Question

Electric Circuits In Exercises 33 and \(34,\) use the differential equation for electric circuits given by \(L \frac{d I}{d t}+R I=E\) . In this equation, \(I\) is the current, \(R\) is the resistance, \(L\) is the inductance, and \(E\) is the electromotive force (voltage). Solve the differential equation for the current given a constant voltage \(E_{0} .\)

Short answer

The solution to this differential equation problem involves rewriting the equation into standard form, identifying and applying the integrating factor, simplifying and integrating, and finally solving for current \(I\). The final expression for current \(I\) is derived and would depend on the values of resistance \(R\), inductance \(L\), and constant voltage \(E_{0}\).

Step-by-step solution

Develop the Differential Equation into Standard Form

First, let's rewrite the given differential equation by isolating the derivative term. Given \(L \frac{d I}{d t}+R I=E\), we can rewrite it into the standard form as \(\frac{d I}{d t}+\frac{R}{L} I=\frac{E}{L}\). Since \(E\) is a constant \(E_{0}\), the equation further simplifies to: \(\frac{d I}{d t}+\frac{R}{L} I=\frac{E_{0}}{L}\).

Identify the Integrating Factor

Once in standard form, a method utilized for solving a linear differential equation is applying the 'Integrating factor' method. The key is to multiply both sides of the equation by an 'integrating factor', in this case \( e^{\int\frac{R}{L} dt}\), to make the left-hand side of the equation a simple product rule for derivatives, allowing for easier integration. Our equation becomes \(e^{\int\frac{R}{L} dt} \frac{d I}{d t} + e^{\int\frac{R}{L} dt} \frac{R}{L} I = e^{\int\frac{R}{L} dt} \frac{E_{0}}{L}\).

Simplify and Integrate

Now our left-hand side of the equation can be simplified as a derivative of the product of \(I\) and integrating factor. Now, \(\frac{d}{dt} (e^{\int\frac{R}{L} dt} I) = e^{\int\frac{R}{L} dt} \frac{E_{0}}{L}\). And we can integrate both sides over time to solve for \(I(t)\). This leads to \(e^{\int\frac{R}{L} dt} I = \int e^{\int\frac{R}{L} dt} \frac{E_{0}}{L} dt = \frac{E_{0}}{L} \int e^{\int\frac{R}{L} dt} dt\).

Solve for I

To get \(I\), we must divide both sides by our integrating factor. Therefore, we'd get \(I = \frac{1}{e^{\int\frac{R}{L} dt}} \frac{E_{0}}{L} \int e^{\int\frac{R}{L} dt} dt\). Finishing the integration and simplification would give us the solution.

Key concepts

Differential Equations

Differential equations play a crucial role in modeling many real-world phenomena, including electric circuits. A differential equation involves derivatives, which represent rates of change. In the context of an electric circuit, the differential equation given in the exercise is \(L \frac{d I}{d t} + R I = E\). Here, \(I\) is the current, \(R\) is the resistance, \(L\) is the inductance, and \(E\) is the electromotive force (voltage). This equation describes how the current \(I\) changes with time \(t\) based on the properties of the circuit.

• The left-hand side of the equation combines the effects of inductance and resistance on the current.
• The right-hand side represents the external voltage source applied to the circuit.

Solving this differential equation helps us understand the current's behavior over time.

Linear Differential Equation

Linear differential equations are a subset of differential equations characterized by their linearity in terms of the unknown function and its derivatives. The given electric circuit equation \(L \frac{d I}{d t} + R I = E\) is linear because it does not involve products, powers, or any nonlinear operations of the current \(I\) or its derivative.

• The coefficients of \(I\) and \(\frac{d I}{d t}\) are constants, making it a first-order linear differential equation.
• The term \(E\) on the right side can be constant \(E_0\), which simplifies the problem further.

Linear equations are simpler to solve compared to nonlinear equations, often involving standard techniques such as the integrating factor method used in the given solution.

Integrating Factor

An integrating factor is a mathematical tool used to simplify the process of solving linear first-order differential equations. It transforms the differential equation into a form that is easier to integrate. For our problem, once the equation is in standard form \(\frac{d I}{d t} + \frac{R}{L} I = \frac{E_0}{L}\), the integrating factor is determined by the expression \(e^{\int \frac{R}{L} dt}\).

• This expression simplifies the differentiation and allows the left side of the equation to be rewritten as the derivative of a product, facilitating integration.
• Multiplying the entire equation by this integrating factor effectively enables us to solve the differential equation by integrating both sides with respect to time.

This method ensures a straightforward path from a complex differential equation to a manageable integral, making it a powerful technique in solving such equations.

Inductance and Resistance

Inductance \(L\) and resistance \(R\) are fundamental properties of electric circuits that describe how currents interact with voltage and magnetic fields.
Inductance:

• Inductance is a measure of a circuit's ability to resist changes in current, often caused by an induced electromotive force.
• It introduces a time-dependent aspect, as seen in the term \(L \frac{d I}{d t}\), where \(L\) influences how quickly the current adjusts to changes in the circuit.

Resistance:

• Resistance is the opposition to the flow of electric current, represented by \(R\).
• In the equation, it appears as \(R I\), directly affecting the amount of current relative to the applied voltage. Higher resistance means less current flows for a given voltage.

Both inductance and resistance critically influence the behavior of electric circuits, impacting how systems stabilize to external voltage changes.