Are differential equations truly necessary? Couldn't this material become more accessible (to lower math backgrounds) if things were introduced in the frequency domain rather than the time domain? Then we would be using only algebra and not calculus. The transformations back to time domain could be introduced later.

I like your style and thinking.Keep up the good thinking and you could be an amazing engineer!

Why was the current in resistor and current in capacitor taken in opposite directions in derivation of RC natural response ?

Good question. This is one of the tricky parts of the analysis that most textbooks skip over. In the section titled "Model the components" I define two current arrows for the resistor and capacitor currents. I use two current arrows because I want to be sure I get the signs right when I use Ohm's Law for the resistor and i = Cdv/dt for the capacitor. Notice that these equations do not have a negative sign. That's because both of them respect the Sign Convention for Passive Components. (Resistor current flows _into_ the positive voltage terminal of the resistor. Capacitor current flows _into_ the positive voltage terminal of the capacitor.) If I had defined a single current arrow, I would have to write one of those equations with a negative sign. I find it awkward to insert negative signs into i-v equations, so I did it the two-arrow way. But now I have two current arrows I have to deal with... In the next section, "Model the circuit", the two currents participate in a KCL equation that gives us the differential equation to be solved. It is a pretty simple KCL equation. The signs on the two terms in the resulting equation came out correct, which is the reward for doing the two-current-arrows-then-KCL method. You should try this: Define just a single current arrow named i. Then model the components with i-v equations, then model the circuit. This time you equate i in the resistor to i in the capacitor. When you derive the differential equation do you get the same thing? It takes some care to get this right. If it works for you, make that your favorite method.

If we are asked to find the current in the resistor of Example 1 . What to do ?

In example 1, we solved the voltage across the resistor, v(t). Since you know the voltage, and you know the resistor value, you can use Ohm's Law to get the current, i(t). Incidentally, that's also the answer for the current in the capacitor. The same current flows in both R and C. Before t=0, the current in the resistor depends on the details of how the "helping" circuit (the switch) was operated. In the case of the specific circuit we started with, the current was 0 before the switch flipped.

How to find equivalent resistance of a complex circuit

There are several different ways to to do this. I would recommend that first you should convert any Inductors and capacitors in your circuit into impedances. For inductors use Z = jwl. (Where j represents the imaginary impedance, similar to i in mathematics). w is the angular frequency. l is the inductance. For capacitors use Z = 1/jwc (also, -j/wc which can be obtained by multiplying both the numerator and the denominator by j). c is the capacitance. Resistors stay the same (the ohms are purely real). Now combine all of these impedances as if they were a large resistor network. You will likely need a calculator that is capable of doing complex numbers. Good Luck and I hope that helps!

In the first part of the lecture, we are asked what is Vc before the switch is flipped up, after the switch is flipped up, and finally after the switch is flipped back down, For the second part (What is Vc after the switch is flipped up) it states that the capacitor voltage(Vc) will rise to the same voltage as the battery. This is incorrect because there is also a voltage drop across the resistor so Vc would be Vbatt - Vr. KVL equation would be Vbatt - Vr - Vc = 0.

As the capacitor is charging, current is flowing and there is some voltage drop across the resistor. But, the longer the current flows, the more charge accumulates on the capacitor. Eventually, the charge on the capacitor reaches the point where the voltage of the capacitor (q/C) is equal and opposite that of Vbatt. At that point (time), current stops flowing. There is no voltage drop across the resistor because there is no current when the capacitor is fully charged. Vr = IR, but I = 0. The charging behavior of capacitors is just the opposite of the discharging behavior that is illustrated so well in this lesson. Another good resource is Prof. Nave's Hyperphysics site. There are pages that show both the charge and discharge behavior of capacitors, along with the math.

