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# RLC natural response - derivation

A formal derivation of the natural response of the RLC circuit. Written by Willy McAllister.

## Introduction

Let's take a deep look at the natural response of a resistor-inductor-capacitor circuit left parenthesis, start text, R, L, C, right parenthesis, end text. This is the last circuit we'll analyze with the full differential equation treatment.
The start text, R, L, C, end text circuit is representative of real life circuits we can actually build, since every real circuit has some finite resistance. This circuit has a rich and complex behavior that finds application in many areas of electrical engineering.
The circuit for the start text, R, L, C, end text natural response.

### What we're building to

We will model the start text, R, L, C, end text circuit with a 2nd-order linear differential equation with current, i, as the independent variable:
start text, L, end text, start fraction, d, squared, i, divided by, d, t, squared, end fraction, plus, start text, R, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, C, end text, end fraction, i, equals, 0
The resulting characteristic equation is:
s, squared, plus, start fraction, start text, R, end text, divided by, start text, L, end text, end fraction, s, plus, start fraction, 1, divided by, start text, L, C, end text, end fraction, equals, 0
We will solve for the roots of the characteristic equation using the quadratic formula:
s, equals, start fraction, minus, start text, R, end text, plus minus, square root of, start text, R, end text, squared, minus, 4, start text, L, end text, slash, start text, C, end text, end square root, divided by, 2, start text, L, end text, end fraction
By substituting variables alpha and omega, start subscript, o, end subscript we can write s a little simpler as:
s, equals, minus, alpha, plus minus, square root of, alpha, squared, minus, omega, start subscript, o, end subscript, squared, end square root
where alpha, equals, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction, and omega, start subscript, o, end subscript, equals, start fraction, 1, divided by, square root of, start text, L, C, end text, end square root, end fraction
alpha is called the damping factor and omega, start subscript, o, end subscript is the resonant frequency.
We will solve an example start text, R, L, C, end text circuit with specific component values and discover what the current and voltages look like.

## Strategy

We follow the same line of reasoning we used for solving the second-order LC circuit in an earlier article.
1. Create a second-order differential equation based on the i-v equations for the start text, R, end text, start text, L, end text, and start text, C, end text components. We will use Kirchhoff's Voltage Law to build the equation.
2. Make an informed guess at a solution. As usual, our guess will be an exponential function of the form K, e, start superscript, s, t, end superscript.
3. Insert the proposed solution into the differential equation. The exponential terms will factor out and leave us with a characteristic equation in variable s.
4. Find the roots of the characteristic equation. This time we will need to use the quadratic formula to solve for the roots.
5. Find the constants by accounting for the initial conditions.
6. Celebrate the solution.

## Model the circuit with a differential equation

Circuit conditions just before the switch closes: The current is 0, and the capacitor is charged up to an initial voltage of start text, V, end text, start subscript, 0, end subscript volts.
When the switch closes, the circuit looks like this (now with voltage labels on the inductor and resistor, v, start subscript, start text, L, end text, end subscript and v, start subscript, start text, R, end text, end subscript).
Circuit conditions just after the switch closes. We still have to find the current and voltage at t, equals, 0, start superscript, plus, end superscript. We work this out in the section titled Find the initial conditions.
For each individual element, we can write i-v equations.
v, start subscript, start text, L, end text, end subscript, equals, start text, L, end text, start fraction, d, i, divided by, d, t, end fraction
v, start subscript, start text, R, end text, end subscript, equals, minus, i, start text, R, end text
v, start subscript, start text, C, end text, end subscript, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, minus, i, d, t
We can write Kirchhoff's Voltage Law (KVL) starting in the lower left corner and summing voltages going around the loop clockwise. The inductor has a voltage rise, while the resistor and capacitor have voltage drops.
plus, v, start subscript, start text, L, end text, end subscript, minus, v, start subscript, start text, R, end text, end subscript, minus, v, start subscript, start text, C, end text, end subscript, equals, 0
Replacing the v terms with the corresponding i terms gives us:
start text, L, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start text, R, end text, i, plus, start fraction, 1, divided by, start text, C, end text, end fraction, integral, i, d, t, equals, 0
If we wanted to, we could attack this equation and try to solve it, but the integral term is awkward to deal with. We can retire the integral if we take the derivative of the entire equation.
start fraction, d, divided by, d, t, end fraction, open bracket, start text, L, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start text, R, end text, i, plus, start fraction, 1, divided by, start text, C, end text, end fraction, integral, i, d, t, equals, 0, close bracket
This gives us the following equation with a second derivative term, a first derivative term, and a plain i term, all still equal to 0.
start text, L, end text, start fraction, d, squared, i, divided by, d, t, squared, end fraction, plus, start text, R, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, C, end text, end fraction, i, equals, 0
This is called a homogeneous second-order ordinary differential equation. It is homogeneous because every term is related to i and its derivatives. It is second order because the highest derivative is a second derivative. It is ordinary because there is only one independent variable (no partial derivatives). Now we set about solving our differential equation.

