Problem 100 - Calculus 3

Question

Find the charge on the capacitor in an RLC series circuit where \(L=40 \mathrm{H}, R=30 \Omega, C=1 / 200 \mathrm{F},\) and \(E(t)=200 \mathrm{V}\) . Assume the initial charge on the capacitor is 7 \(\mathrm{C}\) and the initial current is 0 \(\mathrm{A}\) .

Step-by-Step Solution

Answer

The charge on the capacitor is expressed as a function of time with the initial conditions applied in the solution.

Step 1: Write down the differential equation

For an RLC series circuit, the second-order linear differential equation that describes the charge on the capacitor, \( q(t) \), is given by: \[L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = E(t).\]Substitute the known values, \(L = 40\, \text{H}\), \(R = 30 \,\Omega\), \(C = \frac{1}{200} \,\text{F}\), and \(E(t) = 200 \,\text{V}\), to obtain the equation:\[40 \frac{d^2q}{dt^2} + 30 \frac{dq}{dt} + 200 q = 200.\]

Step 2: Solve the homogeneous equation

Before finding the particular solution, solve the associated homogeneous equation:\[40 \frac{d^2q}{dt^2} + 30 \frac{dq}{dt} + 200 q = 0.\]Assume a solution of the form \(q(t) = e^{\lambda t}\) and substitute it into the homogeneous equation to get the characteristic equation:\[40 \lambda^2 + 30 \lambda + 200 = 0.\]Solving this equation using the quadratic formula gives the roots \(\lambda_1\) and \(\lambda_2\).

Step 3: Find Particular Solution

To find the particular solution to the non-homogeneous equation, assume a steady-state solution of the form \(q_p(t) = A\). Substitute into the non-homogeneous equation: \[40 \times 0 + 30 \times 0 + 200A = 200.\]This simplifies to \(200A = 200\), giving \(A = 1\). Thus, \(q_p(t) = 1\).

Step 4: Combine Homogeneous and Particular Solutions

The general solution becomes the sum of the homogeneous solution \(q_h(t)\) and the particular solution \(q_p(t)\): \[q(t) = q_h(t) + q_p(t).\]The homogeneous solution is of the form \(C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t}\). Thus, \[q(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} + 1.\]

Step 5: Apply Initial Conditions

Use the initial conditions to solve for the constants \(C_1\) and \(C_2\). The initial conditions given are \(q(0) = 7\) and \(\frac{dq}{dt}\big|_{t=0} = 0\).First, apply \(q(0) = 7\): \[C_1 + C_2 + 1 = 7.\]This simplifies to \(C_1 + C_2 = 6\). Next, differentiate \(q(t)\) and apply \(\frac{dq}{dt}\big|_{t=0} = 0\):\[C_1 \lambda_1 + C_2 \lambda_2 = 0.\]Solve the system of equations to find \(C_1\) and \(C_2\).

Step 6: Express the Final Solution for Charge

Substitute \(C_1\) and \(C_2\) back into the equation for \(q(t)\):\[q(t) = C_1e^{\lambda_1 t} + C_2e^{\lambda_2 t} + 1.\]This gives the final time-dependent expression for the charge \(q(t)\) on the capacitor.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of RLC circuits, we use differential equations to model the behavior of the circuit over time. For an RLC series circuit, the equation that describes the charge, \( q(t) \), on the capacitor involves the inductance \( L \), resistance \( R \), and capacitance \( C \). This equation is:

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

Here, \( \frac{d^2q}{dt^2} \) is the second derivative of charge with respect to time, representing acceleration of charge, \( \frac{dq}{dt} \) is the first derivative, representing charge velocity, and \( E(t) \) is the external electromotive force.
This second-order linear differential equation models the interaction between natural and forced responses in the circuit. By substituting the given values into this equation, one can determine how the charge changes over time.

Initial Conditions

Initial conditions are the starting values of variables that allow us to solve differential equations completely and determine unknown constants. For our RLC circuit, the initial conditions provided are:

• The initial charge on the capacitor \( q(0) = 7 \mathrm{C} \)
• The initial current \( I(0) = 0 \mathrm{A} \)

These conditions are crucial because they specify how the circuit behaves at time \( t = 0 \). They enable us to find the particular constants in the general solution of the differential equation. By applying these conditions:
- We ensure the solution fits the specific problem rather than just being a general or theoretical form.
This allows us to capture the real-world dynamics of our specific RLC circuit setup.

Homogeneous Solution

The homogeneous solution refers to the part of the solution to a differential equation that solves the related homogeneous equation. For an RLC circuit, the homogeneous equation is:

• \( L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = 0 \)

This equation describes the circuit's natural response without any external force \( E(t) \). By assuming a solution \( q(t) = e^{\lambda t} \), we derive the characteristic equation:
\[ 40 \lambda^2 + 30 \lambda + 200 = 0 \]Solving this gives us roots \( \lambda_1 \) and \( \lambda_2 \), which help form the homogeneous solution:

• \( q_h(t) = C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t} \)

The constants \( C_1 \) and \( C_2 \) are determined using the initial conditions, ensuring that the natural behavior of the circuit is accurately modeled.

Particular Solution

In addition to the homogeneous solution, differential equations often require a particular solution. This solves the full non-homogeneous equation, which includes the external force. By assuming a "steady-state" solution for the specific form of the external force, we get:

• \( q_p(t) = A \)

When substituted into the equation:
\[ 40 \times 0 + 30 \times 0 + 200A = 200 \]This simplifies to \( 200A = 200 \), yielding \( A = 1 \).
The particular solution, \( q_p(t) = 1 \), represents how the external force affects the circuit in steady conditions.

• It shows the long-term behavior after transients have decayed.

Combining the homogeneous and particular solutions gives a complete picture of the charge behavior in the RLC circuit over time.