Regarding the circut with the battery (constant voltage source). How can I calculate the *time* it takes to charge a capacitor? Just rearange the final RC response equation to find t? What about in a circuit without a resistor? What role does a resistor play when charging a capacitor, anyway (max. current perhaps)? Thanks

Hello Nejc, Yes, you can solve by manipulating the equation as you stated. You could use a rule of thumb and assume 66% charge in one time constant and fully charged in about 5 time constants. In the real world there is no such thing as "no resistance." All wires have some resistance, the capacitor has an internal resistance, the voltage source (battery) has an internal resistance. You are correct about the max current. As an example consider a lithium ion battery storage system connected to a DC to AC inverter. In many of these systems you will find a capacitor on the DC side between the battery and the inverter. Connecting this capacitor directly to the batter would cause an excessive current flow. A "current limiting resistor" is used to charge the capacitor. After the capacitor is fully charged to the DC batter voltage it can be directly connected to the battery. Please leave a comment below if you would like to continue the conversation. Regards, APD

Main content

Electrical engineering

Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response

RC natural response

Google Classroom

Natural response of an RC circuit. The product of R and C is called the time constant. Written by Willy McAllister.

The Resistor-Capacitor left parenthesis, start text, R, C, end text, right parenthesis circuit is one of the first interesting circuits we can create and analyze. Understanding the behavior of this circuit is essential to learning electronics. Forms of this circuit can be found everywhere. Sometimes you will create this circuit on purpose, and other times it appears all on its own.

[On its own?]

This is one of the first circuits we come across where we have to account for time. To develop a precise understanding requires methods from calculus. We use derivatives to describe the start text, R, C, end text circuit.

We want to understand the natural response of this circuit.

What we're building to

A resistor-capacitor circuit, where the capacitor has an initial voltage V, start subscript, 0, end subscript, the voltage will diminish exponentially according to:

$v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

Where start text, V, end text, start subscript, 0, end subscript is the voltage at time t, equals, 0. This is called the natural response.

The time constant for an start text, R, C, end text circuit is tau, equals, start text, R, end text, dot, start text, C, end text

The circuit we will study is a resistor in series with a capacitor. How does this circuit respond to an applied voltage?

First, use intuition to predict what happens

The circuit we explore in this section is:

We want to know what happens to v, start subscript, start text, C, end text, end subscript, the capacitor voltage, when we flip the switch back and forth.

[naming voltages and currents]

We’ll analyze these questions one at a time:

What is v, start subscript, start text, C, end text, end subscript, the voltage across the capacitor,

before the switch flips up?
after the switch flips up?
after the switch flips back down?

Before the switch flips up

We begin our analysis by determining the initial state of the circuit, before anything changes. With the switch in the down position we can draw the following equivalent circuit. v, start subscript, i, n, end subscript is 0 volts, and the left end of start text, R, end text is connected to the bottom of start text, C, end text.

Let's assume for the moment the circuit has been sitting in this state for a long time, so any charge that may have been stored on the capacitor in the past has long since drained away through the resistor, leaving q, start subscript, start text, C, end text, end subscript, equals, 0. From this we know the voltage across the capacitor must be 0 volts, because v, start subscript, start text, C, end text, end subscript, equals, q, slash, start text, C, end text, equals, 0, slash, start text, C, end text, equals, 0.

Since the capacitor has 0 volts across it, so must the resistor, so the current through start text, R, end text (and the current through the capacitor) must be 0 amps. The circuit is said to be ‘in steady-state’ or ‘quiescent’ or ‘at an equilibrium’. We've answered the first question, "What is the voltage across start text, C, end text before the switch flips."

After the switch flips up

Now flip the switch up. The voltage v, start subscript, i, n, end subscript becomes start text, V, end text, start subscript, start text, B, A, T, end text, end subscript, and something is about to change.

Current begins flowing out of the positive terminal of the battery, through start text, R, end text and start text, C, end text. Charge accumulates on the capacitor. The accumulating charge generates a rising voltage across the capacitor (v, start subscript, start text, C, end text, end subscript, equals, q, slash, start text, C, end text). The time period where voltage v, start subscript, C, end subscript is changing is called a transient period.

What keeps v, start subscript, start text, C, end text, end subscript from rising forever? Charge accumulates on the capacitor until v, start subscript, start text, C, end text, end subscript rises to the same value as the battery voltage: v, start subscript, start text, C, end text, end subscript, equals, start text, V, end text, start subscript, start text, B, A, T, end text, end subscript. At that point, the voltage across the resistor is 0 volts, so current in the resistor stops flowing (Ohm's Law). That also means current (charge) stops flowing into the capacitor. The amount of charge on the capacitor stops changing and therefore the capacitor voltage becomes constant: v, start subscript, start text, c, end text, end subscript, equals, start text, V, end text, start subscript, start text, B, A, T, end text, end subscript. The transient period has ended.