## Propose a solution

Just like we did with previous natural response problems ( RC, RL, LC), we assume a solution with an exponential form. Exponential functions have the wondrous property that the derivatives look a lot like the original function. When you have multiple derivatives participating in a differential equation, it is really nice when they look alike. We assume a solution with this form:
i, left parenthesis, t, right parenthesis, equals, K, e, start superscript, s, t, end superscript
K is an adjustable parameter representing the amplitude of the current.
s is up in the exponent next to t, so it must represent some kind of frequency (s has to have units of 1, slash, t). We call this the natural frequency.

## Try the proposed solution

Next, substitute the proposed solution into the differential equation. If the equation turns out to be true, then our solution is a winner.
start text, L, end text, start fraction, d, squared, divided by, d, t, squared, end fraction, K, e, start superscript, s, t, end superscript, plus, start text, R, end text, start fraction, d, divided by, d, t, end fraction, K, e, start superscript, s, t, end superscript, plus, start fraction, 1, divided by, start text, C, end text, end fraction, K, e, start superscript, s, t, end superscript, equals, 0
Now let's work on the terms with derivatives.
Middle term: The first derivative of the start text, R, end text term is
start text, R, end text, start fraction, d, divided by, d, t, end fraction, K, e, start superscript, s, t, end superscript, equals, s, start text, R, end text, K, e, start superscript, s, t, end superscript
Leading term: We take the derivative if the leading start text, K, end text, e, start superscript, s, t, end superscript term two times:
start fraction, d, divided by, d, t, end fraction, K, e, start superscript, s, t, end superscript, equals, s, K, e, start superscript, s, t, end superscript
start fraction, d, divided by, d, t, end fraction, s, K, e, start superscript, s, t, end superscript, equals, s, squared, K, e, start superscript, s, t, end superscript
start text, L, end text, start fraction, d, squared, divided by, d, t, squared, end fraction, K, e, start superscript, s, t, end superscript, equals, s, squared, start text, L, end text, K, e, start superscript, s, t, end superscript
Plug these back into the differential equation:
s, squared, start text, L, end text, K, e, start superscript, s, t, end superscript, plus, s, start text, R, end text, K, e, start superscript, s, t, end superscript, plus, start fraction, 1, divided by, start text, C, end text, end fraction, K, e, start superscript, s, t, end superscript, equals, 0
Now we can factor out the common K, e, start superscript, s, t, end superscript term:
K, e, start superscript, s, t, end superscript, left parenthesis, s, squared, start text, L, end text, plus, s, start text, R, end text, plus, start fraction, 1, divided by, start text, C, end text, end fraction, right parenthesis, equals, 0

## Make the equation true

Now let's figure out how many ways we can make this equation true.
We could set K equal to 0. That means i, equals, 0 and we are putting nothing into the circuit and getting nothing out. Pretty boring.
The term e, start superscript, s, t, end superscript never becomes 0 unless we wait until t goes to infinity. That's a long time from now. That leaves us with one interesting way to make the equation true: if the term with all the s's is zero.
s, squared, start text, L, end text, plus, s, start text, R, end text, plus, start fraction, 1, divided by, start text, C, end text, end fraction, equals, 0
This is called the characteristic equation of the start text, L, R, C, end text circuit.

