Confusion with Resistance and Current

Tags: confusion, current, resistance
 P: 2 1. The problem statement, all variables and given/known data This is my question not from homework or anything. If a capacitor has stored a charge in a circuit and is discharging, and a resistor is attached to this circuit, what affect does this have on the discharge time? If the resistor makes it harder for a current to flow, does this mean the current takes longer to be used up? Also I can't differentiate between charge and voltage in a capacitor. 2. Relevant equations V=IR C= Q/V Q=It Any help please and thanks, detailed as.
 HW Helper P: 3,394 You are completely correct about the capacitor current. For a resistor connected across a charged capacitor, the current as a function of time is an exponential decay: I = V/R*e^(-t/(RC)). The product RC is the time to fall to 63% of the initial value. RC is called the "time constant" http://en.wikipedia.org/wiki/Time_constant The charge and potential on the capacitor are directly related by one of your formulas, Q = C*V. As the charge increases, so does the voltage.
HW Helper
P: 6,679
 Quote by AsksQuestions 1. The problem statement, all variables and given/known data This is my question not from homework or anything. If a capacitor has stored a charge in a circuit and is discharging, and a resistor is attached to this circuit, what affect does this have on the discharge time?
The time required to discharge to 1/e of its original charge is t=RC seconds. So as the resistance increases, the discharge time increases proportionately.
 If the resistor makes it harder for a current to flow, does this mean the current takes longer to be used up?
Yes. See above.
 Also I can't differentiate between charge and voltage in a capacitor.
The voltage between the capacitor plates depends on the distance between the plates, the plate area and permittivity of the material between the plates as well as the charge on the plates.

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