We've answered the second question, "What is the voltage across start text, C, end text after the switch flips up?" After a transient period, the circuit assumes a new steady state with v, start subscript, start text, C, end text, end subscript, equals, start text, V, end text, start subscript, start text, B, A, T, end text, end subscript. It remains there until something comes along to disturb its bliss.

After the switch flips back down

Now we’ll flip the switch again, returning it back to the negative terminal of the battery (v, start subscript, i, n, end subscript, equals, 0). What happens now?

This is the same circuit we started with, but this time start text, C, end text is storing some charge, so there's a starting voltage across it. Because of this, start text, R, end text now has a voltage difference across its terminals. The voltage is v, start subscript, start text, C, end text, end subscript, equals, start text, V, end text, start subscript, start text, B, A, T, end text, end subscript at the moment the switch flips down. Therefore, a current must start to flow in start text, R, end text (so says Ohm's Law). The charge providing this current comes from the charge stored in start text, C, end text. Charge will continue to flow until all the charge originally stored in start text, C, end text is depleted. v, start subscript, start text, C, end text, end subscript gradually falls to zero volts. The voltage difference across start text, R, end text also falls to zero. The circuit has returned to its original equilibrium state. And finally, we've answered the third question, "What is the voltage across start text, C, end text after the switch flips back down?"

Summary

Using just our intuition, we know the capacitor voltage, v, start subscript, start text, C, end text, end subscript, starts at 0 volts, rises to start text, V, end text, start subscript, start text, B, A, T, end text, end subscript, and then goes back to 0 volts again. Said another way, v, start subscript, start text, C, end text, end subscript goes from an initial steady state, through a transient to a new steady state, then through a second transient back to its original state. We know the exact starting and ending voltage of each transient. Not bad, but ... What do we not know? We don’t know how long the transients last or their shape. It’s time to break out some calculus to get a precise and useful solution.

Formal derivation of the start text, R, C, end text natural response

We start with the simplest possible case. The circuit is just start text, R, end text and start text, C, end text connected together. By "find the response" we mean find start color #e07d10, v, end color #e07d10 and start color #11accd, i, end color #11accd as a function of time.

To make the circuit do something (other than just sit there), we place an initial charge on the capacitor. This is done by an external unseen circuit. After adding this energy, we let go and see what the circuit does naturally. Imagine the capacitor was charged to some initial voltage start text, V, end text, start subscript, 0, end subscript by an external circuit, which was disconnected just a moment ago.

The result we are about to derive is called the natural response of an start text, R, C, end text circuit. The natural response is what the circuit does when there is an initial condition, but nothing else is driving the circuit.

Model the components

The start text, R, end text and start text, C, end text components in the circuit can be described by their characteristic voltage-current equations.

For the resistor, we pick a form of Ohm’s Law:

i, start subscript, start text, R, end text, end subscript, equals, start fraction, v, divided by, start text, R, end text, end fraction

The corresponding voltage-current relationship for the capacitor is:

i, start subscript, start text, C, end text, end subscript, equals, start text, C, end text, start fraction, d, v, divided by, d, t, end fraction

[Where did this equation come from?]

Model the circuit

We can write an equation using Kirchhoff's Current Law for the two currents flowing out of the top node.

i, start subscript, start text, C, end text, end subscript, plus, i, start subscript, start text, R, end text, end subscript, equals, 0

start text, C, end text, start fraction, d, v, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, R, end text, end fraction, v, equals, 0

Solve the circuit

The previous equation is a first-order ordinary differential equation (ODE). We have the math skills to solve this kind of equation.

[What do these terms mean?]

The solution to a differential equation is some sort of function, in our case, some function of voltage with respect to time, v, left parenthesis, t, right parenthesis. v, left parenthesis, t, right parenthesis is a solution if it makes the differential equation true.

start text, C, end text, start fraction, d, v, divided by, d, t, end fraction, plus, start fraction, 1, divided by, start text, R, end text, end fraction, v, equals, 0

(differential equation)

Where do ODE solutions come from? One way is to make an informed guess at a solution, and try it out.

[Are there other ways?]

While gazing at the differential equation, consult your mental trash bin of knowledge about functions.

The two terms in the equation have to add up to zero. This suggests the first derivative of the function needs to have the same form or shape as the function itself. Search your memory for any function whose first derivative looks just like the function itself. Hmm...