### Solve for the roots of the characteristic equation

Let's find values of s that make the characteristic equation true. (We want to find the roots of the characteristic equation.)
We have exactly the right tool for this, the quadratic formula:
For any quadratic equation: a, x, squared, plus, b, x, plus, c, equals, 0,
the quadratic formula gives us the roots (the zero-crossings):
x, equals, start fraction, minus, b, plus minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction
Looking back at the characteristic equation, we can plug in our circuit component values to get the roots. a, equals, start text, L, end text, b, equals, start text, R, end text, and c, equals, 1, slash, start text, C, end text.
s, equals, start fraction, minus, start text, R, end text, plus minus, square root of, start text, R, end text, squared, minus, 4, start text, L, end text, slash, start text, C, end text, end square root, divided by, 2, start text, L, end text, end fraction
That's the answer for s, the natural frequency. We need to break this down a little more to get a feeling for the meaning of this solution.
We can make the notation a little more compact by replacing parts of the expression with two new variables, alpha and omega, start subscript, o, end subscript.
alpha, equals, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction
omega, start subscript, o, end subscript, equals, start fraction, 1, divided by, square root of, start text, L, C, end text, end square root, end fraction
Let me write the characteristic equation this way left parenthesisdividing through by start text, L, end text, right parenthesis:
s, squared, plus, start fraction, start text, R, end text, divided by, start text, L, end text, end fraction, s, plus, start fraction, 1, divided by, start text, L, C, end text, end fraction, equals, 0
If we use alpha and omega, start subscript, o, end subscript, the characteristic equation can be written:
s, squared, plus, 2, alpha, s, plus, omega, start subscript, o, end subscript, squared, equals, 0
We can revise the quadratic formula by mashing the 2, start text, L, end text denominator up into each term of the numerator:
s, equals, minus, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction, plus minus, square root of, left parenthesis, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction, right parenthesis, squared, minus, left parenthesis, start fraction, 4, start text, L, end text, slash, start text, C, end text, divided by, 4, start text, L, end text, squared, end fraction, right parenthesis, end square root
The second term under the square root reduces to:
left parenthesis, start fraction, 4, start text, L, end text, slash, start text, C, end text, divided by, 4, start text, L, end text, squared, end fraction, right parenthesis, equals, left parenthesis, start fraction, start cancel, 4, end cancel, start cancel, start text, L, end text, end cancel, slash, start text, C, end text, divided by, start cancel, 4, end cancel, start text, L, end text, start superscript, start cancel, 2, end cancel, end superscript, end fraction, right parenthesis, equals, start fraction, 1, divided by, start text, L, C, end text, end fraction
And this lets us write s in terms of alpha and omega, start subscript, o, end subscript as:
s, equals, minus, alpha, plus minus, square root of, alpha, squared, minus, omega, start subscript, o, end subscript, squared, end square root
We know s is some sort of frequency (it has to have units of 1, slash, t). That means the two terms making up s are also some sort of frequency.
• alpha is called the damping factor. It will determine how quickly the overall signal fades to zero.
• omega, start subscript, o, end subscript is called the resonant frequency. It will determine how fast the system swings back and forth. This is the same resonant frequency we found in the start text, L, C, end text natural response.

The quadratic formula gave us two solutions for s, we'll call these s, start subscript, 1, end subscript and s, start subscript, 2, end subscript. We need to include both of these in the proposed solution, so we update our proposed solution to be a linear combination (the superposition) of two separate exponential terms with four adjustable parameters:
i, equals, K, start subscript, 1, end subscript, e, start superscript, s, start subscript, 1, end subscript, t, end superscript, plus, K, start subscript, 2, end subscript, e, start superscript, s, start subscript, 2, end subscript, t, end superscript
s, start subscript, 1, end subscript and s, start subscript, 2, end subscript are natural frequencies,
K, start subscript, 1, end subscript and K, start subscript, 2, end subscript are amplitude terms.

