Problem 102 - Calculus Volume 3

Question

A series circuit consists of a device where \(L=1\) \(\mathrm{H}, \quad R=20 \Omega, \quad C=0.002 \mathrm{~F},\) and \(E(t)=12 \mathrm{~V} .\) If the initial charge and current are both zero, find the charge and current at time \(t\).

Step-by-Step Solution

Answer

Charge: \(q(t) = 0.024\), Current is derived by differentiating \(q(t)\).

Step 1: Understand the Components

In a series circuit where the inductance \(L\), resistance \(R\), and capacitance \(C\) are given along with the electromotive force \(E(t)\), we need to find the charge \(q(t)\) and the current \(i(t)\) over time. The initial conditions are \(q(0) = 0\) and \(i(0) = 0\).

Step 2: Define the Differential Equation

Use Kirchhoff's law as \(L\frac{di}{dt} + Ri + \frac{1}{C}\int{i}\, dt = E(t)\). Substitute the component values to get \(\frac{d^2q}{dt^2} + 20\frac{dq}{dt} + 500q = 12\). This represents the second-order linear differential equation of the charge \(q(t)\).

Step 3: Solve the Homogeneous Equation

Solve the homogeneous part \(\frac{d^2q}{dt^2} + 20\frac{dq}{dt} + 500q = 0\). Assume a solution of the form \(q(t) = e^{rt}\) and substitute into the equation, leading to the characteristic equation \(r^2 + 20r + 500 = 0\). Solving this gives complex roots.

Step 4: Find the Particular Solution

For the inhomogeneous part \(\frac{d^2q}{dt^2} + 20\frac{dq}{dt} + 500q = 12\), assume a constant particular solution \(q_p = A\). Substituting back, we find \(A = \frac{12}{500} = 0.024\).

Step 5: Complete Solution for Charge

Combine the homogeneous and particular solutions to get \(q(t) = C_1e^{-10t} \cos(20t) + C_2e^{-10t} \sin(20t) + 0.024\). Use the initial conditions \(q(0) = 0\) and \(i(0) = \frac{dq}{dt}\big|_{t=0} = 0\) to find \(C_1\) and \(C_2\).

Step 6: Derive Current From Charge

The current is the derivative of the charge, \(i(t) = \frac{dq}{dt}\). Differentiate \(q(t)\) to find \(i(t)\) by applying the product and chain rules to account for all terms, enforcing initial conditions to solve any constants in the expression.

Key Concepts

Kirchhoff's Law

Kirchhoff's Law is a fundamental principle in electrical circuits. It relates the sum of the electromotive forces (emfs) and potential differences around a closed loop. In the context of this exercise, Kirchhoff's Voltage Law is applied to a series circuit. This law states that the total sum of the emf is equal to the sum of all the voltage drops across the components in the circuit.
In mathematical form, for our series circuit, it is expressed as:

• \(L\frac{di}{dt} + Ri + \frac{1}{C}\int{i}\, dt = E(t)\),

where:

• \(L\) is the inductance,
• \(R\) is the resistance,
• \(C\) is the capacitance, and
• \(E(t)\) is the electromotive force.

By applying Kirchhoff's Law, we form the basis of analyzing the behavior of circuits over time.

Series Circuit

A series circuit is an electrical circuit in which components are connected end-to-end, forming a single path for the electric current to flow. This means that the same current flows through each component. A series circuit arrangement ensures that all components are electrically connected along a single loop.
In our exercise, the components considered are:

• One inductor (1 H),
• A resistor (20 Ω), and
• A capacitor (0.002 F).

The behavior of electrical components in a series emphasizes the sum of all voltages around a loop (as expressed by Kirchhoff's Law). The total resistance, capacitance, and inductance determine how the circuit responds to the applied electromotive force (emf), including how current and charge vary over time.

Second-Order Linear Differential Equation

Differential equations describe how quantities change over time. In this exercise, the relation between the charge, current, and other variables is represented by a second-order linear differential equation. This is because both the first and second derivatives of charge with respect to time are involved.
The differential equation given is:

• \(\frac{d^2q}{dt^2} + 20\frac{dq}{dt} + 500q = 12.\)

Here, \(q(t)\) represents the charge at a given time \(t\). The left side of the equation deals with transformations in the charge and its rate of change, and the right side is the constant electromotive force from the power source.
This type of equation is called 'linear' because each term is either a constant or a product of a constant and the first power of the variable(s). Solving such an equation involves finding a homogeneous and a particular solution, as demonstrated in the exercise.

Electromotive Force

Electromotive Force (emf) is the energy provided by a power source to move a charge through a circuit. It is a crucial concept that drives the flow of electric current. In this exercise, the emf is constant and given as \(E(t) = 12 \text{ V}\).
The electromotive force is, essentially, the cause of current in the circuit. It pushes charges through the circuit, overcoming resistances and other oppositional forces. In the differential equation provided, \(E(t)\) represents this constant energy source, influencing how charge and current vary over time.
Understanding emf helps to grasp how different elements in the circuit behave and interact with each other under the influence of energy sources, and forms a core part of analyzing and solving circuit problems.