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An electromotive force $$E(t)=\left\\{\begin{array}{lr}120, & 0 \leq t \leq 20 \\\0, & t>20\end{array}\right.$$\ is applied to an \(L R\) -series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current \(i(t)\) if \(i(0)=0\)

Short Answer

Expert verified
The current is \(i(t) = -60 e^{-t/10} + 60\) for \(0 \leq t \leq 20\), and \(i(t) = (-60 e^{-2} + 60) e^{2-t/10}\) for \(t > 20\).

Step by step solution

01

Set up the Differential Equation

The given electromotive force \(E(t)\) is a piecewise function. For \(0 \leq t \leq 20\), \(E(t) = 120\), and for \(t > 20\), \(E(t) = 0\). The differential equation for an \(LR\)-circuit is \( L \frac{di(t)}{dt} + Ri(t) = E(t) \). Here, \(L = 20\), \(R = 2\). Substitute these values:\[ 20 \frac{di(t)}{dt} + 2i(t) = E(t) \]
02

Simplify the Differential Equation

Simplify the equation by dividing every term by 2:\[ 10 \frac{di(t)}{dt} + i(t) = \frac{E(t)}{2} \]For \(0 \leq t \leq 20\), \(E(t) = 120\), so:\[ 10 \frac{di(t)}{dt} + i(t) = 60 \]
03

Solve for Homogeneous Solution

Set up the homogeneous equation by setting \(E(t) = 0\):\[ 10 \frac{di(t)}{dt} + i(t) = 0 \]The solution for this differential equation is:\[ i_h(t) = C e^{-\frac{t}{10}} \]
04

Solve for Particular Solution

Assume a constant particular solution \(i_p(t) = A\). Substitute into the simplified equation:\[ A = 60 \].Thus, the particular solution is a constant: \(i_p(t) = 60\).
05

Solve for Total Solution

Combine the homogeneous and particular solutions:\[ i(t) = i_h(t) + i_p(t) = C e^{-\frac{t}{10}} + 60 \].
06

Apply Initial Condition

Use the initial condition \(i(0) = 0\) to find \(C\):\[ i(0) = C e^{0} + 60 = 0 \implies C + 60 = 0 \implies C = -60 \].Thus, the solution becomes:\[ i(t) = -60 e^{-\frac{t}{10}} + 60 \] for \(0 \leq t \leq 20\).
07

Solve for t > 20

For \(t > 20\), the electromotive force \(E(t) = 0\), and the equation becomes homogeneous:\[ 10 \frac{di(t)}{dt} + i(t) = 0 \].The general solution is \(i(t) = D e^{-\frac{t}{10}}\).
08

Determine Constant for t > 20

Use continuity of current at \(t = 20\):\[ i(20) = -60 e^{-2} + 60 \].Set this equal to the expression for \(i(t)\) when \(t > 20\):\[ i(20) = D e^{-2} \].Solve for \(D\):\[ D = (-60 e^{-2} + 60) e^{2} \].Now for \(t > 20\),\[ i(t) = (-60 e^{-2} + 60) e^{2 - \frac{t}{10}} \].

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Key Concepts

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

LR Circuit
An LR circuit consists of an inductor (L) and a resistor (R) connected in series with each other. These circuits are important in electrical engineering due to their ability to filter signals and manage current flow. The inductor stores energy in a magnetic field when electrical current passes through it, while the resistor dissipates energy in the form of heat.

When analyzing an LR circuit, it's essential to understand that the behavior of current over time can be modeled using differential equations. The key components that dictate the circuit's response are the values of inductance (measured in henries) and resistance (measured in ohms). Together, these influence how fast the current builds up or decays over time.
  • Inductor (L): stores energy and opposes changes in current.
  • Resistor (R): controls the rate of current flow and energy dissipation.
The fundamental equation for an LR circuit is: \( L \frac{di(t)}{dt} + Ri(t) = E(t) \), where \( E(t) \) is the electromotive force applied to the circuit. By solving this equation, you can determine the current \( i(t) \) at any time \( t \).
Piecewise Function
A piecewise function is a mathematical expression defined by different formulas or rules over different intervals of its domain. In the context of this problem, the electromotive force \( E(t) \) is a piecewise function. It is defined as 120 volts for the interval \(0 \leq t \leq 20\) and 0 volts for \(t > 20\).

Piecewise functions are useful when modeling situations where a system behaves differently at different times or under different conditions. In our problem, this behavior can be due to switching the power supply on or off at specified times.
  • Defined for specific domains: Different behaviors based on the value of \(t\).
  • Practical in real-world applications: Reflects situations where output changes suddenly.
Piecewise functions allow differential equations to consider these changes, enabling accurate modeling and predictions.
Initial Condition
In differential equations, an initial condition is a given value that specifies the state of a function or its derivative at a particular time. These conditions are essential for determining the unique solution to a differential equation.

For the LR circuit problem, we have an initial condition of \(i(0) = 0\). This tells us that the current at time \(t = 0\) is zero amperes. Without this initial condition, the solution to the differential equation would not be unique, as there would be an arbitrary constant left undetermined.
  • Defines the initial state: Specifies current flow at \(t = 0\).
  • Ensures a unique solution: Essential for finding precise results.
By applying initial conditions, we can solve for unknown constants and ensure that our solution correctly models the system's dynamics over time.
Homogeneous and Particular Solutions
When solving linear differential equations like those seen in LR circuits, it often involves finding both the homogeneous and particular solutions.

The homogeneous solution is the solution to the differential equation when the external force or input \(E(t) = 0\). This means we're only considering the circuit's response due to its properties without any driving source. In our case, the homogeneous solution \( i_h(t) \) is formulated as \( C e^{-\frac{t}{10}} \).
  • Homogeneous solution: Models natural circuit response without external force.
  • Exponential decay: Characterizes how current naturally dissipates over time.
The particular solution, on the other hand, satisfies the non-homogeneous equation, where an external force or function \(E(t)\) is applied. Here, the particular solution \( i_p(t) \) is a constant value of 60, reflecting the steady state reached due to sustained external input.
  • Particular solution: Captures the circuit's response to an external force.
  • Steady state: Constant solution arising from continuous external force over time.
By combining these, we obtain the general solution which reflects both the inherent behavior of the circuit and its response to external influences.

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Most popular questions from this chapter

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3 when friction and contraction of water at the hole are ignored, the height \(h\) of water in the tank is described by $$\frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h},$$ where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve the DE if the initial height of the water is \(H . \mathrm{By}\) hand, sketch the graph of \(h(t)\) and give its interval \(I\) of definition in terms of the symbols \(A_{w}, A_{h},\) and \(H\) Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\) (b) Suppose the tank is 10 feet high and has radius 2 feet and the circular hole has radius \(\frac{1}{2}\) inch. If the tank is initially full, how long will it take to empty?

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The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time \(t\). After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?

(a) A simple model for the shape of a tsunami is given by $$\frac{d W}{d x}=W \sqrt{4-2 W}.$$ where \(W(x) > 0\) is the height of the wave expressed as a function of its position relative to a point offshore. By inspection, find all constant solutions of the DE. (b) Solve the differential equation in part (a). A CAS may be useful for integration. (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition \(W(0)=2\).

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