor Behavior

The behavior of an inductor in an RL circuit is pivotal in shaping the circuit’s overall response. An inductor resists changes in current due to its stored energy. Initially, when a circuit is powered, the inductor generates a back voltage that opposes the increase in current.

This behavior can be understood in terms of \[\frac{di}{dt} = \frac{\varepsilon}{L} - \frac{iR}{L}\]which shows how the rate of change of current is directed by the inductor and resistor's combined effects. At the start, since the current is zero, the rate of increase is only limited by the inductance (i.e., only affected by \(\frac{\varepsilon}{L}\)).

As time progresses, the increasing current develops more voltage across the resistor (\(iR\)), slowing the rate of increase in current. Eventually, when the current stops changing, the inductor has no more opposing voltage to generate, reaching a point of no further opposition to steady current flow.

Steady-State Current

The **steady-state current** is the current that flows through the circuit once it has stabilized after the transient phase. In an RL circuit, once it reaches steady state, the current becomes constant, and any changes introduced will result again in another transient response.

Specifically, for our example, the steady-state current can be calculated simply by Ohm’s Law: \[i = \frac{\varepsilon}{R} = \frac{6.00}{8.00} = 0.75 \ A\]At this steady state, the inductor no longer poses any reactive impedance to the current flow. It's as if the circuit behaves like a resistor-only circuit (with the excepted value of direct current). For analysis purposes, this is the end goal of the current after all resistive and reactive changes have settled.
The steady-state current is critical, especially for devices powered by RL circuits, as this is the amount of current they will operate on for most of their functioning life.