## Example circuit

At this point it's helpful to do a specific example with some actual component values, to see how one particular solution plays out. Here is our example circuit:
start text, R, L, C, end text natural response example. The capacitor has an initial voltage of 10 volts. There is no current flowing in the inductor at the moment the switch is closed.
The differential equation for the start text, R, L, C, end text circuit is
start text, L, end text, start fraction, d, squared, i, divided by, d, t, squared, end fraction, plus, start text, R, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, C, end text, end fraction, i, equals, 0
With real component values it becomes:
1, start fraction, d, squared, i, divided by, d, t, squared, end fraction, plus, 2, start fraction, d, i, divided by, d, t, end fraction, plus, 5, i, equals, 0
As we always do, assume a solution of the form: i, left parenthesis, t, right parenthesis, equals, K, e, start superscript, s, t, end superscript
We perform the analysis we did above, which results in this characteristic equation:
s, squared, plus, 2, s, plus, 5, equals, 0
Solving for the roots of the characteristic equation with the quadratic formula:
s, equals, start fraction, minus, start text, R, end text, plus minus, square root of, start text, R, end text, squared, minus, 4, start text, L, end text, slash, start text, C, end text, end square root, divided by, 2, start text, L, end text, end fraction
With real component values:
s, equals, start fraction, minus, 2, plus minus, square root of, 2, squared, minus, 4, dot, 1, dot, 5, end square root, divided by, 2, end fraction
s, equals, start fraction, minus, 2, plus minus, square root of, 4, minus, 20, end square root, divided by, 2, end fraction
s, equals, minus, 1, plus minus, start fraction, square root of, minus, 16, end square root, divided by, 2, end fraction
s, equals, minus, 1, plus minus, j, 2
(Electrical engineers use the letter j for the imaginary square root of, minus, 1, end square root, since we use i as the symbol for current.)
We get a complex answer, just as we did with the start text, L, C, end text natural response, only this time the complex answer includes both a real part and an imaginary part.
Solving for the roots of the characteristic equation gave us two possible answers for s, so the proposed solution for i is now written as the superposition of two different exponential terms:
i, equals, K, start subscript, 1, end subscript, e, start superscript, left parenthesis, minus, 1, plus, j, 2, right parenthesis, t, end superscript, plus, K, start subscript, 2, end subscript, e, start superscript, left parenthesis, minus, 1, minus, j, 2, right parenthesis, t, end superscript
The terms up in the exponents are complex conjugates. Let's fuss around with the way this is written. We can tease apart the real and imaginary parts of the exponents:
i, equals, K, start subscript, 1, end subscript, e, start superscript, minus, 1, t, end superscript, e, start superscript, plus, j, 2, t, end superscript, plus, K, start subscript, 2, end subscript, e, start superscript, minus, 1, t, end superscript, e, start superscript, minus, j, 2, t, end superscript, comma
and factor out the common e, start superscript, minus, 1, t, end superscript term:
i, equals, e, start superscript, minus, t, end superscript, left parenthesis, K, start subscript, 1, end subscript, e, start superscript, plus, j, 2, t, end superscript, plus, K, start subscript, 2, end subscript, e, start superscript, minus, j, 2, t, end superscript, right parenthesis
Notice how the real part of s came through the factoring process to give us the leading term, a decaying exponential, e, start superscript, minus, t, end superscript.
The terms in the parentheses are a sum of two imaginary exponentials where the exponents are complex conjugates. This looks just like what we saw in the start text, L, C, end text natural response. As we did then, we call on Euler's formula to help us with these terms.

## Euler's formula

Using Maclaurin series expansions for e, start superscript, j, x, end superscript, sine, j, x, and cosine, j, x, it is possible to derive Euler's formula:
e, start superscript, plus, j, x, end superscript, equals, cosine, x, plus, j, sine, x
and
e, start superscript, minus, j, x, end superscript, equals, cosine, x, minus, j, sine, x
In the linked video, whenever Sal says i, we say j.
These formulas let us turn e, start superscript, i, m, a, g, i, n, a, r, y, end superscript into a normal complex number.

