How can i measure current ( with multimetar ) of 35V capasitor with 100uF

capasity ?

Should i connect resitor and how ?

- posted
15 years ago

- posted
15 years ago

How can i measure current ( with multimetar ) of 35V capasitor with 100uF

capasity ?

Should i connect resitor and how ?

capasity ?

Should i connect resitor and how ?

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- posted
15 years ago

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Hard to know what you are asking!

Capacitors, unless they are very large are not normally used to provide current. They can store electricity for short periods and thereby help to remove ripples and variations of voltage at a certain point in a circuit. Also they can store electricity so that it can be released more slowly to create a timing circuit.

Let's say you charge the 100mfd capacitor to 35 volts. Then you disconnect it and then immediately attach your multimeter. The multimeter being set to a suitable voltage scale, say 50 volts? You will read a voltage that is 35 at first and which will slowly/quickly decay down to zero. The rate at which the voltage will decay will depend on the resistance of the meter you are using. A typical meter 'might' have an internal resistance of 10 million ohms (10 megs).

This is simple explanation without getting into the amount of electrical energy stored in the capacitor (coulombs etc.) and or the time constant to determine how long it will take the voltage to fall to various fractions of the initial 35 volts.

The amount of current from the capacitor through the meter will obviously vary. At the moment you attach the meter the current will be (Ohm's law) 35 divided by the internal resistance of the meter; for example 35/10,000,000 = 35 micro amps. When the capacitor has finished discharging the current will be 0/10,000,000 = 0 micro amps. At some point in time in between attaching the meter the voltage will be half of the initial 35 volts, then one third then a quarter etc. .......

Any help? Can you redefine your question?

Capacitors, unless they are very large are not normally used to provide current. They can store electricity for short periods and thereby help to remove ripples and variations of voltage at a certain point in a circuit. Also they can store electricity so that it can be released more slowly to create a timing circuit.

Let's say you charge the 100mfd capacitor to 35 volts. Then you disconnect it and then immediately attach your multimeter. The multimeter being set to a suitable voltage scale, say 50 volts? You will read a voltage that is 35 at first and which will slowly/quickly decay down to zero. The rate at which the voltage will decay will depend on the resistance of the meter you are using. A typical meter 'might' have an internal resistance of 10 million ohms (10 megs).

This is simple explanation without getting into the amount of electrical energy stored in the capacitor (coulombs etc.) and or the time constant to determine how long it will take the voltage to fall to various fractions of the initial 35 volts.

The amount of current from the capacitor through the meter will obviously vary. At the moment you attach the meter the current will be (Ohm's law) 35 divided by the internal resistance of the meter; for example 35/10,000,000 = 35 micro amps. When the capacitor has finished discharging the current will be 0/10,000,000 = 0 micro amps. At some point in time in between attaching the meter the voltage will be half of the initial 35 volts, then one third then a quarter etc. .......

Any help? Can you redefine your question?

- posted
15 years ago

Terry, you have helped me very much! Thank you ! Sorry for long replay time.
************************************************************************************************************************
This is the redefined question, an the problem:
Im making some small test project where i want to control the discharge of
fully charged capasitor ( this is the end part ).

For start i need to understand this situation: Let's take stronger capasitor : 6000 uF at 18V

- Capasitor will be charged thru 2x9V battery. - battery is common eu 9V battery (varta) and it is non rechargeable.

1.) When i connect capasitor to battery and when capasitor is full he should have 18V

Question: 1.) How much current is on this capasitor ? 2.) If i would use some different DC input ( 18V ~ 2mA ) to charge capasitor, how much current will be then, on this capasitor ?

Basicly, i want to find out how to calculate current on the capasitor when you have constant defined voltage input (with low or very low current).

Thank you in advance !

For start i need to understand this situation: Let's take stronger capasitor : 6000 uF at 18V

- Capasitor will be charged thru 2x9V battery. - battery is common eu 9V battery (varta) and it is non rechargeable.

1.) When i connect capasitor to battery and when capasitor is full he should have 18V

Question: 1.) How much current is on this capasitor ? 2.) If i would use some different DC input ( 18V ~ 2mA ) to charge capasitor, how much current will be then, on this capasitor ?

Basicly, i want to find out how to calculate current on the capasitor when you have constant defined voltage input (with low or very low current).

Thank you in advance !

- posted
15 years ago

First you need to understand better how to use the word 'current'.

- posted
15 years ago

I agree. I have used this term from the last post.
But you know what im asking and why :-)

- posted
15 years ago

Capacitor is spelled C-A-P-A-C-I-T-O-R.

I don't know what you are asking. A capacitor stores energy. Are you trying to calculate that energy?

When you charge a capacitor, a charging current flows until the capacitor is fully charged. then it stops.

The charge stays in the capacitor until there is a circuit between the plates.

Then a discharge current flows until there is no charge left.

Please explain what you mean.

When a capacitor is "full" we say it is charged.

We measure the charge in coulombs.

In a capacitor of value C Farads, fully charged to voltage V volts:

(a) the charge Q = C x V coulombs. (b) the energy stored is 0.5 x CV^2 joules

In this case

0.006 farads x 18 volts = 0.108 coulombs. 0.5 x 0.006 x (18 x 18) = 0.972 joules

The same charge. The same energy.

I don't know what you are asking. A capacitor stores energy. Are you trying to calculate that energy?

When you charge a capacitor, a charging current flows until the capacitor is fully charged. then it stops.

The charge stays in the capacitor until there is a circuit between the plates.

Then a discharge current flows until there is no charge left.

Please explain what you mean.

When a capacitor is "full" we say it is charged.

We measure the charge in coulombs.

In a capacitor of value C Farads, fully charged to voltage V volts:

(a) the charge Q = C x V coulombs. (b) the energy stored is 0.5 x CV^2 joules

In this case

0.006 farads x 18 volts = 0.108 coulombs. 0.5 x 0.006 x (18 x 18) = 0.972 joules

The same charge. The same energy.

- posted
15 years ago

Also, a current of 1 ampere means 1 coulomb per second.

The rest is arithmetic for you to do, my friend.

The rest is arithmetic for you to do, my friend.

- posted
15 years ago

Hi !
Thank you (and Terry ) for your time and help.

Joja

p.s. sorry for my typing errors.

Joja

p.s. sorry for my typing errors.

- posted
15 years ago

Good luck, and happy Christmas!!!

- posted
15 years ago

Happy Christmas to you too and to all your family and friends !

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