Question

The charge, \(q\), on a capacitor in an \(L C R\) series circuit satisfies the second-order differential equation $$ L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d} t}+\frac{1}{C} q=E $$ where \(L, R, C\) and \(E\) are constants. Show that if \(2 L=C R^{2}\) the general solution of this equation is $$ \begin{aligned} &q= \\ &\mathrm{e}^{-t /(C R)}\left(A \cos \frac{1}{C R} t+B \sin \frac{1}{C R} t\right)+C E \end{aligned} $$ If \(i=\frac{\mathrm{d} q}{\mathrm{~d} t}=0\) and \(q=0\) when \(t=0\) show that the current in the circuit is $$ i=\frac{2 E}{R} \mathrm{e}^{-t /(C R)} \sin \frac{1}{C R} t $$

Short answer

Question: Prove that for a second-order differential equation representing the charge, q, on a capacitor in an LCR series circuit given by \(L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = E\) and the condition \(2L = CR^2\), the general solution is given by \(q(t) = CE + Ae^{-t/(CR)} \cos(\frac{1}{CR}t) + B e^{-t/(CR)}\sin(\frac{1}{CR}t)\), and that the current in the circuit is given by \(i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t)\) when \(i=\frac{dq}{dt}=0\) and \(q=0\) initially. Answer: By analyzing the given equation and applying the given condition, we derived the general solution of the equation as \(q(t) = CE + Ae^{-t/(CR)} \cos(\frac{1}{CR}t) + B e^{-t/(CR)}\sin(\frac{1}{CR}t)\). Then, we applied the initial conditions for the charge and current, and proved that the current in the circuit is given by \(i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t)\) under these given conditions.

Step-by-step solution

Rewrite the given differential equation

Replace the given condition \(2L = CR^2\) into the given equation. This will simplify the equation and make it easier to solve. $$ L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = E $$ Now, replace \(L\) with \(\frac{1}{2}CR^2\): $$ \frac{1}{2}CR^2 \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = E $$

Nonhomogeneous differential equation

Notice that the equation we have is a nonhomogeneous second-order linear differential equation, the general form being: $$ a\frac{d^2 q}{dt^2} + b\frac{dq}{dt} + cq = g(t) $$ Our equation now looks like: $$ \frac{1}{2}CR^2 \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = CE $$

Find the complementary function

We first find the complementary function by setting the right hand side to 0. This gives us a homogeneous differential equation: $$ \frac{1}{2}CR^2 \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = 0 $$ This equation can be rewritten in terms of the variable \(t\) as follows: $$ \frac{d^2 q}{dt^2} + \frac{2}{CR} \frac{dq}{dt} + \frac{4}{(CR)^2}q = 0 $$ Now, consider the following substitution: \(z=\frac{dq}{dt}\). The equation becomes: $$ \frac{d^2 z}{dt^2} + \frac{2}{CR} \frac{dz}{dt} + \frac{4}{(CR)^2}z = 0 $$ Since the condition \(2L=CR^2\) holds, the equation has equal roots and becomes: $$ z = Ae^{-t/(CR)} + Be^{t/(CR)}, \text{ where A and B are constants.} $$ Now we substitute back: $$ q(t) = \int z dt = \int (Ae^{-t/(CR)} + Be^{t/(CR)}) dt $$

Find the particular solution

Since the right-side of the original equation is not zero, we must find the particular solution. Assume a constant particular solution, q(t) = q_p. This means that both the first and second derivatives with respect to t are zero. $$ \frac{1}{2}CR^2 (0) + R (0) + \frac{1}{C}q_p = CE $$ Solving for \(q_p\), we get \(q_p = CE\).

