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...

Do we have to do all this calculus everytime we make a RLC circuit, And this will get more complicated with complicated RLC circuits. There is no general formula for the RLC like-LR or RC circuits? I wanted to make a function generator using a buck-boost converter, To predict the values of current and voltage at the given continuously changing PWM signal will be hell lot of work!

Most RLC circuits are analyzed with Laplace Transform theory. That topic is beyond the scope of this introductory KA class, but it is the standard method for working with frequency filters and control systems. The RLC circuit analyzed here is the parallel form. The solution to the natural response emerges from this long analysis. The answer is not a simple single exponential equation like we get for RC, but rather a choice of three different responses ("variations") depending on the value of R, L , and C. I did an improved writeup of this at spinningnumbers dot com. Look for the RLC articles in the Natural and Forced Response section.

What role does s(natural frequency) play here ? What it determines, can you please explain about it ?

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.

Main content

Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response

RLC natural response - derivation

Q: How do you get the simpler s equation (the one with the variables α and ωnaught) from the quadratic formula?

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.

Google Classroom

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

(RLC)

. This is the last circuit we'll analyze with the full differential equation treatment.

The

RLC

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.

What we're building to

We will model the

RLC

circuit with a

2

nd-order linear differential equation with current,

i

, as the independent variable:

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 0

The resulting characteristic equation is:

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

We will solve for the roots of the characteristic equation using the quadratic formula:

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

By substituting variables

α

and

ω_{o}

we can write

s

a little simpler as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

where

α = \frac{R}{2 L}

, and

ω_{o} = \frac{1}{\sqrt{LC}}

α

is called the damping factor and

ω_{o}

is the resonant frequency.

We will solve an example

RLC

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.

Create a second-order differential equation based on the $i$ ‍- $v$ ‍ equations for the $R$ ‍, $L$ ‍, and $C$ ‍ components. We will use Kirchhoff's Voltage Law to build the equation.
Make an informed guess at a solution. As usual, our guess will be an exponential function of the form $K e^{s t}$ ‍.
Insert the proposed solution into the differential equation. The exponential terms will factor out and leave us with a characteristic equation in variable $s$ ‍.
Find the roots of the characteristic equation. This time we will need to use the quadratic formula to solve for the roots.
Find the constants by accounting for the initial conditions.
Celebrate the solution.

Model the circuit with a differential equation

There's a bit of cleverness going on where I assign the voltage directions. I looked ahead a little in the analysis and arranged the voltage polarities to get some positive signs where I want them. This is for aesthetic value. Read on to see how I respected the sign convention for passive components.

Capacitor voltage: I want to initialize the capacitor with a positive charge on the top plate, which means the positive sign for

v_{C}

is also the top plate. It is nice when positive voltages are drawn towards the top of the page.

Inductor current: When the switch closes, the initial surge of current flows from the capacitor across to the inductor, in a counter-clockwise direction, down through the inductor. I want the current to have a positive sign in this direction, so when I plot the current, the first surge is in the upward direction, not down. I think this makes the plot looks nicer.

The direction of

i

is chosen to flow into the inductor from the top. The sign convention for the passive inductor lets me assign

v_{L}

with the positive voltage sign at the top, towards the top of the page.

Resistor voltage: The resistor voltage makes no artistic contribution, so it can be assigned to match either the capacitor or the inductor. I happened to match it to the capacitor, but you could do it either way.

The voltage direction I prefer for

v_{C}

(positive at the top) conflicts with the direction of

i

in terms of the passive sign convention. Current

i

flows out of the

+

capacitor terminal instead of in. I can account for that reversal when I write the

i

v

equation for the capacitor, by slipping in a

-

i

The voltage on the resistor has the same issue as the capacitor. I will handle it the same way when I write Ohm's law for the resistor, by including a

-

i

Notice how I achieved my artistic intent while respecting the passive sign convention.

Most textbooks just write out the upcoming integro-differential equation without this long explanation, letting the you work out the signs yourself. I thought it would be helpful walk through it in detail.

When the switch closes, the circuit looks like this (now with voltage labels on the inductor and resistor,

v_{L}

and

v_{R}

For each individual element, we can write

i

v

equations.

v_{L} = L \frac{d i}{d t}

v_{R} = - i R

v_{C} = \frac{1}{C} \int - 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.

