That's depending of its capacity

Here is a question.

I connect a 12VDC battery across a 6800μF/25V electrolytic capacitor (connecting with the polarity matched).

How long does it take to reach 100% charge?

Be warned: This is a trick question.
If you can answer this question correctly then it shows us how much you really know about capacitors and electrical circuits.

 

Answer: 100% charge? Capacitor never charge with all 100%!!! Some of the load transform into the heat or something else!

 

Ooooo wait I forgot diodes! You said right the AC can destroy the cap! I didn't knew that in reverse polarity the capacitor will make boom :-D

 

Ooooo wait I forgot diodes! You said right the AC can destroy the cap! I didn't knew that in reverse polarity the capacitor will make boom :-D

I did not mention any thing about diodes or AC.

 

Answer: 100% charge? Capacitor never charge with all 100%!!! Some of the load transform into the heat or something else!

Has nothing to do with heat or something else.

 

Oooh I forgot to explain why I ask this question about capacitors! I planing to make a power supply that increase the 12V/1A to 12V/7A! Someone in my class said me that if I will put many capacitors in parallel position I can increase the capacity and the current keeping the voltage constant! Is that true?

So good night guys and thank you for all! I hope I will get the answer !

Decidedly not. You absolutely cannot do that.

Why? 12V @ 1A is 12 watts in, and 12V @ 7A is 84 watts. You can't get out more than you put in.

You can't even get out what you put in, there are always losses.

Even with a straight piece of wire... unless it is superconductor, of course. That makes my point even stronger.

 

You can find out why they explode on this thread I posted.
A.H.W.

 

Oooh I forgot to explain why I ask this question about capacitors! I planing to make a power supply that increase the 12V/1A to 12V/7A! Someone in my class said me that if I will put many capacitors in parallel position I can increase the capacity and the current keeping the voltage constant! Is that true?

So good night guys and thank you for all! I hope I will get the answer !

If you mean a continuous load on the power supply, then no, putting more capacitors in parallel doesn't increase the current that the power supply can deliver. More capacitors will only reduce ripple of the power supply.

More capacitors in parallel might allow you to very briefly, perhaps for one second or less, produce 12V at 7A.

You could test your idea in the following way-
1. Measure the voltage of the power supply with no load, call this Vnl.
2. Measure the voltage of the power supply with a load resulting in say 1A through the load, call this Vfl.
3. Then where Ro is the equivalent internal resistance of the power supply,

Ro = (Vfl- Vnl)/ 1A

Then your power supply is modeled as a resistor with resistance equal to Ro in series with the no load voltage Vnl of the power supply. Depending on what the output current is, then that current is through Ro and the voltage drop across Ro subtracts from the no load voltage of the power supply telling you what the full load voltage will be.

What you can further conclude is that if the power supply is delivering 12V/ 2A continuously to a load, and the idea of adding capacitors to boost output is correct, then adding capacitors in parallel to that operating power supply must cause it to produce a greater than 12V/ 2A output.

Regards,
Pete

 

You can find out why they explode on this thread I posted.
A.H.W.

The electrolyte that is inside of the capacitor can work perfectly if the cap is connected in match polarity! If you connect the cap in reverse, after a few moment, the electrolyte that is inside of the cap will boil, with the result it will release gases with pressure, and then, booom!

 

Guys I'm talking about this AC that shown on the picture that I send! The AC that I'm talking is the down schematic, with cap and the "?"

 

The waveform in the lower picture is full wave rectified AC which is DC with a lot of ripple on it. If it was the output of a bridge rectifier (Not a source that could also pull the voltage down.) then if a capacitor (Which could be electrolytic.) was connected to it it would charge to the peak voltage of the waveform. If a load was also connected then the capacitor would partly discharge between peaks.

Les.

 

The waveform in the lower picture is full wave rectified AC which is DC with a lot of ripple on it. If it was the output of a bridge rectifier (Not a source that could also pull the voltage down.) then if a capacitor (Which could be electrolytic.) was connected to it it would charge to the peak voltage of the waveform. If a load was also connected then the capacitor would partly discharge between peaks.

Les.

So what is the result of this?

 

Do you want 12Volts A.C or D.C?? Have a look at the Education sections vol.1 direct current, and vol 2 alternating current.
Study what it says about capacitors and you should have a better understanding.

 

Do you want 12Volts A.C or D.C?? Have a look at the Education sections vol.1 direct current, and vol 2 alternating current.
Study what it says about capacitors and you should have a better understanding.

I need 12V/7A DC

 

I need 12V/7A DC

Then buy a 12V/7A power supply and stop trying to make energy out of thin air.

 

OK!

 

Someone said me that if I join 2 capacitors 6800uF in parallel provision and charge them with 12V/1A, then the current will be increased when the voltage is the same, 12V! Is this wrong? If it is not then could I join some capacitors so as to increase the current of a source, that produce 12V/1A AC, to 12V/7A AC? I just asked I have not idea what exactly happens! I just want to make a power supply that produce 12V/7A! What can I do in case if the theory of capacitors is wrong?

I suspected what you were trying to say a long time ago but wanted to make sure you are learning how capacitors behave in DC and AC since you are in technical lyceum.

Someone told you that if you double the capacitance, the current will double in an AC circuit.
This is true if taken in its proper context.

The context is a AC rectified DC power supply.

If the load is drawing 1A @ 12VDC, the peak current through the rectifier diodes is much larger than 1A. It could be 10-20A!
If you double the capacitance of the filter capacitor, the peak diode current could double to 20-40A. You have to do the proper calculation to get a more accurate value.

The ripple voltage will be different. The current to the load remains the same.

All of this will be revealed when you have a better understanding of how capacitors behave.
If you want to learn more, go back and answer the question in post #21.

 

Post #21? Where is that?

Aaaaa you mean the comment #21 of this thread!! OK I will give it a try!

 

So the formula which helps us to calculate a capacitor is:

τ=RC

Thus, we need to know the resistor which is applied so as to charge it! Lets take for example 3KΩ resistor! The capacitor charges 5 times constant! A capacitor charges to 63% of the supply voltage that is charging it after one time period. After 5 time periods, a capacitor charges up to over 99% of its supply voltage! We know that! So let's calculate:

One time period to charge the capacitor for power supply of 12V is:

τ=R*C

so we make this calculation: τ=(3*10³)Ω*(6800*10^-6)F=20.4 sec.

So we know that the capacitor need 5 time period so as to be charged up to 99% so 5RC

Thus 5*20,4=102 seconds

 

So the formula which helps us to calculate a capacitor is:

τ=RC

Thus, we need to know the resistor which is applied so as to charge it! Lets take for example 3KΩ resistor! The capacitor charges 5 times constant! A capacitor charges to 63% of the supply voltage that is charging it after one time period. After 5 time periods, a capacitor charges up to over 99% of its supply voltage! We know that! So let's calculate:

One time period to charge the capacitor for power supply of 12V is:

τ=R*C

so we make this calculation: τ=(3*10³)Ω*(6800*10^-6)F=20.4 sec.

So we know that the capacitor need 5 time period so as to be charged up to 99% so 5RC

Thus 5*20,4=102 seconds

Excellent! You get full marks.

Now calculate the ripple voltage in a full-wave rectified DC power supply if the output voltage is 12VDC, the load resistor is 12Ω and the filter capacitor is 6800μF.