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Chapter 3: Problem 19

An \(R L C\) circuit with input voltage \(E(t)\) is described. Find the current \(I(t)\) using the given initial current and charge on the capacitor $$ \begin{aligned} &R=60 \Omega, L=2 \mathrm{H}, C=0.0025 \mathrm{~F} \\ &E(t)=100 e^{-10 t} \mathrm{~V} ; \quad I(0)=0, Q(0)=1 \end{aligned} $$

Short Answer

Expert verified
Solve the differential equation and apply initial conditions to find \(I(t)\).

Step by step solution

01

Write the Differential Equation

For an RLC circuit, the differential equation is given by \( L\frac{dI}{dt} + RI + \frac{1}{C}Q = E(t) \). Substituting the given values, we get \( 2\frac{dI}{dt} + 60I + 400Q = 100e^{-10t} \).
02

Express Q in Terms of I

Since \( \frac{dQ}{dt} = I \), then \( Q = \int I \, dt \). Thus, the equation becomes \( 2\frac{dI}{dt} + 60I + 400 \int I \, dt = 100e^{-10t} \).
03

Solve the Homogeneous Equation

Consider the homogeneous part: \( 2\frac{dI}{dt} + 60I + 400 \int I \, dt = 0 \). Assume \( I(t) = Ae^{rt} \), then solve the characteristic equation to find roots.
04

Solve the Particular Solution

As the particular solution is similar to the input function, assume \( I_p(t) = Be^{-10t} \). Substitute into the main equation to find \( B \).
05

Apply Initial Conditions

Using \( I(0) = 0 \) and \( Q(0) = 1 \), substitute into the found general solution to determine constants.

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Key Concepts

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

Differential Equations
In electrical engineering, differential equations are fundamental for modeling and analyzing circuit behavior over time. They describe how circuit variables such as current or voltage change. In the context of an RLC circuit, this understanding helps us see how the circuit responds to a voltage input and how current evolves. For this scenario, the differential equation is given by:
  • \( L\frac{dI}{dt} + RI + \frac{1}{C}Q = E(t) \)
where:
  • \( L \) is inductance, \( R \) is resistance, \( C \) is capacitance, \( Q \) is charge, and \( E(t) \) is the voltage supply.
This equation essentially captures the balance between energy stored in the inductor, dissipated in the resistor, and stored in the capacitor. Solving this equation helps determine how the current \( I(t) \) behaves.
Initial Conditions
Initial conditions are the starting values required to solve any differential equation completely. They inform us about the state of the system at a specific point in time, usually at the beginning. For circuits, these typically involve current and charge:
  • \( I(0) \) represents the initial current, which is given here as 0.
  • \( Q(0) \) represents the initial charge, here given as 1.
These values are important because they allow us to calculate any unknown constants when we solve the differential equation. Initial conditions adjust the general solution of a differential equation to a specific application case. Without them, the solution remains too abstract and doesn't reflect a real-world scenario.
Characteristic Equation
To solve the homogeneous part of a differential equation, we use the characteristic equation. This involves assuming a solution of the form \( I(t) = Ae^{rt} \), where \( A \) and \( r \) are constants to be determined. For our RLC circuit:
  • First, substitute this assumed form into the homogeneous equation derived from step 3 of the original solution.
  • This will yield a polynomial equation in terms of \( r \) called the characteristic equation.
Solving the characteristic equation gives the roots which determine the form of the homogeneous solution. These roots can be real or complex and tell us about the nature of the circuit鈥檚 natural response, such as oscillations or damping.
Particular Solution
The particular solution satisfies the non-homogeneous part of the differential equation, which is driven by the external input to the system. In our example, we seek a form of \( I_p(t) \) that mimics the input voltage function, like \( Be^{-10t} \).
  • By substituting this form back into the original differential equation, we solve for the constant \( B \).
      • Combining the homogeneous and particular solutions gives us the complete description of current over time. The particular solution specifically accounts for the RLC circuit's response to the input voltage \( E(t) \). Finding this enables us to combine it with other solutions to get the total current.

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