+ v_{L} - v_{R} - v_{C} = 0

Replacing the

v

terms with the corresponding

i

terms gives us:

L \frac{d i}{d t} + R i + \frac{1}{C} \int i d t = 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.

\frac{d}{d t} [L \frac{d i}{d t} + R i + \frac{1}{C} \int i d t = 0]

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

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 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 (t) = K e^{s t}

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 / 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.

L \frac{d^{2}}{d t^{2}} K e^{s t} + R \frac{d}{d t} K e^{s t} + \frac{1}{C} K e^{s t} = 0

Now let's work on the terms with derivatives.

Middle term: The first derivative of the

R

term is

R \frac{d}{d t} K e^{s t} = s R K e^{s t}

Leading term: We take the derivative if the leading

K e^{s t}

term two times:

\frac{d}{d t} K e^{s t} = s K e^{s t}

\frac{d}{d t} s K e^{s t} = s^{2} K e^{s t}

so the leading term becomes:

L \frac{d^{2}}{d t^{2}} K e^{s t} = s^{2} L K e^{s t}

Plug these back into the differential equation:

s^{2} L K e^{s t} + s R K e^{s t} + \frac{1}{C} K e^{s t} = 0

Now we can factor out the common

K e^{s t}

term:

K e^{s t} (s^{2} L + s R + \frac{1}{C}) = 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 = 0

and we are putting nothing into the circuit and getting nothing out. Pretty boring.

The term

e^{s t}

never becomes

0

unless we wait until

t

goes to

\infty

. 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^{2} L + s R + \frac{1}{C} = 0

This is called the characteristic equation of the

LRC

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^{2} + b x + c = 0

the quadratic formula gives us the roots (the zero-crossings):

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Looking back at the characteristic equation, we can plug in our circuit component values to get the roots.

a = L

b = R

, and

c = 1 / C

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

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,

α

and

ω_{o}

α = \frac{R}{2 L}

ω_{o} = \frac{1}{\sqrt{LC}}

Let me write the characteristic equation this way

(

dividing through by

L)

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

If we use

α

and

ω_{o}

, the characteristic equation can be written:

s^{2} + 2 α s + ω_{o}^{2} = 0

We can revise the quadratic formula by mashing the

2 L

denominator up into each term of the numerator:

s = - \frac{R}{2 L} \pm \sqrt{{(\frac{R}{2 L})}^{2} - (\frac{4 L / C}{4 L^{2}})}

The second term under the square root reduces to:

(\frac{4 L / C}{4 L^{2}}) = (\frac{4 L / C}{4 L^{2}}) = \frac{1}{LC}

And this lets us write

s

in terms of

α

and

ω_{o}

as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

We know

s

is some sort of frequency (it has to have units of

1 / t

). That means the two terms making up

s

are also some sort of frequency.

$α$ ‍ is called the damping factor. It will determine how quickly the overall signal fades to zero.
$ω_{o}$ ‍ 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 $LC$ ‍ natural response.

Proposed solution, upgraded

The quadratic formula gave us two solutions for

s

, we'll call these

s_{1}

and

s_{2}

. 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 = K_{1} e^{s_{1} t} + K_{2} e^{s_{2} t}

s_{1}

and

s_{2}

are natural frequencies,

K_{1}

and

K_{2}

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:

The differential equation for the

RLC

circuit is

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 0

With real component values it becomes:

1 \frac{d^{2} i}{d t^{2}} + 2 \frac{d i}{d t} + 5 i = 0

As we always do, assume a solution of the form:

i (t) = K e^{s t}

We perform the analysis we did above, which results in this characteristic equation:

s^{2} + 2 s + 5 = 0

Solving for the roots of the characteristic equation with the quadratic formula:

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

With real component values:

s = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot 5}}{2}

s = \frac{- 2 \pm \sqrt{4 - 20}}{2}

s = - 1 \pm \frac{\sqrt{- 16}}{2}

s = - 1 \pm j 2

(Electrical engineers use the letter

j

for the imaginary

\sqrt{- 1}

, since we use

i

as the symbol for current.)

