Problem 29 - Calculus: Early Transcendentals

Question

The figure shows a circuit containing an electromotive force, a capacitor with a capacitance of \( C \) farads \( (F), \) and a resistor with resistance of \( R \) ohms \( (\Omega). \) The voltage drop across is \( Q/C, \) where \( Q \) is the charge (in coulombs), so in this case Kirchhoff's Law gives \( RI + \frac {Q}{C} = E(t) \) But \( I = dQ/dt \) (see Examples 3.7.3), so we have \( R \frac {dQ}{dt} + \frac {1}{C}Q = E(t) \) Suppose the resistance is \( 5 \Omega, \) the capacitance is 0.05 F, a battery gives a constant voltage of 60 V, and the initial charge is \( Q(0) = 0 C. \) Find the charge and the current at time \( t. \)

Step-by-Step Solution

Answer

The charge is \( Q(t) = -3 e^{-4t} + 3 \) and the current is \( I(t) = 12e^{-4t} \).

Step 1: Identify Given Values and Setup Differential Equation

We have a resistor with resistance \( R = 5 \Omega \), a capacitor with capacitance \( C = 0.05 \text{ F} \), a constant voltage \( E(t) = 60 \text{ V} \), and an initial charge \( Q(0) = 0 \text{ C} \). The differential equation is \( R \frac{dQ}{dt} + \frac{1}{C}Q = E(t) \) which simplifies to \( 5 \frac{dQ}{dt} + 20Q = 60 \).

Step 2: Solve the Homogeneous Equation

First, solve the homogeneous equation \( 5 \frac{dQ}{dt} + 20Q = 0 \). Separating variables gives \( \frac{dQ}{Q} = -4 \, dt \). Integrating both sides yields \( \ln |Q| = -4t + C_1 \), where \( C_1 \) is a constant of integration. Thus, \( Q = e^{-4t + C_1} = C_2 e^{-4t} \), where \( C_2 = e^{C_1} \).

Step 3: Find the Particular Solution

For the particular solution, assume a constant solution of the form \( Q_p = A \). Substitute into the equation: \( 5 \cdot 0 + 20A = 60 \). Solving for \( A \) gives \( A = 3 \). Thus, the particular solution is \( Q_p = 3 \).

Step 4: Combine Solutions and Apply Initial Condition

The general solution is \( Q = C_2 e^{-4t} + 3 \). Apply the initial condition \( Q(0) = 0 \): \( 0 = C_2 + 3 \). Thus, \( C_2 = -3 \). The solution becomes \( Q(t) = -3 e^{-4t} + 3 \).

Step 5: Calculate the Current

Current \( I(t) = \frac{dQ}{dt} \). Differentiate \( Q(t) = -3 e^{-4t} + 3 \) with respect to \( t \) gives \( I(t) = (-3)(-4)e^{-4t} = 12e^{-4t} \).

Key Concepts

Kirchhoff's Law

Kirchhoff's Law is a fundamental principle used to analyze electrical circuits. It helps us understand how voltage and current behave around circuit loops and nodes. In our given exercise, Kirchhoff's Law is used to create a differential equation that describes the circuit in question.

In simple terms, Kirchhoff's Law states that the sum of the electromotive force (EMF) and the voltage drops across each component of a closed loop, like resistors and capacitors, is equal to the voltage supplied. For the circuit described in the exercise, it takes the form:

• Voltage from the battery: given by the constant function, 60 volts.
• Voltage drop across the resistor: given by Ohm's Law, and expressed as \( RI \), where \( R \) is resistance, and \( I \) is the current.
• Voltage drop across the capacitor: given by \( Q/C \), where \( Q \) is the charge stored on the capacitor, and \( C \) is the capacitance.

By applying Kirchhoff's Law, we derive the differential equation: \( R \frac{dQ}{dt} + \frac{1}{C}Q = E(t) \). This equation is pivotal for finding both the charge \( Q \) and the current \( I \) in the circuit.

Homogeneous Equation

The homogeneous equation is a simplified version of the differential equation where the right-hand side is equal to zero. In this exercise, it helps us determine the behavior of the circuit when there is no external voltage applied (ignoring the battery, for example).

For the given circuit, the homogeneous equation is derived as:\[5 \frac{dQ}{dt} + 20Q = 0\]This equation indicates that the charge \( Q(t) \) will diminish over time if no external voltage is applied. Solving this equation involves separating variables and integrating to find the general form of the solution. The steps are as follows:

• Separate the variables: \( \frac{dQ}{Q} = -4 \, dt \).
• Integrate both sides: \( \ln |Q| = -4t + C_1 \) (where \( C_1 \) is an integration constant).
• Solve for \( Q \): \( Q = e^{-4t + C_1} = C_2 e^{-4t} \); here, \( C_2 = e^{C_1} \).

The general solution \( Q_h(t) = C_2 e^{-4t} \) shows the inherent damping nature of the circuit, where the charge decays exponentially with time.

Particular Solution

A particular solution is obtained by considering the effect of an external force or input, such as a constant voltage from a battery, on the system. In solving differential equations like in this exercise, you look for a solution that satisfies the non-homogeneous equation.

In this context, let's assume the function \( Q_p = A \) to find constants for the particular solution. Substituting this into the differential equation:

• Insert \( Q_p = A \) into the equation: \( 5 \cdot 0 + 20A = 60 \).
• Solve for \( A \): \( A = 3 \).

Therefore, the particular solution is \( Q_p = 3 \). This solution represents a steady-state charge in the circuit when the battery is applying a constant voltage across it. Essentially, it shows how the circuit behaves under external influence over time, which is paramount for reaching an accurate total solution.

Initial Condition

The initial condition provides a snapshot of the system at time \( t = 0 \) and helps us find specific constants in the solution of a differential equation.

In our exercise, the initial condition is given as \( Q(0) = 0 \). This means that at the starting point, there is no charge on the capacitor. We utilize this condition to find the value of the constant in the general solution:

Starting from the general solution \( Q(t) = C_2 e^{-4t} + 3 \), apply the initial condition:

• Substitute \( t = 0 \), \( Q(0) = 0 \): \( 0 = C_2 + 3 \).
• Solve for \( C_2 \): \( C_2 = -3 \).

Substituting \( C_2 \) back into the general form gives us \( Q(t) = -3 e^{-4t} + 3 \). Thus, the initial condition sets the stage for determining the exact nature of how the system evolves over time. It's vital in ensuring that our solutions fit the real-world scenario properly, considering initial circumstances.