Main content

## Electrical engineering

### Course: Electrical engineering > Unit 2

Lesson 4: Natural and forced response- Capacitor i-v equations
- A capacitor integrates current
- Capacitor i-v equation in action
- Inductor equations
- Inductor kickback (1 of 2)
- Inductor kickback (2 of 2)
- Inductor i-v equation in action
- RC natural response - intuition
- RC natural response - derivation
- RC natural response - example
- RC natural response
- RC step response - intuition
- RC step response setup (1 of 3)
- RC step response solve (2 of 3)
- RC step response example (3 of 3)
- RC step response
- RL natural response
- Sketching exponentials
- Sketching exponentials - examples
- LC natural response intuition 1
- LC natural response intuition 2
- LC natural response derivation 1
- LC natural response derivation 2
- LC natural response derivation 3
- LC natural response derivation 4
- LC natural response example
- LC natural response
- LC natural response - derivation
- RLC natural response - intuition
- RLC natural response - derivation
- RLC natural response - variations

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# Capacitor i-v equation in action

Demonstrates the capacitor i-v equation by deriving the voltage on a capacitor driven by a current source. Written by Willy McAllister.

The capacitor is one of the ideal circuit elements. Let's put a capacitor to work to see the relationship between current and voltage. The two forms of the capacitors's i-v equation are:

i, equals, start text, C, end text, start fraction, d, v, divided by, d, t, end fraction v, equals, start fraction, 1, divided by, start text, C, end text, end fraction, integral, start subscript, 0, end subscript, start superscript, T, end superscript, i, start text, d, end text, t, plus, v, start subscript, 0, end subscript

start text, C, end text is the

start text, C, end text is the scale factor for the relationship between i and d, v, slash, d, t.

start text, C, end text determines how much i gets generated for a given amount of d, v, slash, d, t.

*capacitance*, a physical property of the capacitor.start text, C, end text is the scale factor for the relationship between i and d, v, slash, d, t.

start text, C, end text determines how much i gets generated for a given amount of d, v, slash, d, t.

v, start subscript, 0, end subscript is the initial voltage across the capacitor, at t, equals, 0.

In this article we'll work with the integral form of the capacitor equation. Our example circuit is a current source connected to a 1, mu, start text, F, end text capacitor.

## Voltage before, during, and after the current pulse

Suppose we apply a 2, start text, space, m, A, end text pulse of current to the 1, mu, start text, F, end text capacitor for 3 milliseconds. We'll assume the initial voltage across the capacitor is zero.

**What is the capacitor voltage, v, left parenthesis, t, right parenthesis?**

We use the integral form of the capacitor equation to solve for v, left parenthesis, t, right parenthesis in three separate chunks: before, during, and after the current pulse.

### Before the pulse

Before the current pulse left parenthesis, t, is less than, 0, right parenthesis, no current is flowing, so no charge accumulates on start text, C, end text. Therefore, v, start subscript, left parenthesis, t, is less than, 0, right parenthesis, end subscript, equals, 0. We didn't even have to use the equation.

### During the pulse

For any time T during the current pulse left parenthesis, 0, \lt, t, \lt, 3, start text, m, s, end text, right parenthesis, charge accumulates on start text, C, end text and the voltage rises. We can apply the capacitor equation to find out how v changes,

Since i is constant during this time, we can take it outside the integral. We can also ignore v, start subscript, 0, end subscript, since it's zero.

This is the equation of a line with slope i, slash, start text, C, end text, valid any time during the current pulse. The slope is:

By the end of the pulse, T, equals, 3, start text, m, s, end text, the voltage across the capacitor rises to:

### After the pulse

After the pulse left parenthesis, 3, start text, m, s, end text, \lt, t, right parenthesis - The current falls to 0, so charge stops accumulating on the capacitor. Since no charge is moving, we should expect the voltage not to change. We can confirm this by applying the capacitor equation at starting time t, equals, 3, start text, m, s, end text, and starting voltage v, start subscript, 3, start text, m, s, end text, end subscript, equals, 6, start text, V, end text.

The current has stopped, so the charge stays put, and the capacitor voltage remains at 6, start text, V, end text.

Assembling the three chunks together gives us v, left parenthesis, t, right parenthesis,

Try it yourself. Adjust the size and duration of the current pulse (start color #1fab54, start text, g, r, e, e, n, end text, end color #1fab54 dot).

**How many ways can you achieve an ending voltage of 4, start text, V, end text?****What happens to v, left parenthesis, t, right parenthesis if the current pulse goes negative?**

This circuit configuration (current source driving a capacitor) has a nickname, it is called an

*integrator*.## Want to join the conversation?

- Would you walk us through the UNITS issues for this example? Specifically, how Amps per Farad becomes volts per second. I keep getting lost somewhere in the middle.(3 votes)
- Here's one path to an answer. I'll start with the basic capacitor equation: q = Cv.

(q = charge, C = capacitance, v = voltage)

Now convert the variables to unit names. The units of q are Coul (Coulombs), units of capacitance are F (farads), v stays volts.

Substitute in the unit names and you get the definition of a farad,

F = Coul/V

Now modify Coul so we can talk about current. Multiply the right side by sec/sec.

F = (Coul/sec) / (V/sec)

Coul/sec is the definition of current, so Coul/sec has units of Amps,

F = A / (V/sec)

Now one last step to rearrange this into the form you asked for,

A/F = V/sec

(By the way, congratulations on 4 million points on Khan Academy. You have possibly learned nearly everything!)(12 votes)

- How could you know what is the current value that the capacitor is charging at(which is represented by i in the equations) when you don't have a current source but rather a voltage source?(3 votes)
- Good question. You have curiosity, a great trait for a learner.