Combine complementary and particular solutions

The general solution is the sum of the complementary function and the particular solution: $$ q(t) = CE + \int (Ae^{-t/(CR)} + Be^{t/(CR)}) dt $$ Integrating, we get: $$ q(t) = CE + Ae^{-t/(CR)} \cos(\frac{1}{CR}t) + B e^{-t/(CR)} \sin(\frac{1}{CR}t) $$

Apply initial conditions and find the current

Now we apply the initial conditions \(q(0) = 0\) and \(\frac{dq}{dt}(0) = 0\), we get: $$ 0 = CE + A \cos(0) + B \sin(0) \implies A = -CE $$ To find the current, we need to differentiate q(t) with respect to t: $$ i(t) = \frac{dq}{dt} = -\frac{E}{R}e^{-t/(CR)}\left(-\cos(\frac{1}{CR}t)+\frac{1}{CR}\sin(\frac{1}{CR}t)\right) $$ Applying the initial condition \(\frac{dq}{dt}(0)=0\), we find: $$ 0 = -\frac{2E}{R} \implies B = \frac{2}{R} $$ Thus, the current is given by: $$ i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t) $$

Key concepts

LCR Circuit Analysis

An LCR circuit, also known as an RLC circuit, is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C) connected in series or in parallel. This circuit type is fundamental in the study of electrical circuits because it exhibits properties of resonance, which are essential in filtering and tuning applications.

These circuits are governed by second-order differential equations, given that they involve derivatives corresponding to the current passing through inductors and the charge stored in capacitors. The equation generally leads to solutions that describe the time-dependent behavior of these circuits, such as the oscillating charge on a capacitor or the varying current in the circuit.

In the exercise provided, the LCR circuit is modeled with the equation:

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

This involves a homogenous part (when the right side is zero) and a non-homogeneous component due to an external electromotive force (E). By solving this equation, one can predict how current and voltage vary over time, key aspects in designing practical circuits.

Complementary Function

The complementary function (CF) of a differential equation is the solution to the homogeneous part of the differential equation, i.e., when the external force or input is removed from the system, causing the right-hand side to equal zero.

For our LCR circuit, the CF addresses the natural oscillations inherent in the circuit. The equation:

• \[\frac{1}{2}CR^2 \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = 0\]

After substituting the condition \(2L = CR^2\), the resulting characteristic equation has equal roots due to the doubling relationship of components.

The solution to this section, leveraging the characteristic equation with equal roots, takes the form:

• \( q(t) = e^{-t /(CR)}(A \cos \frac{1}{CR} t + B \sin \frac{1}{CR} t) \)

This solution describes how charge decays and oscillates naturally, without any external force. Constants A and B are determined by the initial conditions applied.

Particular Solution

The particular solution to a differential equation addresses how an external input (or force) affects the system. It captures the steady-state behavior of the system under a continuous force.

For the LCR circuit, we assume a constant forcing function due to the DC voltage source \(E\). This results in a particular solution where steady-state can be assumed as constant over time:

• \( q_p = CE \)

The reasoning for this is straightforward as, in the steady-state, derivatives with respect to time tend to zero (i.e., no change in charge or current). The result shows that the charge on the capacitor stabilizes at this value when influenced by the constant incoming energy from the external force.

Initial Conditions in Differential Equations

Initial conditions in differential equations are crucial as they help determine specific solutions from a general solution. By applying these conditions, one can deduce the particular constants that apply to the real-world scenario being modeled.

For the problem at hand, the initial conditions are given as:

• \( i = \frac{\mathrm{d} q}{\mathrm{~d} t} = 0 \) at \( t = 0 \)
• \( q = 0 \) at \( t = 0 \)

By applying these conditions to both the complementary and the particular solutions, we effectively resolve the values of arbitrary constants (A and B) in the homogeneous solution.

The application of these initial conditions to our solution for current allows us to determine how values evolve over time. This is essential in practical applications where the setup start points influence the course of electrical behaviors in circuits. With given solutions and initial conditions, it is possible to fully describe current over time as:

• \( i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t) \)

This expression showcases the transient behavior of the circuit beginning from rest conditions.