Capacitor: RMS value or double RMS value?

[Dcus]

Hi! I have the following question:
A capacitor is connected to a transformer(sine wave) with 5v of output voltage(RMS). What final voltage is applied to the capacitor plates? And should I consider the current applied to the capacitor equal to RMS value or double RMS value? Thanks.

 

[chroot]

You told us what voltage is applied to the capacitor.

The current can be found by using the basic equation:

i = C \frac{dV}{dt}

- Warren

 

[berkeman]

Hi! I have the following question:
A capacitor is connected to a transformer(sine wave) with 5v of output voltage(RMS). What final voltage is applied to the capacitor plates? And should I consider the current applied to the capacitor equal to RMS value or double RMS value? Thanks.

You told us what voltage is applied to the capacitor.

The current can be found by using the basic equation:

i = C \frac{dV}{dt}

- Warren

Since the AC waveform is 5Vrms, the voltage waveform you would see on an oscilloscope is 2 * SQRT(2) * Vrms, measured peak-to-peak differentially. So the + amplitude is SQRT(2) * Vrms, and the - amplitude is -SQRT(2) * Vrms.

Now put your numbers into the general AC sine wave equation:

V(t) = A sin(\omega t)

What is the value of A when your RMS value is 5V?

Once you have that equation for V(t), then use the equation chroot reminded you of to calculate I(t).

Note that when you are picking a capacitor to handle 5Vrms across it, it has to be a non-polar capacitor (to handle the AC), with a voltage rating at least how much bigger than 5V? Capacitor voltage ratings are not generally in Vrms...


EDIT -- And note that the current I(t) will depend on the angular frequency \omega of the sine wave...

 

[Dcus]

Since the AC waveform is 5Vrms, the voltage waveform you would see on an oscilloscope is 2 * SQRT(2) * Vrms, measured peak-to-peak differentially. So the + amplitude is SQRT(2) * Vrms, and the - amplitude is -SQRT(2) * Vrms.

Thank you. I understand all the things you've mentioned above. But you see the capacitor is an energy accumulating system, i.e. do I understand you correctly that I should consider peak-to-peak voltage amplitude value* 0.707 when calculating? And as I see it, the ripple current will be peak-to-peak current amplitude value * 0.707 too, will it not?
For your convenience I've attached the circuite diagram I need to calculate though it differs from the one I mentioned in the first post.

 

[Redbelly98]

It appears that C1 and C2 form a voltage divider of sorts, so the "AC voltage" output would be 1/2 the overall voltage out of the transformer (to the extent that C1 and C2 really are equal).

But what is the box at the bottom with the "+" and "-" labels?

 

[Dcus]

As for the box at the bottom it is a power supply (DC bias). Concerning the voltage across the two equal capacitors, it goes without saying that the overall voltage will be divided in two. But this is not the question. I need to know what current there will be: peak-to-peak current amplitude value * 0.707 or peak current amplitude value *0.707?

 

[hxtasy]

are those ac waves power sources? what is the voltage of the battery, and the transformer? I am thinking that there is going to be a DC offset?

 

[Redbelly98]

DCus, you are free to choose how to express currents and voltages: in terms of amplitude, rms, or peak-to-peak. It doesn't matter, as long as you make it clear which you are using.

 

[Phrak]

Thank you. I understand all the things you've mentioned above. But you see the capacitor is an energy accumulating system, i.e. do I understand you correctly that I should consider peak-to-peak voltage amplitude value* 0.707 when calculating? And as I see it, the ripple current will be peak-to-peak current amplitude value * 0.707 too, will it not?
For your convenience I've attached the circuite diagram I need to calculate though it differs from the one I mentioned in the first post.

Maybe you could explain what you are attempting to do with this circuit. Do you think your capacitors are going to accumulate energy?

 

[Dcus]

Thank you for your advice, but my English (I am not from an Englishspeaking country) is rather poor and it is difficult for me to explain my problem. Thank you for the assistance once again. Bye! :)