We get a complex answer, just as we did with the

LC

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 = K_{1} e^{(- 1 + j 2) t} + K_{2} e^{(- 1 - j 2) t}

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 = K_{1} e^{- 1 t} e^{+ j 2 t} + K_{2} e^{- 1 t} e^{- j 2 t},

and factor out the common

e^{- 1 t}

term:

i = e^{- t} (K_{1} e^{+ j 2 t} + K_{2} e^{- j 2 t})

Notice how the real part of

s

came through the factoring process to give us the leading term, a decaying exponential,

e^{- t}

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

LC

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^{j x}

\sin j x

, and

\cos j x

, it is possible to derive Euler's formula:

e^{+ j x} = \cos x + j \sin x

and

e^{- j x} = \cos x - j \sin x

In the linked video, whenever Sal says

i

, we say

j

These formulas let us turn

e^{i m a g i n a r y}

into a normal complex number.

Use Euler's formula

We can use Euler's formula to transform the sum

K_{1} e^{+ j 2 t} + K_{2} e^{- j 2 t}

into

K_{1} (\cos 2 t + j \sin 2 t) + K_{2} (\cos 2 t - j \sin 2 t) .

Multiply through the constants

K_{1}

and

K_{2}

K_{1} \cos 2 t + j K_{1} \sin 2 t + K_{2} \cos 2 t - j K_{2} \sin 2 t,

and gather the cosine terms and sine terms:

(K_{1} + K_{2}) \cos 2 t + j (K_{1} - K_{2}) \sin 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_{1} = (K_{1} + K_{2})

, and

A_{2} = j (K_{1} - K_{2}

The previous expression becomes:

A_{1} \cos 2 t + A_{2} \sin 2 t

And now put this back into our proposed solution:

i = e^{- t} (A_{1} \cos 2 t + A_{2} \sin 2 t)

So far so good. Next, we need to figure out

A_{1}

and

A_{2}

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 / d t

If we can figure out

i

and

d i / d t

at a specific time, we can find

A_{1}

and

A_{2}

Finding the initial conditions for

RLC

is pretty much the same as for the LC circuit. We just have account for the resistor.

Here's what we know about

t = 0^{-}

(the moment before the switch closes):

The switch is open, so $i (0^{-}) = 0$ ‍
The starting capacitor voltage is specified: $v_{C} (0^{-}) = V_{0}$ ‍

t = 0^{+}

is the moment just after the switch closes, our goal is to find

i (0^{+})

and

d i / d t (0^{+})

. We know some properties of inductors and capacitors that tell us what happens when the switch closes, going from

t = 0^{-}

t = 0^{+}

Inductor current does not change instantly, so $i (0^{+}) = i (0^{-}) = 0$ ‍
Capacitor voltage does not change instantly, so $v_{C} (0^{+}) = v_{C} (0^{-}) = V_{0}$ ‍

Now we know one initial condition,

i (0^{+}) = 0

, and we know something about the voltage, but we don't know

d i / d t (0^{+})

Let's go after the second initial condition,

d i / d t (0^{+})

. Any time I see

d i / 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 / d t

. Let's do that by process of elimination.

Kirchhoff's Voltage Law around the loop is:

+ v_{L} - v_{R} - v_{C} = 0

Since

i (0^{+}) = 0

, that means the voltage across the resistor,

v_{R}

, has to be

0

. We also know the voltage across the capacitor is

v_{C} = V_{0}

. Let's fill in the KVL equation with these values:

v_{L} - 0 - V_{0} = 0

And now we know the voltage across the inductor at

t = 0^{+}

v_{L} = V_{0}

We can use this to derive

d i / d t

using the inductor

i

v

equation.

v_{L} (0^{+}) = L \frac{d i}{d t} (0^{+})

10 = 1 \frac{d i}{d t} (0^{+})

\frac{d i}{d t} (0^{+}) = 10 A / sec

The moment just after the switch closes, the current in the inductor has an initial slope of

10

amperes per second.

