capacitors - cant understand !

[mihir08]

how to determine current a charged capacitor will provide ?
what will be the time taken to charge and dischage a 1 F 5.5 v capcitor ( give formula i will calculate )

 

[svhb]

The 'law' of capacitors is I = C.(dV/dt)

For discharging:
The current provided by the capacitor is determined by the load resistor you apply, and the voltage on it at that moment (because the voltage will change according above formula)

For charging:
If you inject a constant current, you will have a constant rate of change of the voltage (dV/dt).

 

[penrico]

mihir08,

The time to charge o discharge a capacitor depends in how much current you apply to it. But, its most easy. If you use a resistor in series to charge o discharge that formula is:

V -----/\/\/\------||---- GND

t = 1 / (RxC)

It means the time required to the voltage drop from 0 to half of value applied. It's the same for discharge. You can (but is not 100% correct), assume that the time from 0 to 100% of V is twice that value (remember that is not 100% correct, but you can use it as approximate)

 

[praveensss]

Look it as a barrel of wine. . The flow (current) from the barrel depends on how much you open the tap, the less you open it, the longer is the flow, the more you open it, greater the flow and shorter the flow.

So the question you should ask is how much wine (Charge) can the barrel hold?

The formulas you require are:
Current, I = V / R. (Voltage / Resistance)
Charge, Q = 1/2(CV^2) (Capacitance * Voltage * Voltage)
Q = I*t (Current * Time)

 

[svhb]

The formulas you require are:
Current, I = V / R. (Voltage / Resistance)
Charge, Q = 1/2(CV^2) (Capacitance * Voltage * Voltage)
Q = I*t (Current * Time)

The 'corrected' formulas you require are:
Current, I = V / R. (Voltage / Resistance)
Energy stored in the capacitor, W = 1/2(CV^2) (Capacitance * Voltage * Voltage)
Q = I*t (Current * Time) = C*V (Capacitance * Voltage)

 

[treitmey]

You may ultimately be looking for something like this
estimating time of a supercap.
**broken link removed**

 

[mihir08]

thnksssssssss u allll

 

[threephase]

In regard of time to charge, the voltage across a capacitor in series with a resistor will reach 63.2% of the supply voltage in a a time of R x C seconds.

For discharging (capacitor and resistor now in parallel), it takes the same time (R x C seconds) for the voltage across the capacitor to fall to 36.8% (i.e. 100% - 63.2%) of its starting value.

RC is referred to as the time constant for the circuit.

Whilst this value of 63.2% seems odd, it is 1 - (e to power of -1). The exponent e is involved here because charging/dicharging is exponential.

3phase.