If you replace the current source with a voltage source in the schematic at the top of this article you end up a circuit that does not make much sense. Say the voltage source produces a voltage pulse with very very sharp transitions. The capacitor equation says i = C dv/dt. The sharp transition means dv/dt will be a very large value for a very short time. If the voltage transition is instantaneous the equation predicts a pulse of infinite current in zero time. The reason this simple circuit is troublesome is because no resistor is involved.

When you add a resistor in series with the capacitor you get what is called the Natural Response of an RC circuit. That is covered in great detail in the sequence of videos and articles here,

https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/v/ee-rc-natural-response-intuition

or try this revised version: https://spinningnumbers.org/a/rc-natural-response-intuition.html

Simulate this circuit by copy/pasting this big fat URL into a browser:

https://spinningnumbers.org/circuit-sandbox/index.html?value=[["r",[224,72,1],{"name":"","r":"0","*json*":0},["3","1"]],["c",[256,88,0],{"name":"","c":"1u","*json*":1},["2","0"]],["w",[256,72,256,88]],["g",[200,136,0],{"*json*":3},["0"]],["w",[256,136,200,136]],["a",[232,72,0],{"color":"magenta","offset":"0","*json*":5},["3","2"]],["w",[224,72,232,72]],["w",[256,72,248,72]],["w",[200,136,88,136]],["v",[88,88,0],{"name":"Vin","value":"pulse(0,1,0,1n,1n,10m,20m)","*json*":9},["1","0"]],["w",[176,72,88,72]],["w",[88,72,88,88]],["view",38.928,33.1336,3.0517578125,"50","10","1G",null,"100","0.029","1000"]]

Click on**TRAN**. The current spikes are huge. Double-click on the resistor and change it to 1000 ohms. Do**TRAN**again and you are looking at the Natural Response of an RC.(4 votes)

- The said the Capacitance they were going to use was one micro farad, but in the calculation 1*10^-6 micro farads was used. Is this a mistake?(2 votes)
- I am confused between t and T. What is difference between both of them, what quantites are specified by them ? Aren't both time itself ?(1 vote)
- I admit this is a tricky notation. Little t time represents the time axis. Big T a specific time, it is the duration of the current pulse.

You may want to check out this revised version of the article: https://spinningnumbers.org/a/capacitor-iv-equation-in-action.html(3 votes)

- Why would no charges flow after pulse ? Capacitor has voltage of 6 volts, and won't it force electrons to flow around wire ? How can we decide that current is zero. We stopped current from current source, but won't capacitor cause current ?(1 vote)
- Think of a capacitor as a bucket. It can hold charge, it can let charge flow in, or flow out. If the current pulse goes to a value of zero, that forces the value of current everywhere in the circuit to zero. The charge on the capacitor does not "have to" flow out. It just sits on the cap.

The analogy is filling a bucket with a garden hose. If the hose is on water current flows into the bucket and the water level (voltage) goes up and up. If you shut off the hose what happens? Easy! The water sits in the bucket. All the water is static, it does not flow anywhere because there is no path for water current. It's exactly the same for a capacitor.(3 votes)

- Isn't it i=Cdv/dt because the it says di instead of dv(1 vote)
- Alejandro - Thank you for catching the error. It has been fixed. - W(2 votes)

**Voltage before the pulse**:

"Before the current pulse (t < 0), no current is flowing, so no charge accumulates on C. Therefore, v_(t<0) = 0. We didn't even have to use the equation."

Isn't this wrong? Wouldn't the fact that no current is flowing just mean the voltage is*constant*and not necessarily zero? Wouldn't it be more correct to say that we*assume*that v_0 = 0, and calculate v(T)*relative*to v_0?

C*(dv/dt) = i

C*(dv/dt) = 0

For i to be equal to zero, then (dv/dt) must be equal to zero, thus the voltage must be constant, but not necessarily zero?(1 vote)- There is an assumption, stated at the beginning of the section, "We'll assume the initial voltage across the capacitor is zero."

If we don't make that assumption then yes, you are correct. The voltage starts at some constant v_o.(2 votes)

- Can someone please post the answer for the end question? I feel it's kind hard and I wish I can check with the answers...thank you!(1 vote)
- Have you grabbed the green dot in the last plot and dragged it around?(1 vote)

- If we have a pulse that is something other than zero after a certain time, how do we represent that in the limits of the integral? That is, suppose you had a pulse that was 20mA for 1 ms, then -20mA for the next 1ms. We can find the voltage between 0 and 1 ms by integrating from 0 to t, what would we put for the limits when finding the voltage between 1 and 2 ms?(1 vote)
- Both limits on the integral are values of time, so your second integral has limits of 1ms and 2ms. During that interval of time, the expression inside the integral is -20ma dt.(1 vote)

- How do the equations work if you have a resistor in series with the capacitor and a voltage source but you do not know the current? How do you figure the current with the capacitor in the series? Vs=10, R1=10k, C1=100uF . Do i assume dt to equal 0 and use the formula for i=C(dv/dt), i cant divide by 0 can i?(1 vote)
- You are asking about the "RC step response". That is covered in a sequence of videos, starting here: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/v/ee-rc-step-response-intuition(1 vote)