A Capacitor with initial charge

In summary: Definition" section and see the example with the "time" variable.In summary, the problem is asking for the time it takes for a capacitor with initial charge q0 to lose a specific fraction of its charge when discharged through a resistor. By using the equation q(t) = q0e^(-t/RC), we can set up two equations for (a) the first 1/8-th of the charge and (b) 7/8-th of the charge, and solve for time by taking the natural log of both sides and isolating the time variable.
  • #1
Onur
2
0

Homework Statement



A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the capacitor takes to lose

(a) the first 1/8-th of its charge
(b) 7/8-th of its charge?

Homework Equations

The Attempt at a Solution


since ı know q0.e-t/rc but ı could not do it
 
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  • #2
Onur said:

Homework Statement



A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the capacitor takes to lose

Homework Equations


(a) the first 1/8-th of its charge
(b)7/8-th of its charge?

The Attempt at a Solution


since ı know q0.e-t/rc but ı could not do it
Welcome to the PF.

Just set up those exponential equations and solve for time. Do you know how to take the natural log of both sides of an exponential equation to be able to pull the time variable out of the exponential?
 
  • #3
berkeman said:
Welcome to the PF.

Just set up those exponential equations and solve for time. Do you know how to take the natural log of both sides of an exponential equation to be able to pull the time variable out of the exponential?
no i don't know
 
  • #4
Last edited:

1. What is a capacitor with initial charge?

A capacitor with initial charge refers to a type of electrical component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric. The plates are initially charged with opposite charges, creating a potential difference between them.

2. How does a capacitor with initial charge work?

When a capacitor with initial charge is connected to a circuit, it allows the flow of current until the voltage across the plates reaches the same as the initial charge. The capacitor then acts as an open circuit, blocking any further flow of current. This process of charging and discharging allows the capacitor to store and release electrical energy.

3. What factors affect the initial charge of a capacitor?

The initial charge of a capacitor is affected by the capacitance of the capacitor, the voltage applied, and the type of dielectric used. The capacitance is determined by the size and distance between the plates, while the voltage is determined by the power source connected to the capacitor. Different dielectrics have different permittivity, which affects the amount of charge that can be stored.

4. Can the initial charge of a capacitor be changed?

Yes, the initial charge of a capacitor can be changed by adjusting the voltage applied to the capacitor. Increasing the voltage will result in a higher initial charge, while decreasing the voltage will result in a lower initial charge. Additionally, the initial charge can also be changed by connecting the capacitor to a different circuit with a different capacitance.

5. What are some common uses for a capacitor with initial charge?

Capacitors with initial charge have various applications in electronics, such as smoothing power supply voltages, filtering signals, and storing energy in flash cameras, defibrillators, and electronic circuits. They are also used in radio frequency circuits, power factor correction, and energy harvesting systems.

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