Find constants $A_{1}$ ‍ and $A_{2}$ ‍ using the initial conditions

As a reminder, our proposed solution is:

i = e^{- t} (A_{1} \cos 2 t + A_{2} \sin 2 t)

and the initial conditions are:

i (0^{+}) = 0

\frac{d i}{d t} (0^{+}) = 10

If we evaluate

i

t = 0

, we can find one of the

A

constants. Insert

t = 0

and

i = 0

into the proposed solution:

0 = e^{- 0} (A_{1} \cos 2 \cdot 0 + A_{2} \sin 2 \cdot 0)

0 = 1 (A_{1} \cos 0 + A_{2} \sin 0)

0 = (A_{1} \cdot 1 + A_{2} \cdot 0)

A_{1} = 0

A_{1} = 0

, so the cosine term drops out of the solution. One down, one to go. Our proposed solution now looks like:

i = A_{2} e^{- t} \sin 2 t

Let's chase down

A_{2}

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?

\frac{d i}{d t} = \frac{d}{d t} (A_{2} e^{- t} \sin 2 t)

The proposed solution is the product of two functions. To take its derivative we use the product rule:

{(f g)}^{'} = f^{'} g + f g^{'}

Identify the two parts of the product and their derivatives:

f = A_{2} e^{- t} g = \sin 2 t

f^{'} = - A_{2} e^{- t} g^{'} = 2 \cos 2 t

Assemble the parts according to the product rule:

\frac{d i}{d t} = - A_{2} e^{- t} \sin 2 t + A_{2} e^{- t} 2 \cos 2 t

\frac{d i}{d t} = A_{2} e^{- t} (2 \cos 2 t - \sin 2 t)

We can evaluate this expression at

t = 0^{+}

10 = A_{2} e^{- 0} (2 \cos 0 - \sin 0)

10 = A_{2} \cdot 1 \cdot (2 - 0) = 2 A_{2}

A_{2} = 5

Solution for current

And finally, after a bunch of hard work, the solution for current is:

i = 5 e^{- t} \sin 2 t

The graph of

i

as a function of time looks like this:

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:

(

there's a

-

sign because

i

is backwards relative to

v_{R})

v_{R} = - i R

v_{R} = - 5 e^{- t} \sin 2 t \cdot 2 Ω

v_{R} = - 10 e^{- t} \sin 2 t

Inductor voltage

The inductor voltage emerges from the inductor

i

v

equation:

v_{L} = L \frac{d i}{d t}

v_{L} = 1 \cdot \frac{d}{d t} (5 e^{- t} \sin 2 t)

v_{L} = - 5 e^{- t} (\sin 2 t - 2 \cos 2 t)

Capacitor voltage

To find the capacitor voltage we can use the integral form of the capacitor

i

v

equation:

(

another extra

-

sign because of the direction of

i

is reversed relative to

v_{C})

v_{C} = \frac{1}{C} \int - i d t

v_{C} = \frac{1}{1 / 5} \int - 5 e^{- t} \sin 2 t d t

v_{C} = - 25 \int e^{- t} \sin 2 t d t

\frac{v_{C}}{- 25} = \int e^{- t} \sin 2 t d t

We will solve the integral on the right, and account for the

- 25

at the end.

The integral is the product of two functions. To solve this, we'll use integration by parts:

\int f^{'} g = f g - \int f g^{'}

f^{'} = e^{- t} g = \sin 2 t

f = - e^{- t} g^{'} = 2 \cos 2 t

First integration by parts:

\int f^{'} g = f g - \int f g^{'}

\int e^{- t} \sin 2 t d t = - e^{- t} \sin 2 t - \int - e^{- t} 2 \cos 2 t d t

Second integration by parts of the integral on the far right:

f^{'} = - e^{- t} g = 2 \cos 2 t

f = e^{- t} g^{'} = - 4 \sin 2 t

\int e^{- t} \sin 2 t d t = - e^{- t} \sin 2 t - [e^{- t} 2 \cos 2 t - \int e^{- t} (- 4 \sin 2 t) d t]

\int e^{- t} \sin 2 t d t = - e^{- t} \sin 2 t - e^{- t} 2 \cos 2 t - 4 \int e^{- t} \sin 2 t d t

Aha! Notice the integral on the left side appears again over on the right side. Let's merge those two terms on the left side. (plus the obligatory extra constant):