## Use Euler's formula

We can use Euler's formula to transform the sum
K, start subscript, 1, end subscript, e, start superscript, plus, j, 2, t, end superscript, plus, K, start subscript, 2, end subscript, e, start superscript, minus, j, 2, t, end superscript
into
K, start subscript, 1, end subscript, left parenthesis, cosine, 2, t, plus, j, sine, 2, t, right parenthesis, plus, K, start subscript, 2, end subscript, left parenthesis, cosine, 2, t, minus, j, sine, 2, t, right parenthesis, point
Multiply through the constants K, start subscript, 1, end subscript and K, start subscript, 2, end subscript:
K, start subscript, 1, end subscript, cosine, 2, t, plus, j, K, start subscript, 1, end subscript, sine, 2, t, plus, K, start subscript, 2, end subscript, cosine, 2, t, minus, j, K, start subscript, 2, end subscript, sine, 2, t, comma
and gather the cosine terms and sine terms:
left parenthesis, K, start subscript, 1, end subscript, plus, K, start subscript, 2, end subscript, right parenthesis, cosine, 2, t, plus, j, left parenthesis, K, start subscript, 1, end subscript, minus, K, start subscript, 2, end subscript, right parenthesis, sine, 2, t
Without messing up the equation, we can simplify how this appears if we replace the unknown K's with different unknown A's. Let A, start subscript, 1, end subscript, equals, left parenthesis, K, start subscript, 1, end subscript, plus, K, start subscript, 2, end subscript, right parenthesis, and A, start subscript, 2, end subscript, equals, j, left parenthesis, K, start subscript, 1, end subscript, minus, K, start subscript, 2, end subscript).
The previous expression becomes:
A, start subscript, 1, end subscript, cosine, 2, t, plus, A, start subscript, 2, end subscript, sine, 2, t
And now put this back into our proposed solution:
i, equals, e, start superscript, minus, t, end superscript, left parenthesis, A, start subscript, 1, end subscript, cosine, 2, t, plus, A, start subscript, 2, end subscript, sine, 2, t, right parenthesis
So far so good. Next, we need to figure out A, start subscript, 1, end subscript and A, start subscript, 2, end subscript using the initial conditions.

## Find the initial conditions

For a second-order equation, you need two initial conditions to get a complete solution: one for the independent variable, i, and another for its first derivative, d, i, slash, d, t.
If we can figure out i and d, i, slash, d, t at a specific time, we can find A, start subscript, 1, end subscript and A, start subscript, 2, end subscript.
Finding the initial conditions for start text, R, L, C, end text is pretty much the same as for the LC circuit. We just have account for the resistor.
Here's what we know about t, equals, 0, start superscript, minus, end superscript (the moment before the switch closes):
Circuit conditions just before the switch closes. At t, equals, 0, start superscript, minus, end superscript
the current is 0 and the initial voltage on the capacitor is v, start subscript, start text, C, end text, end subscript, equals, start text, 1, end text, 0, start text, V, end text.
• The switch is open, so i, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, 0
• The starting capacitor voltage is specified: v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, start text, V, end text, start subscript, 0, end subscript
If t, equals, 0, start superscript, plus, end superscript is the moment just after the switch closes, our goal is to find i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis and d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis. We know some properties of inductors and capacitors that tell us what happens when the switch closes, going from t, equals, 0, start superscript, minus, end superscript to t, equals, 0, start superscript, plus, end superscript:
• Inductor current does not change instantly, so i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, i, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, 0
• Capacitor voltage does not change instantly, so v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, minus, end superscript, right parenthesis, equals, start text, V, end text, start subscript, 0, end subscript
Circuit conditions just after the switch closes, at t, equals, 0, start superscript, plus, end superscript. i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0 and v, start subscript, start text, C, end text, end subscript, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 10, start text, V, end text.
Now we know one initial condition, i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0, and we know something about the voltage, but we don't know d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis.
Let's go after the second initial condition, d, i, slash, d, t, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis. Any time I see d, i, slash, d, t it makes me think of the inductor i-v equation. If we can figure out the voltage across the inductor, we can figure out d, i, slash, d, t. Let's do that by process of elimination.
Kirchhoff's Voltage Law around the loop is:
plus, v, start subscript, start text, L, end text, end subscript, minus, v, start subscript, start text, R, end text, end subscript, minus, v, start subscript, start text, C, end text, end subscript, equals, 0
Since i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0, that means the voltage across the resistor, v, start subscript, start text, R, end text, end subscript, has to be 0. We also know the voltage across the capacitor is v, start subscript, start text, C, end text, end subscript, equals, start text, V, end text, start subscript, 0, end subscript. Let's fill in the KVL equation with these values:
v, start subscript, start text, L, end text, end subscript, minus, 0, minus, start text, V, end text, start subscript, 0, end subscript, equals, 0
And now we know the voltage across the inductor at t, equals, 0, start superscript, plus, end superscript:
v, start subscript, start text, L, end text, end subscript, equals, start text, V, end text, start subscript, 0, end subscript
We can use this to derive d, i, slash, d, t using the inductor i-v equation.
v, start subscript, start text, L, end text, end subscript, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, start text, L, end text, start fraction, d, i, divided by, d, t, end fraction, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis
10, equals, 1, start fraction, d, i, divided by, d, t, end fraction, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis
start fraction, d, i, divided by, d, t, end fraction, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 10, start text, A, end text, slash, start text, s, e, c, end text
The moment just after the switch closes, the current in the inductor has an initial slope of 10 amperes per second.

