Question

A circuit has in series an electromotive force given by \(E=20 \sin 5 t \vee\), a capacitor with capacitance \(0.02 \mathrm{~F}\), and a resistor of \(8 \Omega\). If the initial charge is \(4 \mathrm{C}\), find the charge at time \(t>0\)

Short answer

The charge at time \( t > 0 \) is given by \( q(t) = 4e^{-6.25t} - \frac{4}{13} \sin(5t) \).

Step-by-step solution

Understanding the Basics

The circuit described consists of an electromotive force connected in series with a capacitor and a resistor. The applied electromotive force is given by the function \( E(t) = 20 \sin(5t) \vee \). The capacitor has a capacitance of 0.02 F, and the resistor has a resistance of 8 \( \Omega \). Additionally, the initial charge on the capacitor is 4 C, and we need to find the charge at any time \( t > 0 \).

Set Up the Differential Equation

The charge \( q(t) \) on the capacitor in an RLC circuit satisfies the differential equation \[ L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{q}{C} = E(t) \]. However, there is no inductor present (\( L=0 \)), so the equation simplifies to \[ R \frac{dq}{dt} + \frac{q}{C} = E(t) \]. Substituting known values: \( R=8 \), \( C=0.02 \), and \( E(t)=20 \sin(5t) \). Thus, the differential equation becomes \[ 8 \frac{dq}{dt} + 50q = 20 \sin(5t) \].

Solve the Differential Equation - Homogeneous Part

First, solve the homogeneous equation \( 8 \frac{dq}{dt} + 50q = 0 \). Solving this gives \( q_h(t) = Ce^{-\frac{50}{8}t} = Ce^{-6.25t} \), where \( C \) is a constant.

Solve the Differential Equation - Particular Solution

To find a particular solution, assume \( q_p(t) = A \sin(5t) + B \cos(5t) \). Differentiating gives \( \frac{dq_p}{dt} = 5A \cos(5t) - 5B \sin(5t) \). Substitute these into the non-homogeneous differential equation and solve for \( A \) and \( B \). Equating coefficients, we find \( A = -\frac{4}{13}, \ B = 0 \). Thus, \( q_p(t) = -\frac{4}{13} \sin(5t) \).

Combine Solutions and Apply Initial Condition

The general solution is \( q(t) = Ce^{-6.25t} - \frac{4}{13} \sin(5t) \). Apply the initial condition: \( q(0) = 4 \). This yields \( C - 0 = 4 \), so \( C = 4 \). Therefore, \( q(t) = 4e^{-6.25t} - \frac{4}{13} \sin(5t) \).

Final Solution

We have determined the complete expression for the charge in the circuit as a function of time: \[ q(t) = 4e^{-6.25t} - \frac{4}{13} \sin(5t) \]. This expression provides the charge at any time \( t > 0 \).

Key concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. They play a crucial role in describing physical phenomena in various fields, such as engineering, physics, and economics. In the context of an RLC circuit, differential equations are used to model the behavior of charge, voltage, or current over time within the circuit components.

In this exercise, the differential equation models the change in charge over time in a resistor-capacitor circuit. The general form for an RLC circuit is given by:

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

However, since this circuit lacks an inductor, the equation simplifies to:

• \( R \frac{dq}{dt} + \frac{q}{C} = E(t) \)

This equation considers the resistance \( R \), capacitance \( C \), and applied electromotive force \( E(t) \). Such representations enable the determination of how electrical quantities evolve given initial and boundary conditions.

Particular Solution

The particular solution to a differential equation is a solution that satisfies not only the differential equation but also the specific conditions or external inputs applied to the system. Finding a particular solution often requires assuming a form that reflects the non-homogeneous part of the equation or any external forces.

In our circuit problem, the input is \( E(t) = 20 \sin(5t) \), a sinusoidal function, prompting the assumption of the form \( q_p(t) = A \sin(5t) + B \cos(5t) \) for the solution since trigonometric functions often repeat similarly in response. By substituting this assumed form into the simplified differential equation, we equate coefficients to determine the constants \( A \) and \( B \).

For this circuit, the calculated particular solution is \( q_p(t) = -\frac{4}{13} \sin(5t) \), where \( A = -\frac{4}{13} \) and \( B = 0 \). It specifically accounts for the sinusoidal electromotive force influencing the system.

Initial Condition

Initial conditions are essential in solving differential equations as they allow us to determine the constants of integration in a general solution. They represent the state of the system at a specific starting point in time.

For the given electrical circuit, we have the initial charge on the capacitor as 4 Coulombs when time \( t = 0 \). This initial condition is applied to the general solution of the differential equation after combining the homogeneous and particular solutions to find any undetermined constant.

In our example, this leads us to solve for \( C \) in the expression \( q(t) = Ce^{-6.25t} - \frac{4}{13} \sin(5t) \) with the condition \( q(0) = 4 \). Substituting \( t = 0 \) into the equation gives \( C - 0 = 4 \), thus \( C = 4 \). The initial condition ensures that the solution precisely reflects the starting state of the circuit.

Homogeneous Equation

Homogeneous equations form an integral part of solving differential equations. They represent the part of the equation that equates to zero, revealing the inherent behavior of the system without external inputs.

In the context of our circuit, the homogeneous equation derived from the differential equation is:

• \( 8 \frac{dq}{dt} + 50q = 0 \)

Solving this equation involves separating variables or recognizing it as a standard first-order linear differential equation. The solution comes out as \( q_h(t) = Ce^{-6.25t} \), where \( C \) is a constant yet to be determined.

This component of the solution reflects the natural decay of charge in the system due to the resistor and capacitor, without considering the effects of the external electromotive force. It describes how the charge on the capacitor would diminish over time if the circuit were isolated from external sinusoidal forces.