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Ic=C x dv/dt

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Justin311

Apr 18, 2017
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I am currently trying to earn my associates in electronics and this is my frist time dealing with capacitors and I am stuck on this one part for current through a capacitor.

We need to find IC and are given a 5Vrms source at 100Hz in a circuit with one Capacitor rated at 0.47uF

Now am confused on where to find the dv and dt. Can anyone help me on this..

Thank you
 

Ratch

Mar 10, 2013
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I am currently trying to earn my associates in electronics and this is my frist time dealing with capacitors and I am stuck on this one part for current through a capacitor.

We need to find IC and are given a 5Vrms source at 100Hz in a circuit with one Capacitor rated at 0.47uF

Now am confused on where to find the dv and dt. Can anyone help me on this..

Thank you

Although you don't say so, I assume you are talking about a sinusoidal voltage source. Do you know calculus? If so, then you can easily find the derivative (dv/dt) of the sinusoidal wave. If not, you are in the wrong class. Don't forget to turn the cycle frequency (100 Hz) into radian frequency (omega) before differentiating.

By the way, current does not pass through a capacitor, The charge accumulates on one plate and depletes on the opposite plate when a voltage is applied across the capacitor. An ammeter in series with the capacitor will read a current while the charge is accumulating and depleting, but none of this current should leak through the capacitor.

Ratch
 

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How your maths?

50Hz means 50 cycles per second. Sin is a trig functions, and you have a multiplier to give a peak magnitude of V. Can you create the equation describing the waveform in terms of t?

Having V in terms of t, can you derive dV/dt?
 

Justin311

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This is what we are given for the problem. In the book they didn't go into any detail on how to find the derivatives of dv/dt all it gives is the formula of Ic=C x dv/dt and I have never heard of this formula or the dv/dt before.
 

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Bluejets

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First line says "watch the videos".
I'm guessing the answer may be in there.
 

Justin311

Apr 18, 2017
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After doing a lot of reading online I think I have found that it might be written as

I=0.47uf x 5sin/100t

Can anyone verify if this is correct or not..

Thank you
 

Ratch

Mar 10, 2013
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After doing a lot of reading online I think I have found that it might be written as

I=0.47uf x 5sin/100t

Can anyone verify if this is correct or not..

Thank you

It's not. OK, let's start at the beginning. Do you know what dv/dt means? Can you write out the voltage source equation?

Ratch
 

Justin311

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It's not. OK, let's start at the beginning. Do you know what dv/dt means? Can you write out the voltage source equation?

Ratch

From what I have read DV refers to a small portion of voltage and DT refers to a small portion of time.... That's about all I have been able to find out... I haven't been able to find how to insert the numbers to get the correct output.
 

ElectronicPotatoe

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From what I have read DV refers to a small portion of voltage and DT refers to a small portion of time.... That's about all I have been able to find out... I haven't been able to find how to insert the numbers to get the correct output.

Ok I suck at electronics, but I'm good at calculus. Derivatives are the division of 2 very small numbers related by a function. Someone mentioned V in terms of t. You have to have something that relates the two, in the form of:

V=f(t), the when you solve the derivative, you get dV/dt=f´(t).

I don't know if that helps.
 

Justin311

Apr 18, 2017
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Ok I suck at electronics, but I'm good at calculus. Derivatives are the division of 2 very small numbers related by a function. Someone mentioned V in terms of t. You have to have something that relates the two, in the form of:

V=f(t), the when you solve the derivative, you get dV/dt=f´(t).

I don't know if that helps.


So Voltage = Frequency x time? where would I get the time from?
 

ElectronicPotatoe

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So Voltage = Frequency x time? where would I get the time from?

Ok, I didn't help :p

dV/dt is just a measure of how the voltage variates when there is a very small ("infinitesimal") increase on the time. What kind of function is the voltage? Is it sinusoidal?

Edit: You need to read a calculus book.

Edit 2: I assure you, it's not as hard as you probably think it is.
 
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Ratch

Mar 10, 2013
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From what I have read DV refers to a small portion of voltage and DT refers to a small portion of time.... That's about all I have been able to find out... I haven't been able to find how to insert the numbers to get the correct output.

You just have to look further. See https://en.wikipedia.org/wiki/Derivative. Once you know the concept of a derivative, we can go from there.

Ratch
 

ElectronicPotatoe

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There is a very good Calculus book I used. Not too complicated if you are pacient. It's recommended in the best universities in my country. It was written by Spivak and it's called Calculus. It's seriously a great book to buy if you are gonna go down that road. I've read many other Calculus books but that one was my favourite.
 

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But first you need to get your equation correct, and probably in terms of radians per second.
 

Laplace

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I am stuck on this one part for current through a capacitor.
The differential equation in the title of this thread is the usual starting point for the transformation of networks to the steady-state frequency domain. But if you had read further in the book it would have undoubtedly explained how to calculate the value of capacitive reactance, and how to apply Ohm's Law in the steady-state frequency domain with the calculated value of capacitive reactance, Xc, in ohms and constant RMS values for AC voltage and current at a single given frequency.

Somewhere in that book it will give the formula for capacitive reactance as Xc=1/(2πf⋅C)Ω or it may use ω in place of 2πf. Then just apply Ohm's Law. It really is just that simple.

Also, regarding the flow of current through a capacitor: It is unquestionably true that for every electron that flows into one terminal of a capacitor, there is a corresponding electron that flows out of the other terminal. If you want to call that current flowing through the capacitor, then feel free to do so. Everyone else does, except for one known curmudgeon.
 

Ratch

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The differential equation in the title of this thread is the usual starting point for the transformation of networks to the steady-state frequency domain. But if you had read further in the book it would have undoubtedly explained how to calculate the value of capacitive reactance, and how to apply Ohm's Law in the steady-state frequency domain with the calculated value of capacitive reactance, Xc, in ohms and constant RMS values for AC voltage and current at a single given frequency.

Somewhere in that book it will give the formula for capacitive reactance as Xc=1/(2πf⋅C)Ω or it may use ω in place of 2πf. Then just apply Ohm's Law. It really is just that simple.

If he could just learn to differentiate the voltage equation, then he would not have to get involved with reactance. By the way, Ohm's law is not R=V/I and all its variations, like most folks believe. That formula is the definition of resistance or impedance. Ohm's law is a property of a material, specifically its resistive linearity. A metal wire obeys Ohm's law, but a junction diode does not. I can show you quotes in a couple of good physics books that show that my statement to be true.

Also, regarding the flow of current through a capacitor: It is unquestionably true that for every electron that flows into one terminal of a capacitor, there is a corresponding electron that flows out of the other terminal. If you want to call that current flowing through the capacitor, then feel free to do so. Everyone else does, except for one known curmudgeon.

Anyone who says current exists through a capacitor is talking technical trash (TTT). A capacitor is not a frequency dependent resistor. A capacitor accumulates and depletes charge at the same time in a circuit. By doing so, it stores energy. If a capacitor allowed current to pass, it would instead be a resistor. Everyone should describe it correctly.

Ratch
 

Ratch

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Is there an equation I can use to find DV/DT?

First, write the equation for V. Then, differentiate with respect to "t" it to get dV/dt . Then multiply dV/dt by 0.47 uf to get the current that exists when the voltage changes. If you do not understand what dV/dt means, then the answer to the problem will be for nothing.

Ratch
 
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