Confusion with Resistance and Current

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SUMMARY

The discussion focuses on the relationship between a resistor and a capacitor during the discharge process in an electrical circuit. When a resistor is connected to a charged capacitor, the discharge current follows an exponential decay described by the equation I = V/R * e^(-t/(RC)), where RC is the time constant. As resistance increases, the discharge time also increases proportionately, meaning the current takes longer to deplete. Additionally, the voltage across the capacitor is directly related to the charge stored, as expressed by the formula Q = C * V.

PREREQUISITES
  • Understanding of basic electrical concepts such as voltage, current, and resistance.
  • Familiarity with capacitor behavior in circuits.
  • Knowledge of the time constant in RC circuits.
  • Ability to apply Ohm's Law (V = IR) and capacitor charge equations (C = Q/V).
NEXT STEPS
  • Study the concept of exponential decay in electrical circuits.
  • Learn about the implications of the time constant (RC) in circuit design.
  • Explore the relationship between charge, voltage, and capacitance in greater detail.
  • Investigate practical applications of capacitors in timing circuits and filters.
USEFUL FOR

Electrical engineering students, hobbyists working with circuits, and anyone interested in understanding capacitor discharge dynamics and their implications in circuit design.

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Homework Statement


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.

Homework Equations


V=IR
C= Q/V
Q=It


Any help please and thanks, detailed as.
 
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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.
 
AsksQuestions said:

Homework Statement


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.

AM
 

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