A function that fits the bill is some form of the exponential, e, start superscript, x, end superscript, because the derivative of an exponential is another exponential.

start fraction, d, divided by, d, t, end fraction, e, start superscript, alpha, t, end superscript, equals, alpha, e, start superscript, alpha, t, end superscript

[Review: derivative of an exponential]

To solve our differential equation, we are going to make a bold proposal for the form of the solution. (This part takes courage.) Then we will plug our solution into the equation and work out a few constants specific to the circuit. (This part takes math.) If we find constants that make the equation true, then the proposed function is a solution to the equation, and we win.

Our proposed solution is an exponential function decorated with adjustable parameters, K and s.

v, left parenthesis, t, right parenthesis, equals, K, e, start superscript, s, t, end superscript

t is time
v, left parenthesis, t, right parenthesis is voltage as a function of time
K and s are constants we have to figure out
- K is an amplitude factor that makes the voltage bigger or smaller.
- s is in the exponent. It has to have units that cancel out time. So the units of s are 1, slash, t.

Let’s check to see if our proposed solution works...

Substitute v, left parenthesis, t, right parenthesis, equals, K, e, start superscript, s, t, end superscript, into the differential equation:

start text, C, end text, start fraction, d, divided by, d, t, end fraction, left parenthesis, K, e, start superscript, s, t, end superscript, right parenthesis, plus, start fraction, 1, divided by, start text, R, end text, end fraction, left parenthesis, K, e, start superscript, s, t, end superscript, right parenthesis, equals, 0

Work out the derivative in the first term

start fraction, d, divided by, d, t, end fraction, left parenthesis, K, e, start superscript, s, t, end superscript, right parenthesis, equals, s, start text, K, end text, e, start superscript, s, t, end superscript

Put s, start text, K, end text, e, start superscript, s, t, end superscript back into the differential equation:

s, start text, C, end text, K, e, start superscript, s, t, end superscript, plus, start fraction, 1, divided by, start text, R, end text, end fraction, K, e, start superscript, s, t, end superscript, equals, 0

Now we can factor out K, e, start superscript, s, t, end superscript

left parenthesis, s, start text, C, end text, plus, start fraction, 1, divided by, start text, R, end text, end fraction, right parenthesis, K, e, start superscript, s, t, end superscript, equals, 0

This equation represents our specific circuit, with the proposed solution. Almost there. Next, we work out two constants and see if the equation is true.

How many ways can we make the left side equal zero? Three ways: any of the three terms could be zero, K or e, start superscript, s, t, end superscript or left parenthesis, s, start text, C, end text, plus, 1, slash, start text, R, end text, right parenthesis.

A trivial solution is K, equals, 0. That’s equivalent to setting the initial charge on the capacitor to 0 and the circuit just sits there doing nothing. That’s so boring.

Another trivial solution is to make e, start superscript, s, t, end superscript, equals, 0. Set s to any negative value and let t go to plus, infinity. The exponential e, start superscript, minus, infinity, end superscript completely dies out, which means we sit around for infinity time waiting for the capacitor to fully discharge. Again, not too interesting.

A more thought-provoking solution comes from the third choice:

s, start text, C, end text, plus, start fraction, 1, divided by, start text, R, end text, end fraction, equals, 0

This equation is true if:

s, equals, minus, start fraction, 1, divided by, start text, R, C, end text, end fraction

[magic moment]

So far, our proposed solution looks like:

$v(t) = K\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

Nearly done. All that's left is to figure out K. Examine the initial conditions of the circuit. Recall the capacitor was initially charged to voltage start text, V, end text, start subscript, 0, end subscript. If we call that moment t, equals, 0, then

$v(0) = \text V_0 = Ke\^{\displaystyle{-\frac 0 {\text{RC}}}}$

Which evaluates to K, equals, start text, V, end text, start subscript, 0, end subscript.

We found an s and K to make the differential equation true. We are all done. Drum roll, please...

The general solution for the natural response of an start text, R, C, end text circuit is,

$\large v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

Time Constant

An exponent cannot have units. This means the product start text, R, C, end text in e, start superscript, minus, t, slash, start text, R, C, end text, end superscript has to have units of time, to cancel time t in the numerator. This means start text, o, h, m, s, end text, dot, start text, f, a, r, a, d, s, space, =, space, s, e, c, o, n, d, s, end text, something you might not have guessed.

