What Resistor Value Discharges a Capacitor 99% in 5 Seconds?

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SUMMARY

The discussion centers on calculating the resistor value needed to discharge a 0.50 µF capacitor to 99% of its charge within 5 seconds when initially connected to a 6.0V battery. The user initially attempted to use voltage instead of charge, leading to an incorrect calculation of resistance. The correct approach involves understanding that 99% discharge corresponds to the capacitor voltage dropping to 1% of its initial value, which is 0.06V. The formula used should be based on the time constant τ = R × C, where R is the resistance and C is the capacitance.

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Homework Statement


.50uf capacitor is charged connecting it to a 6.0v battery then discharged through 100k ohm resistor.
. Question: Estimate the resistance of the resistor that you would use in place of the 100k ohm resistor if the discharge is to be 99% completed within about 5 secs.

Homework Equations



cf8fbc5ea72dc057fbc63976e4083353.png


The Attempt at a Solution


I used voltage instead of charge.

99% of 6V in 5.94V which is discharged in 5 seconds.

5.94/6= e^-5/CR

Then make the R the subject gave 995x10^5 ohm. The answer is wrong, help me please.
 
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Kajan thana said:

Homework Statement


.50uf capacitor is charged connecting it to a 6.0v battery then discharged through 100k ohm resistor.
. Question: Estimate the resistance of the resistor that you would use in place of the 100k ohm resistor if the discharge is to be 99% completed within about 5 secs.

Homework Equations



cf8fbc5ea72dc057fbc63976e4083353.png


The Attempt at a Solution


I used voltage instead of charge.

99% of 6V in 5.94V which is discharged in 5 seconds.

5.94/6= e^-5/CR

Then make the R the subject gave 995x10^5 ohm. The answer is wrong, help me please.
The discharge is 99% complete when the charge or voltage on the capacitor is 1% of the initial value.
 
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