\int 5 e^{- t} \sin 2 t d t = - e^{- t} \sin 2 t - e^{- t} 2 \cos 2 t

\int e^{- t} \sin 2 t d t = \frac{1}{5} (- e^{- t} \sin 2 t - e^{- t} 2 \cos 2 t)

Referring back to our starting integral, the term on the left is equal to

\frac{v_{C}}{- 25}

\frac{v_{C}}{- 25} = \int e^{- t} \sin 2 t d t = \frac{1}{5} (- e^{- t} \sin 2 t - e^{- t} 2 \cos 2 t)

The result is (including the obligatory constant):

v_{C} = 5 e^{- t} (\sin 2 t + 2 \cos 2 t) + C

We can deal with the constant by evaluating the equation at a time when we know

v_{C}

. A good time to pick is

t = 0

because that's when we know

v_{C} = + 10 V

+ 10 V = 5 e^{- 0} (\sin 2 \cdot 0 + 2 \cos 2 \cdot 0) + C

+ 10 V = 5 \cdot 1 \cdot (0 + 2 \cdot 1) + C

C = 0

v_{C} = 5 e^{- t} (\sin 2 t + 2 \cos 2 t)

Here are all three voltages plotted together:

There's another way we could have found the capacitor voltage,

v_{C}

. Since we already figured out the resistor and inductor voltages, we can use Kirchhoff's Voltage Law (KVL) to find the capacitor voltage.

KVL around the loop, starting from the bottom left:

v_{L} - v_{R} - v_{C} = 0

v_{C} = v_{L} - v_{R}

Now fill in the voltages we know:

v_{L} = - 5 e^{- t} (\sin 2 t - 2 \cos 2 t)

v_{R} = - 10 e^{- t} \sin 2 t

v_{C} = - 5 e^{- t} (\sin 2 t - 2 \cos 2 t) - (- 10 e^{- t} \sin 2 t)

Combining the like terms gives us

v_{C}

v_{C} = 5 e^{- t} (\sin 2 t + 2 \cos 2 t)

It's reassuring the answer is the same as we found by integration. This KVL method worked because we knew all the other voltages, which might not always be the case. So there might be times when you have to do the integration. Who doesn't enjoy a nice integration by parts every now and then?

Summary

The

RLC

circuit is the electronic equivalent of a swinging pendulum with friction. The circuit can be modeled by this

2

nd-order linear differential equation:

L \frac{d^{2} i}{d t^{2}} + R \frac{d i}{d t} + \frac{1}{C} i = 0

The resulting characteristic equation is:

s^{2} + \frac{R}{L} s + \frac{1}{LC} = 0

We solved for the roots of the characteristic equation using the quadratic formula:

s = \frac{- R \pm \sqrt{R^{2} - 4 L / C}}{2 L}

By substituting variables

α

and

ω_{o}

we wrote

s

a little simpler as:

s = - α \pm \sqrt{α^{2} - ω_{o}^{2}}

where

α = \frac{R}{2 L}

and

ω_{o} = \frac{1}{\sqrt{LC}}

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

α

and

ω_{o}

. In the next article, we will describe these three forms in additional detail:

overdamped, $α > ω_{0}$ ‍, leads to the sum of two decaying exponentials
critically damped, $α = ω_{0}$ ‍, leads to $t \cdot$ ‍ decaying exponential
underdamped, $α < ω_{0}$ ‍, leads to a decaying sine

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Mark Freiheit
Posted 7 years ago. Direct link to Mark Freiheit's post “You guys should really ma...”
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
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(25 votes)
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- StarSword
  Posted a year ago. Direct link to StarSword's post “It's the same process for...”
  It's the same process for a parallel, it's just that the differential equation is voltage-based: Cv''(t) + v'(t)/R + v(t)/L = 0.
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  (1 vote)
ksanchezlacroix
Posted 7 years ago. Direct link to ksanchezlacroix's post “Why is the voltage across...”
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 ?
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(5 votes)
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- Ali Jaya Meilio
  Posted 7 years ago. Direct link to Ali Jaya Meilio's post “Because, in this article,...”
  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...
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  (5 votes)
megaultrachickenrawr
Posted 7 years ago. Direct link to megaultrachickenrawr's post “Hi, When finding A2 in d...”
Hi,
When finding A2 in derivative of the proposed solution, di/dt=d(e^(−t)A2sin2t)/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?
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(2 votes)
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- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “Thanks for pointing how w...”
  Thanks for pointing how where I skipped the product rule step when taking the derivative. I will ask to have that fixed. Here's how it should go:
  
  di/dt = d/dt ( e^-t A2 sin2t) = e^-t A2 (2cos2t - sin 2t)
  