The product of start text, R, end text and start text, C, end text is called the time constant of this circuit, and it usually has the Greek letter name tau (tau).

tau, equals, start text, R, C, end text

And we write the solution as:

$v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\tau}}}$

When t is equal to the time constant, the exponent of e becomes minus, 1, and the exponential term is equal to 1, slash, e, or about 0, point, 37. The time constant determines how fast the exponential curve comes down to zero. After 1 time constant has passed, the voltage is down to 37, percent of its initial value.

Example 1

For the natural response circuit,
let start text, R, end text, equals, 3, start text, k, end text, \Omega, start text, C, end text, equals, 1, mu, start text, F, end text, and start text, V, end text, start subscript, 0, end subscript, equals, 1, point, 4, start text, V, end text.

a. Write the expression for v, left parenthesis, t, right parenthesis
b. What is v, left parenthesis, t, right parenthesis when t, equals, start text, R, C, end text ?
c. Plot v, left parenthesis, t, right parenthesis

Solution to example 1

a. Write the expression for v, left parenthesis, t, right parenthesis

$v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

$v(t) = 1.4\,\,e \^{\displaystyle-\dfrac t {3 \,\text{k}\Omega \cdot 1 \,\mu\text{F}}}$

$v(t) = 1.4\,\,e \^{\displaystyle-\dfrac t {3\times10^{3} \cdot 1\: \times10^{-6}}}$

$v(t) = 1.4\,e \^{\displaystyle-\dfrac t {3\times10^{-3}}}$

$v(t) = 1.4\,e \^{\displaystyle-\dfrac t {3 \:\text{ms}}}$

b. What is v, left parenthesis, t, right parenthesis when t, equals, start text, R, C, end text ?

The start text, R, C, end text product has units of seconds.

tau, equals, start text, R, C, end text, equals, 3, times, 10, cubed, dot, 1, times, 10, start superscript, minus, 6, end superscript

tau, equals, 3, times, 10, start superscript, minus, 3, end superscript, equals, 3, start text, m, s, end text

v, left parenthesis, 3, start text, m, s, end text, right parenthesis, equals, 1, point, 4, e, start superscript, minus, start fraction, 3, start text, m, s, end text, divided by, 3, start text, m, s, end text, end fraction, end superscript

v, left parenthesis, 3, start text, m, s, end text, right parenthesis, equals, 1, point, 4, e, start superscript, minus, 1, end superscript

v, left parenthesis, 3, start text, m, s, end text, right parenthesis, equals, 1, point, 4, dot, 0, point, 3679

v, left parenthesis, 3, start text, m, s, end text, right parenthesis, equals, 0, point, 515, start text, v, o, l, t, s, end text (circled in the plot below)

c. Plot v, left parenthesis, t, right parenthesis

The start color #ff8482, start text, c, i, r, c, l, e, end text, end color #ff8482 shows the answer from part b: v, left parenthesis, t, right parenthesis, equals, 0, point, 515, start text, V, end text when t, equals, start text, R, C, end text, equals, 3, start text, m, s, end text.

A useful rule of thumb:

When time equals the time constant, start text, R, C, end text, the voltage is down from its initial value by a factor of 1, slash, e, or down to roughly 37, percent of its starting value. This is true for any initial voltage, and any start text, R, C, end text product.

Example 2

Let start text, R, end text, equals, 1, start text, k, end text, \Omega, start text, C, end text, equals, 1, start text, p, F, end text, and start text, V, end text, start subscript, 0, end subscript, equals, 1, point, 0, start text, V, end text.

a. Write the expression for v(t).
b. What is the time constant?
c. Plot v, left parenthesis, t, right parenthesis.
d. How many times constants does it take the voltage to drop by 95, percent from its initial value?

Solution to example 2

a. Write the expression for v(t).

