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HelicopterMain Discussion › Capacitor banks for NiCd voltage dips?
10-18-2004 02:54 PM  13 years agoPost 21
AirWolfRC

rrProfessor

42½ N, 83½ W

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The point of my 5 V to 4 V discharge at a constant 1 A was to show how time gets in there without involving any resistance values.
Yes, but the numbers 1A, 5V to 4V and 1 second in your example just don't live in the same space-time continuum together at the same time.

Wolfgang

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10-19-2004 04:10 AM  13 years agoPost 22
Phil Cole

rrVeteran

Menlo Park CA

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Yes, but the numbers 1A, 5V to 4V and 1 second in your example just don't live in the same space-time continuum together at the same time.
Yes they do.

The fundamental equation relating the voltage across and current through a capacitor is

I = C dV/dt.

This equation is the equivalent to Ohm's law for resistors. There is a similar relation for inductors:

V = L dI/dt.

If you let dV/dt = 1 V/s and C = 1 F then I = 1 A.

The current in my example is constant as the voltage on the cap changes. This make it valid to apply the above eqn. directly.

If a resistive load is used to discharge the cap, then the current will change as the voltage falls.

If you have a series RC circuit, the T = RC and 63% stuff can be derived from this equation combined with V= IR.

For a resistor in a series RC circuit you end up with

V(t) = Vo * (1 - e ^ (-t/RC)).

V(t) = Voltage on the capacitor as a fn. of time.
Vo = voltage source driving the R and C in series.

R,C = resistance and capacitance of the resistor and capacitor in series.

t = time (t = 0 is when the voltage source is applied to the circuit)

If you let t = RC you get

V(RC) = Vo * (1 - 0.367) = 0.632 Vo.

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HelicopterMain Discussion › Capacitor banks for NiCd voltage dips?
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