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Chapter 3: Problem 6

An \(L R\) series circuit has a variable inductor with the inductance defined by $$ L= \begin{cases}1-\frac{t}{10}, & 0 \leq t<10 \\ 0, & t \geq 10\end{cases} $$ Find the current \(i(t)\) if the resistance is \(0.2 \mathrm{ohm}\), the impressed voltage is \(E(t)=4\), and \(i(0)=0\). Graph \(i(t)\).

Short Answer

Expert verified
i(t) = -8 for 0鈮<10, and i(t) = 20 for t鈮�10.

Step by step solution

01

Write the Differential Equation

Using Kirchoff's voltage law, for an LR circuit, we have the equation \( E(t) = L \frac{di}{dt} + Ri \). Given \( E(t) = 4 \), \( R = 0.2 \), and \( L = 1 - \frac{t}{10} \) for \( 0 \leq t < 10 \), the equation becomes \( 4 = \left(1 - \frac{t}{10}\right) \frac{di}{dt} + 0.2i \).
02

Solve the Differential Equation for 0 鈮� t < 10

Rearrange and solve the differential equation \( \left(1 - \frac{t}{10}\right) \frac{di}{dt} + 0.2i = 4 \):Separate variables by dividing by \( 1 - \frac{t}{10} \):\[ \frac{di}{dt} + \frac{0.2i}{1 - \frac{t}{10}} = \frac{4}{1 - \frac{t}{10}} \]This is a first-order linear ODE. Use an integrating factor \( \mu(t) = e^{\int \frac{0.2}{1 - \frac{t}{10}} dt} = (1 - \frac{t}{10})^{-2} \) to transform it to:\[ \left((1 - \frac{t}{10})^{-2}i\right)' = 4(1 - \frac{t}{10})^{-3} \]Integrating both sides gives:\[ (1 - \frac{t}{10})^{-2}i = \int 4(1 - \frac{t}{10})^{-3} dt + C \]This integral results in:\[ -8(1 - \frac{t}{10})^{-2} = 8 \int (1 - \frac{t}{10})^{-3} dt \]Substituting back, we find:\[ (1 - \frac{t}{10})^{-2}i = (1 - \frac{t}{10})^{-2}i_0 - 8(1 - \frac{t}{10})^{-2} \] Thus, the solution is \( i(t) = i_0 - 8 = -8 \) when \( i(0) = 0 \).
03

Consider the Case When t 鈮� 10

For \( t \geq 10 \), \( L = 0 \) and the circuit reduces to just the resistor. The voltage equation becomes \( 0.2i = 4 \). Solving for \( i \), we find \( i(t) = 20 \).
04

Combine Solutions

The full expression for the current \( i(t) \) is therefore:\[ i(t) = \begin{cases} -8, & 0 \leq t < 10 \ 20, & t \geq 10 \end{cases} \]
05

Graph the Current Function

Graph the function \( i(t) \) with time \( t \) on the x-axis and current \( i \) on the y-axis. The graph will remain constant at \( -8 \) for \( 0 \leq t < 10 \), and at \( t = 10 \), it will jump to \( 20 \) and remain there for \( t \geq 10 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that relate a function to its derivatives. They are critical in modeling how things change under various conditions. In the context of LR circuits like the one in our exercise, a differential equation is used to describe the relationship between voltage, inductance, current, and resistance.
The particular differential equation used here arises from Kirchhoff's voltage law, leading to the equation:
  • \( E(t) = L(t) \frac{di}{dt} + Ri \)
This equation tells us how the current \(i\) in the circuit changes over time given the voltage \(E(t)\) and the components \(L\) and \(R\). By solving this differential equation, we can determine the function \(i(t)\), which gives us the current at any time \(t\).
Kirchhoff's Law
Kirchhoff's laws are foundational in electrical circuit theory, and they help us understand how currents and voltages behave in a circuit. Specifically, Kirchhoff鈥檚 voltage law (KVL) states that for any closed loop in a circuit, the sum of the electromotive forces must equal the sum of potential drops (i.e., voltage).
Using KVL in our LR circuit, we write:
  • \( E(t) = L \frac{di}{dt} + Ri \)
This equation basically asserts that the total voltage supplied by the power source \(E(t)\) is equal to the voltage drop across the inductor \(L \frac{di}{dt}\) plus the voltage drop across the resistor \(Ri\). Understanding and setting up this equation is key to solving circuit problems that involve changing currents, particularly when the inductance varies over time.
First-order Linear ODE
First-order linear ordinary differential equations (ODEs) are equations that involve the first derivative of a function. They are of the form:
  • \( \frac{di}{dt} + P(t)i = Q(t) \)
Our given equation after rearranging becomes a first-order linear ODE:
  • \( \frac{di}{dt} + \frac{0.2i}{1 - \frac{t}{10}} = \frac{4}{1 - \frac{t}{10}} \)
The typical way to solve these equations is by using an integrating factor. For this problem, we use the integrating factor:
  • \( \mu(t) = (1 - \frac{t}{10})^{-2} \)
After applying this integrating factor, the equation simplifies and can be easily integrated, allowing us to find \(i(t)\), the function describing current over time.
Variable Inductance
Variable inductance in an LR circuit means that the inductance \(L\) changes over time. In our exercise, the inductor's inductance is given as:
  • \( L = 1 - \frac{t}{10} \), for \(0 \leq t < 10 \)
  • \( L = 0 \) for \(t \geq 10 \)
Variable inductance can significantly affect the behavior of the circuit, especially the rate at which current changes. Initially, as \(t\) goes from 0 to 10, the inductance decreases linearly from 1 to 0. This reduction implies less opposition to the change in current, influencing the current \(i(t)\) directly.
When \(t\) reaches 10, the inductance drops to 0, meaning the inductor no longer affects the circuit, and only the resistor is present. This results in a sudden change in the current behavior, as reflected in the current solution transitions from \(-8\) to \(20\).

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