## Find constants $A_1$A, start subscript, 1, end subscript and $A_2$A, start subscript, 2, end subscript using the initial conditions

As a reminder, our proposed solution is:
i, equals, e, start superscript, minus, t, end superscript, left parenthesis, A, start subscript, 1, end subscript, cosine, 2, t, plus, A, start subscript, 2, end subscript, sine, 2, t, right parenthesis
and the initial conditions are:
i, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 0
start fraction, d, i, divided by, d, t, end fraction, left parenthesis, 0, start superscript, plus, end superscript, right parenthesis, equals, 10
If we evaluate i at t, equals, 0, we can find one of the A constants. Insert t, equals, 0 and i, equals, 0 into the proposed solution:
0, equals, e, start superscript, minus, 0, end superscript, left parenthesis, A, start subscript, 1, end subscript, cosine, 2, dot, 0, plus, A, start subscript, 2, end subscript, sine, 2, dot, 0, right parenthesis
0, equals, 1, left parenthesis, A, start subscript, 1, end subscript, cosine, 0, plus, A, start subscript, 2, end subscript, sine, 0, right parenthesis
0, equals, left parenthesis, A, start subscript, 1, end subscript, dot, 1, plus, A, start subscript, 2, end subscript, dot, 0, right parenthesis
A, start subscript, 1, end subscript, equals, 0
A, start subscript, 1, end subscript, equals, 0, so the cosine term drops out of the solution. One down, one to go. Our proposed solution now looks like:
i, equals, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, sine, 2, t
Let's chase down A, start subscript, 2, end subscript using the second initial condition:
We need an equation for the derivative of i. Where might we get such a thing? How about we take the derivative of the proposed solution?
start fraction, d, i, divided by, d, t, end fraction, equals, start fraction, d, divided by, d, t, end fraction, left parenthesis, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, sine, 2, t, right parenthesis
The proposed solution is the product of two functions. To take its derivative we use the product rule:
left parenthesis, f, g, right parenthesis, prime, equals, f, prime, g, plus, f, g, prime
Identify the two parts of the product and their derivatives:
f, equals, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, g, equals, sine, 2, t
f, prime, equals, minus, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, g, prime, equals, 2, cosine, 2, t
Assemble the parts according to the product rule:
start fraction, d, i, divided by, d, t, end fraction, equals, minus, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, sine, 2, t, plus, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, 2, cosine, 2, t
start fraction, d, i, divided by, d, t, end fraction, equals, A, start subscript, 2, end subscript, e, start superscript, minus, t, end superscript, left parenthesis, 2, cosine, 2, t, minus, sine, 2, t, right parenthesis
We can evaluate this expression at t, equals, 0, start superscript, plus, end superscript:
10, equals, A, start subscript, 2, end subscript, e, start superscript, minus, 0, end superscript, left parenthesis, 2, cosine, 0, minus, sine, 0, right parenthesis
10, equals, A, start subscript, 2, end subscript, dot, 1, dot, left parenthesis, 2, minus, 0, right parenthesis, equals, 2, A, start subscript, 2, end subscript
A, start subscript, 2, end subscript, equals, 5

## Solution for current

And finally, after a bunch of hard work, the solution for current is:
i, equals, 5, e, start superscript, minus, t, end superscript, sine, 2, t
The graph of i as a function of time looks like this:
Natural response of an start text, R, L, C, end text circuit, start text, R, end text, equals, 2, \Omega, start text, L, end text, equals, 1, start text, H, end text, and start text, C, end text, equals, start fraction, 1, divided by, 5, end fraction, start text, F, end text. The faint curves plot plus minus, 5, e, start superscript, minus, t, end superscript, the envelope of the decaying sine wave.
When the switch closes, the current takes a big surge upwards and takes on the shape of the first hump of a sine wave. The sine wave quickly fades away after a few swings because the energy in the system rapidly dissipates as heat as charge flows back and forth through the resistor.
The role of "friction" in this example, as played by the resistor value, represents a fairly high rate of energy dissipation. The current visibly changes sign only two times before settling to zero.
This is an example of an underdamped solution. We will introduce this descriptive term in the next section.