$v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

$v(t) = 1.0 \,e \^{\displaystyle-\dfrac t {1 \:\text{k}\Omega \cdot 1 \:\text{pF}}}$

$v(t) = 1.0 \,e \^{\displaystyle-\dfrac t {1 \times 10^{-9}}}$

$v(t) = 1.0 \,e \^{\displaystyle-\dfrac t {1 \:\text{ns}}}$

b. What is the time constant?

tau, equals, start text, R, C, end text, equals, 1, start text, k, end text, \Omega, dot, 1, start text, p, F, end text

tau, equals, 1, times, 10, start superscript, plus, 3, end superscript, dot, 1, times, 10, start superscript, minus, 12, end superscript

tau, equals, 1, times, 10, start superscript, minus, 9, end superscript, equals, 1, start text, n, s, end text

c. Plot v, left parenthesis, t, right parenthesis.

The start color #ff8482, start text, c, i, r, c, l, e, end text, end color #ff8482 shows the answer for part d.

d. How many times constants does it take the voltage to drop by 95, percent from its initial value?

Reading the plot above, we see the voltage falling to left parenthesis, 1, minus, 0, point, 95, right parenthesis, dot, 1, start text, V, end text, equals, 0, point, 05 volts at around 3 nsec, which corresponds to 3 time constants. This point is marked by the start color #ff8482, start text, c, i, r, c, l, e, end text, end color #ff8482.

Another rule of thumb

Any start text, R, C, end text transient is just about finished after 3 time constants. This is true for any initial voltage, and any start text, R, C, end text product.

Summary

The natural response of an start text, R, C, end text circuit is an exponential:

$v(t) = \text V_0\,e^{\displaystyle{\raise{.6em}{-t}\raise{.4em}{ /}\raise{.1em}{\text{RC}}}}$

Where start text, V, end text, start subscript, 0, end subscript is the voltage at time t, equals, 0.

The time constant for an start text, R, C, end text circuit is tau, equals, start text, R, C, end text

Epilogue

function e, start superscript, x, end superscript

The function e, start superscript, x, end superscript either grows (x, is greater than, 0) or decays (x, is less than, 0) at some rate, depending on x. There are plenty of other functions with this same general shape. Any function that looks like y, start superscript, x, end superscript has the same curved shape. If we can get the same shape with different values of y, like 2, start superscript, x, end superscript or 10, start superscript, x, end superscript, what's the big deal about this irrational number e? The reason we love e more than any other choice is that e is the one and only number for which the derivative of y, start superscript, x, end superscript is the same as the function. That is, the slope of e, start superscript, x, end superscript at any point x equals the value of e, start superscript, x, end superscript.

start fraction, empty space, start text, d, end text, e, start superscript, x, end superscript, divided by, empty space, start text, d, end text, x, empty space, end fraction, equals, e, start superscript, x, end superscript

No muss, no fuss, exactly the same thing.

Exponentials occur in nature

The problem we just solved, the natural response of an RC circuit, is representative of things that occur often in nature. The exponential function is a very good mathematical model for describing how things grow or decay. Uranium decay, population growth, mortgage payments, heating and cooling, and other real-world processes. In the broadest terms: Exponentials arise in situations where the amount of change is proportional to the amount of stuff. For our RC circuit, the rate of change of voltage is proportional to the voltage. The curve is steep when the voltage is high, and shallows out as voltage drops.

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Jorge R. Martinez Perez-Tejada
Posted 7 years ago. Direct link to Jorge R. Martinez Perez-Tejada's post “Are differential equation...”
Are differential equations truly necessary? Couldn't this material become more accessible (to lower math backgrounds) if things were introduced in the frequency domain rather than the time domain? Then we would be using only algebra and not calculus. The transformations back to time domain could be introduced later.
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(21 votes)
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- Willow
  Posted 7 years ago. Direct link to Willow's post “I like your style and thi...”
  I like your style and thinking.Keep up the good thinking and you could be an amazing engineer!
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  (12 votes)
manavpandya31
Posted 4 years ago. Direct link to manavpandya31's post “Why was the current in re...”
Why was the current in resistor and current in capacitor taken in opposite directions in derivation of RC natural response ?
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(6 votes)
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- Willy McAllister
  Posted 4 years ago. Direct link to Willy McAllister's post “Good question. This is on...”
  Good question. This is one of the tricky parts of the analysis that most textbooks skip over.
  
  In the section titled "Model the components" I define two current arrows for the resistor and capacitor currents. I use two current arrows because I want to be sure I get the signs right when I use Ohm's Law for the resistor and i = Cdv/dt for the capacitor. Notice that these equations do not have a negative sign. That's because both of them respect the Sign Convention for Passive Components. (Resistor current flows into the positive voltage terminal of the resistor. Capacitor current flows into the positive voltage terminal of the capacitor.)
  
