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### Network # Capacitor current

P

#### Patrick Cobb

Jan 1, 1970
0
I need to know how to find the number of amperes a .3 farad capacitor can
put out if it is charged to 7 volts.

R

#### Robert Baer

Jan 1, 1970
0
Patrick said:
I need to know how to find the number of amperes a .3 farad capacitor can
put out if it is charged to 7 volts.

Use I = E / R for initial current, if the resistance is large compared
to the ESR of the capacitor.
Remember that the discharge current drops exponentially.

R

#### Robert Baer

Jan 1, 1970
0
Animesh said:
If a .3 farad capacitor is charged by 7 volts ,it will store charge
(Q) = C*V = .3*7 = 2.1 Coulombs, and the current (I) = Q/time =
2.1/time will flow in the circuit.

Animesh Maurya

Sorry, that is incorrect.
Connect a resistor of 7 meghoms across the capacitor, and read an
initial current of 1 microampere.
Use a 7K resistor, same initial charge, and read 1 milliampere.
You have something missing.

A

#### Animesh Maurya

Jan 1, 1970
0
Robert Baer said:
Sorry, that is incorrect.
Connect a resistor of 7 meghoms across the capacitor, and read an
initial current of 1 microampere.
Use a 7K resistor, same initial charge, and read 1 milliampere.
You have something missing.

Yes you are correct, but I think time factor for both the resistances
will remain the same so same amount of current will flow. What do you

A

#### Animesh Maurya

Jan 1, 1970
0
Yes you are correct, but I think time factor for both the resistances
will remain the same so same amount of current will flow. What do you

Yes you are totally correct I was somehow misleaded. I simulated the
circuit and verified that you are right also my above discussion was
wrong as time of discharge (exponential current decay) is directly
proportional to the value of the resistance.
Although this is a discussion one can not always be right, but I
apologize for my wrong discussion.

Animesh Maurya

R

#### Robert Baer

Jan 1, 1970
0
Animesh said:
Yes you are correct, but I think time factor for both the resistances
will remain the same so same amount of current will flow. What do you

No; 1uA is *not* 1mA.
However, the total *charge* is the same (Q=c*v) but after that
everything else changes.
The 7Meg load will take 1000 times longer to reduce the voltage to 37%
(one time constant, T=R*C) of initial as compared to using a 7K load.

U

#### unitron

Jan 1, 1970
0
Yes you are correct, but I think time factor for both the resistances
will remain the same so same amount of current will flow. What do you

Since the formula for a capacitor's time constant is T=RC any change
in resistance (R) will change the the time constant. As with any
other circuit, the more resistance the smaller the current. The
smaller the current the longer it can flow at that smaller rate before
the capacitor is discharged. Decreasing the resistance will allow
more current to flow, but, since the capacitor holds a finite amount
of charge, the time over which that greater current will flow before
the capacitor is discharged will be shorter.

R

#### Robert Baer

Jan 1, 1970
0
unitron said:
Since the formula for a capacitor's time constant is T=RC any change
in resistance (R) will change the the time constant. As with any
other circuit, the more resistance the smaller the current. The
smaller the current the longer it can flow at that smaller rate before
the capacitor is discharged. Decreasing the resistance will allow
more current to flow, but, since the capacitor holds a finite amount
of charge, the time over which that greater current will flow before
the capacitor is discharged will be shorter.

Aie-Yup!

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