## Solve the voltages

There is only one current in the circuit. Now that we know the natural response of the current, we can find the natural response of the three voltages.

### Resistor voltage

We use Ohm's Law to find the resistor voltage: left parenthesisthere's a minus sign because i is backwards relative to v, start subscript, start text, R, end text, end subscript, right parenthesis
v, start subscript, start text, R, end text, end subscript, equals, minus, i, start text, R, end text
v, start subscript, start text, R, end text, end subscript, equals, minus, 5, e, start superscript, minus, t, end superscript, sine, 2, t, dot, 2, \Omega
v, start subscript, start text, R, end text, end subscript, equals, minus, 10, e, start superscript, minus, t, end superscript, sine, 2, t

### Inductor voltage

The inductor voltage emerges from the inductor i-v equation:
v, start subscript, start text, L, end text, end subscript, equals, start text, L, end text, start fraction, d, i, divided by, d, t, end fraction
v, start subscript, start text, L, end text, end subscript, equals, 1, dot, start fraction, d, divided by, d, t, end fraction, left parenthesis, 5, e, start superscript, minus, t, end superscript, sine, 2, t, right parenthesis
v, start subscript, start text, L, end text, end subscript, equals, minus, 5, e, start superscript, minus, t, end superscript, left parenthesis, sine, 2, t, minus, 2, cosine, 2, t, right parenthesis

### Capacitor voltage

To find the capacitor voltage we can use the integral form of the capacitor i-v equation: left parenthesisanother extra minus sign because of the direction of i is reversed relative to v, start subscript, start text, C, end text, end subscript, right parenthesis
v, start subscript, start text, C, end text, end subscript, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, minus, i, d, t
v, start subscript, start text, C, end text, end subscript, equals, start fraction, 1, divided by, 1, slash, 5, end fraction, integral, minus, 5, e, start superscript, minus, t, end superscript, sine, 2, t, d, t
v, start subscript, start text, C, end text, end subscript, equals, 5, e, start superscript, minus, t, end superscript, left parenthesis, sine, 2, t, plus, 2, cosine, 2, t, right parenthesis
Here are all three voltages plotted together:

## Summary

The start text, R, L, C, end text circuit is the electronic equivalent of a swinging pendulum with friction. The circuit can be modeled by this 2nd-order linear differential equation:
start text, L, end text, start fraction, d, squared, i, divided by, d, t, squared, end fraction, plus, start text, R, end text, start fraction, d, i, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, C, end text, end fraction, i, equals, 0
The resulting characteristic equation is:
s, squared, plus, start fraction, start text, R, end text, divided by, start text, L, end text, end fraction, s, plus, start fraction, 1, divided by, start text, L, C, end text, end fraction, equals, 0
We solved for the roots of the characteristic equation using the quadratic formula:
s, equals, start fraction, minus, start text, R, end text, plus minus, square root of, start text, R, end text, squared, minus, 4, start text, L, end text, slash, start text, C, end text, end square root, divided by, 2, start text, L, end text, end fraction
By substituting variables alpha and omega, start subscript, o, end subscript we wrote s a little simpler as:
s, equals, minus, alpha, plus minus, square root of, alpha, squared, minus, omega, start subscript, o, end subscript, squared, end square root
where alpha, equals, start fraction, start text, R, end text, divided by, 2, start text, L, end text, end fraction and omega, start subscript, o, end subscript, equals, start fraction, 1, divided by, square root of, start text, L, C, end text, end square root, end fraction
We finished up by solving an example circuit whose components produced a current (and voltages) that swing back and forth a few times.
The roots of the characteristic equation can take on both real and complex forms, depending on the relative size of alpha and omega, start subscript, o, end subscript. In the next article, we will describe these three forms in additional detail:
• overdamped, alpha, is greater than, omega, start subscript, 0, end subscript, leads to the sum of two decaying exponentials
• critically damped, alpha, equals, omega, start subscript, 0, end subscript, leads to t, dot decaying exponential
• underdamped, alpha, is less than, omega, start subscript, 0, end subscript, leads to a decaying sine

