How Long to Discharge a Capacitor in a Disconnected Circuit?

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Homework Help Overview

The discussion revolves around the discharge of a capacitor in a disconnected circuit, specifically focusing on the voltage across the capacitor and the time it takes to discharge to a fraction of its initial voltage. The problem involves concepts from circuit theory and capacitor behavior.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster calculates the initial voltage across the capacitor and seeks clarification on the resistance value to use in the discharge time equation. Other participants discuss the configuration of resistors in the circuit after the voltage source is removed and question the series and parallel relationships.

Discussion Status

Participants are actively engaging with the problem, offering insights into the resistor configurations and clarifying the relationships between components. There is a mix of interpretations regarding the circuit setup after disconnection, and some guidance has been provided, but no consensus has been reached on the resistance value or the discharge time calculation.

Contextual Notes

There is a mention of a specific voltage source and the need to determine the effective resistance in the circuit after disconnection, which may influence the calculations. The original poster also references a specific answer that seems coincidental, indicating uncertainty in the approach.

thenewbosco
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this seems to be a simple question
for the diagram at http://img.photobucket.com/albums/v11/biggm/ohm.jpg

a)the circuit has been connected a long time...what is the voltage across the capacitor?

this one i calculated and got the right answer of 6V.

b)if the battery is disconnected how long does it take to discharge the capacitor to 1/10 of its initial voltage.

i have the equation t=-RC ln (q/Q)

where R= resistance, C=capacitance, q= final charge and Q=initial charge, i have calculated the initial and final charges using q=CV and final voltage of 3/5V. Capacitance is given...what value do i use for resistance?
i can see given the answer that it should be 18/5...which is the product of the initial and final voltage although this could be coincidental...any help here??
 
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With the voltage source removed and the capacity supplying the voltage, which capacitors are in series?
 
the 1ohm and 4 ohm are in series, and the 8ohm and 2ohm are in series.
do i just add these as in series and then...if i add them together as two then in parallel it does not work out...
 
The 1 and 4 ohm resistors are in series and the 1 and 8 ohm resistors are with respect to the 10V source.

However when 10V source is removed and the capacitor provides the potential difference, resistors 4 and 2 ohm are then in series and resistors 1 and 8 ohm are in series.
 
ah thanks for that tip...it works out now but i fail to see how the 4 and 2 combination and the 1 and 8 ohm combination are in series when the potential is removed.
 
Last edited:
Some quick photoshopping reveals

http://img299.imageshack.us/img299/1403/untitled14xz.gif
 
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