  If I had defined a single current arrow, I would have to write one of those equations with a negative sign. I find it awkward to insert negative signs into i-v equations, so I did it the two-arrow way.
  
  But now I have two current arrows I have to deal with... In the next section, "Model the circuit", the two currents participate in a KCL equation that gives us the differential equation to be solved. It is a pretty simple KCL equation. The signs on the two terms in the resulting equation came out correct, which is the reward for doing the two-current-arrows-then-KCL method.
  
  You should try this: Define just a single current arrow named i. Then model the components with i-v equations, then model the circuit. This time you equate i in the resistor to i in the capacitor. When you derive the differential equation do you get the same thing? It takes some care to get this right. If it works for you, make that your favorite method.
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Ryan Pate
Posted 7 years ago. Direct link to Ryan Pate's post “how many different parts ...”
how many different parts of a circuit are there
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(7 votes)
Answer
Aaditya Joshi
Posted 7 years ago. Direct link to Aaditya Joshi's post “If we are asked to find t...”
If we are asked to find the current in the resistor of Example 1 . What to do ?
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Answer
- Willy McAllister
  Posted 7 years ago. Direct link to Willy McAllister's post “In example 1, we solved t...”
  In example 1, we solved the voltage across the resistor, v(t). Since you know the voltage, and you know the resistor value, you can use Ohm's Law to get the current, i(t). Incidentally, that's also the answer for the current in the capacitor. The same current flows in both R and C.
  
  Before t=0, the current in the resistor depends on the details of how the "helping" circuit (the switch) was operated. In the case of the specific circuit we started with, the current was 0 before the switch flipped.
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Avinash Kr Singh
Posted 7 years ago. Direct link to Avinash Kr Singh's post “How to find equivalent re...”
How to find equivalent resistance of a complex circuit
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Answer
- Samuel Fuller
  Posted 7 years ago. Direct link to Samuel Fuller's post “There are several differe...”
  There are several different ways to to do this. I would recommend that first you should convert any Inductors and capacitors in your circuit into impedances. For inductors use Z = jwl. (Where j represents the imaginary impedance, similar to i in mathematics). w is the angular frequency. l is the inductance. For capacitors use Z = 1/jwc (also, -j/wc which can be obtained by multiplying both the numerator and the denominator by j). c is the capacitance. Resistors stay the same (the ohms are purely real). Now combine all of these impedances as if they were a large resistor network. You will likely need a calculator that is capable of doing complex numbers. Good Luck and I hope that helps!
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Ken Opheim
Posted 7 years ago. Direct link to Ken Opheim's post “In the first part of the ...”
In the first part of the lecture, we are asked what is Vc before the switch is flipped up, after the switch is flipped up, and finally after the switch is flipped back down, For the second part (What is Vc after the switch is flipped up) it states that the capacitor voltage(Vc) will rise to the same voltage as the battery. This is incorrect because there is also a voltage drop across the resistor so Vc would be Vbatt - Vr. KVL equation would be Vbatt - Vr - Vc = 0.
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Answer
- qrrqtx
  Posted 7 years ago. Direct link to qrrqtx's post “As the capacitor is charg...”
  As the capacitor is charging, current is flowing and there is some voltage drop across the resistor. But, the longer the current flows, the more charge accumulates on the capacitor. Eventually, the charge on the capacitor reaches the point where the voltage of the capacitor (q/C) is equal and opposite that of Vbatt. At that point (time), current stops flowing. There is no voltage drop across the resistor because there is no current when the capacitor is fully charged. Vr = IR, but I = 0. The charging behavior of capacitors is just the opposite of the discharging behavior that is illustrated so well in this lesson. Another good resource is Prof. Nave's Hyperphysics site. There are pages that show both the charge and discharge behavior of capacitors, along with the math.
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nejc.kukenberger
Posted 6 years ago. Direct link to nejc.kukenberger's post “Regarding the circut with...”
Regarding the circut with the battery (constant voltage source).
How can I calculate the time it takes to charge a capacitor? Just rearange the final RC response equation to find t? What about in a circuit without a resistor? What role does a resistor play when charging a capacitor, anyway (max. current perhaps)?
Thanks
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(1 vote)
Answer
- APDahlen
  Posted 6 years ago. Direct link to APDahlen's post “Hello Nejc, Yes, you can...”
  Hello Nejc,
  
  Yes, you can solve by manipulating the equation as you stated. You could use a rule of thumb and assume 66% charge in one time constant and fully charged in about 5 time constants.
  