## Want to join the conversation?

• You guys should really make a video on this topic! Also, it is much easier to find v(0+), i(0+), dv(0+)/dt, di(0+)/dt, v(infinity), i(infinity) right off the bat so you don't have to pound through the problem the long way.

Oh, and the alpha and omega for this only works if they are in series. You need to do a write up to show how to solve if they are in parallel.

Thanks • Why is the voltage across the capacitor (vc(t)) equals to the integral of the opposite of the current ? where does this minus sign come from ? • Because, in this article, it is assumed that the positive sign of the capacitor is on the top, so, the direction of current that it used is different from the current that supposed to be in the circuit... so that's why it got the minus sign...

If you flip the sign of the capacitor, it could be also worked though... so we don't get the minus sign, but in the KVL equation, it will be regarded as voltage rise, and it will get the plus sign... and in the end... it gets the same equation...
• Hi,
When finding A2 in derivative of the proposed solution, ​di​/dt​​=​​​d​​(e​^(−t​​)A​2​​sin2t)/dt
Shouldn't the answer be e^(-t) (4 cos(2 t) - 2 sin(2 t)) since we are using the product rule? With the answer coming out as A2=10/4? • How do you get the simpler s equation (the one with the variables α and ω​naught) from the quadratic formula?
(1 vote) • Review the definitions of alpha and ωo. This happens right after it says, "We can make the notation a little more compact by replacing parts of the expression with two new variables, alpha and ωo:"

Now look at the equation right after the sentence, "We can revise the quadratic formula by mashing the 2L denominator up into each term of the numerator:".

The first part of the equation, before the +/- sign, looks like -alpha.
The second part of the equation, under the square root symbol you find two terms, they look like alpha^2 and ωo^2.

Please let me know if this helped you understand. This is a spot where the article needs to be really clear.
• How can we include imaginary constant (j) in A2 and still get A2 as real (=5)? • When you assume a solution in the exponential form in the "Propose a solution" section, that lets you factor out the exponential and set the other, quadratic factor to 0. Then you use the solutions for s in the "Proposed solution, upgraded" section to make a new form for the solution as a sum of 2 exponentials. But why were you able to do that when you assumed the form of the solution to be in the form of a single exponential to begin with? Wouldn't that be contradicting the assumption? • What role does s(natural frequency) play here ? What it determines, can you please explain about it ?
(1 vote) • In the RLC derivation we found that s could have real or complex values. We described s in terms of alpha and omega, and found there were three general classes of behavior (under-damped, critically-damped, and over-damped). The solutions are rich and wonderful and varied. You get a drooping behavior from the real parts, and a wiggly behavior from the imaginary parts.

If you want to think about the meaning of s, it is a parameter (a complex number) that captures both the natural droopy and wiggly aspects of current and voltage as they change with time. The units of s are 1/t, which is the unit of frequency. The real part of s specifies the rate of droop while the imaginary part of s tells you the rate of wiggle.
• Why do we have 2L as the denominator in the quadratic formula?

Rachel
(1 vote) • Our differential equation for the RLC can be turned into a "characteristic equation". It turns out the characteristic equation follows the general form of a quadratic equation...

ax^2 + bx + c = 0

which can be solved with the quadratic formula...

x = -b +/- sqrt(b^2 -4ac) / 2a

In our case the RLC circuit assigns a, b, and c as...
a = L, b = R, and c = 1/C

The 2 comes from the quadratic formula.
(1 vote)
• Please , how do I find the value of R in terms of L and C that will result in repeated roots to be able to demonstrate an underdamped, overdamped and critically damped response in a RLC combination circuit.
(1 vote) • How did you obtained the second order DE for the RLC in series? Did you apply Kirchhoff's Loop Rule and take a total derivative of the function with respect to current?
(1 vote) 