Problem 49 - College Physics

Question

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistor with a resistance of 8.00\(\Omega\) are connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible internal resistance. Find (a) the initial rate of increase of the current in the circuit, (b) the initial potential difference across the inductor, (c) the current 0.313 s after the circuit is closed, and (d) the maximum current.

Step-by-Step Solution

Answer

(a) 2.40 A/s. (b) 6.00 V. (c) 0.475 A. (d) 0.75 A.

Step 1: Understanding the Circuit Equations

First, we need to understand the components in the circuit. An inductor, a resistor, and a DC voltage source form an RL circuit, for which the behavior of current is governed by the differential equation: \(\varepsilon = L \frac{di(t)}{dt} + Ri(t)\), where \(\varepsilon\) is the EMF, \(L\) is the inductance, \(R\) is the resistance, and \(i(t)\) the current as a function of time.

Step 2: Calculate Initial Rate of Current Increase

To find the initial rate of increase of the current, we consider that initially \(i(0) = 0\). Thus, the equation becomes \(\varepsilon = L \frac{di(0)}{dt}\). Solving for the initial rate of change, we find \(\frac{di(0)}{dt} = \frac{\varepsilon}{L} = \frac{6.00}{2.50}\). This gives \(\frac{di(0)}{dt} = 2.40 \, \mathrm{A/s}\).

Step 3: Initial Potential Difference Across the Inductor

The initial potential difference across the inductor can be calculated using \(V_L = L \frac{di(0)}{dt}\). Substituting the values, \(V_L = 2.50 \times 2.40 = 6.00 \, \mathrm{V}\). This potential difference also equals the EMF at \(t=0\).

Step 4: Calculating Current at a Given Time

The current as a function of time, \(i(t)\), in an RL circuit is given by \(i(t) = \frac{\varepsilon}{R}(1-e^{-\frac{R}{L}t})\). Substituting the known values, \(i(0.313) = \frac{6.00}{8.00}(1-e^{-\frac{8.00}{2.50} \times 0.313})\). Calculating, \(i(0.313) = 0.750 \times (1-e^{-1.000}) \approx 0.475 \, \mathrm{A}\).

Step 5: Finding the Maximum Current

The maximum current in the circuit occurs when \(e^{-\frac{R}{L}t} \to 0\), thus \(i_{max} = \frac{\varepsilon}{R} = \frac{6.00}{8.00} = 0.75 \, \mathrm{A}\).

Key Concepts

Inductance

Inductance is a property of an electrical component, generally a coil or inductor, that measures its ability to store energy in a magnetic field. In an RL circuit, which consists of a resistor and an inductor connected in series with a voltage source, the inductance (denoted by \( L \)) plays a crucial role in determining the behavior of the circuit.
When voltage is applied, current begins to build up, but due to the inductor, this increase is not instantaneous. The inductor resists changes in current flow by generating a magnetic field, which effectively slows the rate of current increase.
The unit of inductance is the henry (\( ext{H} \)). In our exercise, the circuit has an inductance of 2.50 henries. This means the inductor will store and release magnetic energy as the current changes, influencing the overall dynamic of the circuit. The greater the inductance, the more significant the effect of slowing down the change in current.

Resistance

Resistance is a measure of how much a component opposes the flow of electric current. It converts electric energy into heat and is symbolized by the letter \( R \).
In an RL circuit, resistance affects both the rate at which the current builds up and the maximum current that can be achieved. Our problem involves a resistor with a resistance of 8.00 ohms (\( \, \Omega \)). The resistor limits the current flow according to Ohm's Law:

• The relationship between voltage (\( V \)), current (\( I \)), and resistance (\( R \)) is given by \( V = IR \).
• The higher the resistance, the smaller the current for a given voltage.

In the given exercise, we calculate the maximum current using the formula \( i_{max} = rac{\varepsilon}{R} \), where \( \varepsilon \) is the electromotive force (EMF). This resistance plays a crucial role in defining the final current value in the circuit.

Differential Equation

Differential equations are equations that involve the derivatives of a function. They are fundamental in describing systems where change is a function of time, as in the case of RL circuit analysis.
In this exercise, the behavior of the RL circuit is governed by the differential equation:

• \( \varepsilon = L \frac{di(t)}{dt} + Ri(t) \)

Here, \( \varepsilon \) represents the EMF, \( L \) is the inductance, \( \frac{di(t)}{dt} \) is the rate of change of current with respect to time, and \( Ri(t) \) is the voltage drop across the resistor.
This equation helps us understand how current builds up over time when the circuit is closed. It reflects the balance between the induced EMF in the inductor and the voltage across the resistor. Solving this differential equation determines how quickly the current approaches its maximum value, taking into account both inductance and resistance.

Electric Current

Electric current is the rate of flow of electric charge in a circuit. It is usually measured in amperes (A) and is the central focus of analysis in RL circuits.
In the context of our exercise, we look at how the current changes right after the circuit is closed and over time. Initially, when the circuit closes, the inductor opposes any abrupt change in current; hence the current starts at zero and gradually increases. The initial rate of increase of current is defined by the formula \( \frac{di(0)}{dt} = \frac{\varepsilon}{L} \).

• For the circuit in the exercise, this rate was calculated to be 2.40 A/s.
• The current after a certain time can be found with \( i(t) = \frac{\varepsilon}{R}(1-e^{-\frac{R}{L}t}) \).

As time progresses, the current reaches a steady state, where it becomes constant—this is the maximum current calculated as \( 0.75 \text{A} \) in this case. Understanding these changes is crucial in the study of circuits.