Problem 101 - Calculus 3

Question

Find the charge on the capacitor in an RLC series circuit where \(L=2\) H, \(R=24 \Omega, C=0.005\) F, and \(E(t)=12 \sin 10 t\) . Assume the initial charge on the capacitor is 0.001 \(\mathrm{C}\) and the initial current is 0 \(\mathrm{A}\) .

Step-by-Step Solution

Answer

The charge on the capacitor is given by: \[q(t) = e^{-6t}(0.001\cos(8t) + \frac{10}{397}\sin(8t)) + \left(\frac{-3}{49}\right)\sin(10t)\].\]

Step 1: Define the differential equation

The voltage across the circuit is governed by the differential equation \(L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{1}{C}q = E(t)\). Substituting the given values for \(L\), \(R\), \(C\), and \(E(t)\), we get: \[2\frac{d^2q}{dt^2} + 24\frac{dq}{dt} + 200q = 12\sin(10t)\].

Step 2: Solve the homogeneous equation

Firstly, solve the homogeneous equation: \(2\frac{d^2q}{dt^2} + 24\frac{dq}{dt} + 200q = 0\). The characteristic equation is \(2m^2 + 24m + 200 = 0\). Solve for \(m\) to find the roots, \(m = -6 \pm 8i\). Thus, the solution to the homogeneous equation is \(q_h(t) = e^{-6t}(A\cos(8t) + B\sin(8t))\).

Step 3: Solve the particular solution

We guess a particular solution of the form: \(q_p(t) = C\sin(10t)\). Substitute \(q_p(t)\) and its derivatives into the non-homogeneous equation and solve for \(C\). This results in \(C\sin(10t) = \frac{12}{-196}\sin(10t)\), so \(C = \frac{-3}{49}\).

Step 4: Form the general solution

Combine the homogeneous and particular solutions to obtain the general solution: \[q(t) = e^{-6t}(A\cos(8t) + B\sin(8t)) + \left(\frac{-3}{49}\right)\sin(10t)\].

Step 5: Apply initial conditions to solve for constants

Use the initial conditions to solve for constants \(A\) and \(B\). Given \(q(0) = 0.001\) and \(\frac{dq}{dt}(0) = 0\). Substitute these into the general solution and its derivative to find \(A = 0.001\) and \(B = 0.002517\approx \frac{10}{397}\).

Step 6: Write the final solution

Substitute \(A\) and \(B\) back into the general solution to get the specific solution for the charge on the capacitor: \[q(t) = e^{-6t}(0.001\cos(8t) + \frac{10}{397}\sin(8t)) + \left(\frac{-3}{49}\right)\sin(10t)\].

Key Concepts

Differential Equation

In an RLC series circuit, the charge on the capacitor is defined by a differential equation. A differential equation is a mathematical expression that relates a function with its derivatives. Here, it describes how the charge on the capacitor changes over time due to resistive, inductive, and capacitive elements in the circuit.
The standard form of the differential equation for an RLC circuit is given by:

• \( L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = E(t) \)

Where:

• \( L \) is the inductance,
• \( R \) is the resistance,
• \( C \) is the capacitance,
• \( q \) is the charge on the capacitor, and
• \( E(t) \) is the applied electromotive force (EMF) function.

This differential equation is second-order because it involves the second derivative of the charge with respect to time \( \frac{d^2q}{dt^2} \). In our specific example, after substituting the known values \( L = 2 \) H, \( R = 24 \Omega \), \( C = 0.005 \) F, and \( E(t) = 12 \sin(10t) \), the differential equation becomes:

• \( 2\frac{d^2q}{dt^2} + 24\frac{dq}{dt} + 200q = 12\sin(10t) \)

This equation governs the dynamic behavior of the charge in the system.

Characteristic Equation

The characteristic equation arises when solving the homogeneous part of the differential equation associated with the RLC circuit. A homogeneous equation is where the function equals zero, representing the system's response without external forces.
For our circuit, the homogeneous equation is:

• \( 2\frac{d^2q}{dt^2} + 24\frac{dq}{dt} + 200q = 0 \)

The characteristic equation is a polynomial obtained by replacing the function and its derivatives with algebraic terms. By substituting \( q(t) = e^{mt} \) (where \( m \) is a constant) into the homogeneous equation, we get:

• \( 2m^2 + 24m + 200 = 0 \)

Solving this characteristic equation for \( m \) is crucial as it determines the form of the solution. The roots of this equation, \( m = -6 \pm 8i \), imply a complex conjugate solution, which typically results from circuits with inductors and capacitors that can resonate. Thus, the solution to the homogeneous equation is:

• \( q_h(t) = e^{-6t}(A \cos(8t) + B \sin(8t)) \)

This solution reflects the natural response of the circuit to changes in charge.

Particular Solution

To solve the non-homogeneous differential equation in the RLC circuit, we need to find the particular solution. The particular solution accounts for the steady-state response due to the external force \( E(t) \), here given as \( 12\sin(10t) \).
We hypothesize a form for the particular solution based on the form of \( E(t) \), typically involving sinusoidal functions like sine and cosine. We choose:

• \( q_p(t) = C \sin(10t) \)

We then substitute \( q_p(t) \) and its derivatives back into the non-homogeneous differential equation:

• \( 2\frac{d^2(C \sin(10t))}{dt^2} + 24\frac{d(C \sin(10t))}{dt} + 200(C \sin(10t)) = 12\sin(10t) \)

Solve this equation for the coefficient \( C \). Through equating coefficients, \( C \) is determined to be \( -\frac{3}{49} \).
The particular solution thus forms a crucial part of the general solution, reflecting how the system reaches a new steady state when influenced by an external input.