  Evaluate this at t=0,
  
  10 = e^0 A2 (2cos 0 - sin 0)
  
  10 = 1 . A2 . (2 - 0) = 2 . A2
  
  A2 = 5
  Comment on Willy McAllister's post “Thanks for pointing how w...”
  (1 vote)
Emmen Bramasta
Posted 8 years ago. Direct link to Emmen Bramasta's post “How do you get the simple...”
How do you get the simpler s equation (the one with the variables α and ωnaught) from the quadratic formula?
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(1 vote)
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- Willy McAllister
  Posted 8 years ago. Direct link to Willy McAllister's post “Review the definitions of...”
  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.
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  (3 votes)
pushkarm27
Posted 6 years ago. Direct link to pushkarm27's post “How can we include imagin...”
How can we include imaginary constant (j) in A2 and still get A2 as real (=5)?
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(2 votes)
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- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Back in the section Use E...”
  Back in the section Use Euler's Formula, A2 was defined in terms of the K constants.
  
  A2 = j(K1 - K2)
  
  Then we pressed ahead and figured out A2 = 5.
  
  If you go back to the definition of A2 in terms of K's, that means the K's have to be imaginary numbers. They have to include a j term to get rid of the j in A2 = j(K1 - K2).
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  (1 vote)
Frank Gu
Posted 7 years ago. Direct link to Frank Gu's post “When you assume a solutio...”
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?
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(2 votes)
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yssh
Posted a year ago. Direct link to yssh's post “Do we have to do all this...”
Do we have to do all this calculus everytime we make a RLC circuit, And this will get more complicated with complicated RLC circuits. There is no general formula for the RLC like-LR or RC circuits?
I wanted to make a function generator using a buck-boost converter, To predict the values of current and voltage at the given continuously changing PWM signal will be hell lot of work!
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(1 vote)
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- Willy McAllister
  Posted a year ago. Direct link to Willy McAllister's post “Most RLC circuits are ana...”
  Most RLC circuits are analyzed with Laplace Transform theory. That topic is beyond the scope of this introductory KA class, but it is the standard method for working with frequency filters and control systems.
  
  The RLC circuit analyzed here is the parallel form. The solution to the natural response emerges from this long analysis. The answer is not a simple single exponential equation like we get for RC, but rather a choice of three different responses ("variations") depending on the value of R, L , and C. I did an improved writeup of this at spinningnumbers dot com. Look for the RLC articles in the Natural and Forced Response section.
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  (2 votes)
manavpandya31
Posted 5 years ago. Direct link to manavpandya31's post “What role does s(natural ...”
What role does s(natural frequency) play here ? What it determines, can you please explain about it ?
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(1 vote)
Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “In the RLC derivation we ...”
  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.
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  (2 votes)
Rachel
Posted a year ago. Direct link to Rachel's post “Why do we have 2L as the ...”
Why do we have 2L as the denominator in the quadratic formula?

Rachel
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(1 vote)
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- Willy McAllister
  Posted a year ago. Direct link to Willy McAllister's post “Our differential equation...”
  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.
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  (1 vote)
Rachel
Posted a year ago. Direct link to Rachel's post “Please , how do I find th...”
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.
Button navigates to signup pageComment on Rachel's post “Please , how do I find th...”
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Answer

Electrical engineering

Course: Electrical engineering > Unit 2

Introduction

What we're building to

Strategy

Model the circuit with a differential equation

Propose a solution

Try the proposed solution

Make the equation true

Solve for the roots of the characteristic equation

Proposed solution, upgraded

Example circuit

Euler's formula

Use Euler's formula

Find the initial conditions

Find constants A1‍ and A2‍ using the initial conditions

Solution for current

Solve the voltages

Resistor voltage

Inductor voltage

Capacitor voltage

Summary

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Find constants $A_{1}$ ‍ and $A_{2}$ ‍ using the initial conditions