  In the real world there is no such thing as "no resistance." All wires have some resistance, the capacitor has an internal resistance, the voltage source (battery) has an internal resistance.
  
  You are correct about the max current. As an example consider a lithium ion battery storage system connected to a DC to AC inverter. In many of these systems you will find a capacitor on the DC side between the battery and the inverter. Connecting this capacitor directly to the batter would cause an excessive current flow. A "current limiting resistor" is used to charge the capacitor. After the capacitor is fully charged to the DC batter voltage it can be directly connected to the battery.
  
  Please leave a comment below if you would like to continue the conversation.
  
  Regards,
  
  APD
  Comment on APDahlen's post “Hello Nejc, Yes, you can...”
  (2 votes)
JOHANN WEGMANN
Posted 5 years ago. Direct link to JOHANN WEGMANN's post “1.-Can we use an Euler fo...”
1.-Can we use an Euler formula to solve RC natural response?
2.-If a baterry chargig DC Vs,is feeding an C R C circuit,being capacitances,opposed or not opposed polarities,may the circuit starts Resonance?
3.-Any CAP plate metals in LC switches charges and polarity.
4.-Yesterday I wonder deriving Eulers formula,with Taylor or Maclaurinn series expansion.Thanks and regards Johann Wgmann
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Answer
- Willy McAllister
  Posted 5 years ago. Direct link to Willy McAllister's post “1. Euler's Formula is not...”
  1. Euler's Formula is not needed to solve an RC response. The exponents in the RC solution are all real numbers. You use Euler's Formula when there are complex or imaginary exponents, as in the LC or RLC natural response.
  
  2. Resonant circuits are made with at least one capacitor and at least one inductor.
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energeticasish
Posted 6 years ago. Direct link to energeticasish's post “So if I am not wrong, v(t...”
So if I am not wrong, v(t) is voltage across resistor? Am I right or Am i missing something? v(t) is just mentioned as natural response. But I want to know whether this v(t) is voltage across resistor or capacitor? As voltage across capacitor is V(0) (V naught), so I believe v(t) is just voltage across resistor with changing time!
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(1 vote)
Answer
- Willy McAllister
  Posted 6 years ago. Direct link to Willy McAllister's post “Starting with the section...”
  Starting with the section "Formal derivation of the RC natural response", the variable "v" and "v(t)" represents the voltage across both the capacitor and resistor.
  
  V_0 is the value of v(t) when t=0.
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deena wanika
Posted 4 years ago. Direct link to deena wanika's post “can someone help me with ...”
can someone help me with this type of questions ?
Question 1 of EEE20002 Quiz 2 is on the analysis of the first order RL circuit in Figure 2.1, where R1 = R2 = R3 = 10 LaTeX: \Omega Ω and L is an inductor.

Figure21.png

Question 1 is to test students’ knowledge and problem solving in the natural response of the first order RL circuit. A possible question may like:

If L = 10 mH and the initial condition of the current I(t) at t = 0 is I(0) = 1.0 A, the current I1(t) at t = 1.0 ms is ______.
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(1 vote)
Answer
- Willy McAllister
  Posted 4 years ago. Direct link to Willy McAllister's post “The steps for solving the...”
  The steps for solving the RL natural response exactly follows the steps of the RC natural response. The difference is you replace the voltage source with a current source.
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  (1 vote)

Electrical engineering

Course: Electrical engineering > Unit 2

What we're building to

First, use intuition to predict what happens

Before the switch flips up

After the switch flips up

After the switch flips back down

Summary

Formal derivation of the RC\text{RC}RCstart text, R, C, end text natural response

Model the components

Model the circuit

Solve the circuit

Time Constant

Example 1

Solution to example 1

A useful rule of thumb:

Example 2

Solution to example 2

Another rule of thumb

Summary

Epilogue

function exe^xexe, start superscript, x, end superscript

Exponentials occur in nature

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Formal derivation of the start text, R, C, end text natural response

function e, start superscript, x, end superscript