Q: *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?

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.

Question 1

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?

Accepted Answer

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.

Question 2

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.

Accepted Answer

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!)

Question 3

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!

Accepted Answer

Have you grabbed the green dot in the last plot and dragged it around?

Question 4

I thought a capacitor should lose its charge once voltage is taken away.

Accepted Answer

When you "take away" the voltage you disconnect the wire to the capacitor, and it leaves behind a capacitor with no physical connection to anything else. Therefore, there is no current path available for charge to move off the capacitor. The charge remains on the capacitor plates, producing a voltage according to q = Cv, or v = q/C.

It's just like a bucket that you fill with water from a hose. Water happens to be able to flow through the air from the hose end into the bucket. (Charge generally does not jump off wires, so it needs a continuous metal path to flow.) If you fill the bucket half way up with water (charge), and then turn off the hose, what happens to the water? Does it "discharge" over time? No it does not, assuming there's no hole in the bucket. A hole would be called a "leakage path".

If you charge up a capacitor and remove the current path the voltage should remain the same forever. If you observe the voltage going down (it won't ever go up), that means there is some kind of electrical leakage path for charge to escape.

Question 5

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 ?

Accepted Answer

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

Question 6

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 ?

Accepted Answer

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.

Question 7

Isn't it i=C​dv/dt because the it says di instead of dv

Accepted Answer

Alejandro - Thank you for catching the error. It has been fixed. - W

Question 8

*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?

Accepted Answer

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.

Course: Electrical engineering > Unit 2

Capacitor i-v equation in action

Voltage before, during, and after the current pulse

Before the pulse

During the pulse

After the pulse

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