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

  • Thread starter Kajan thana
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Using the equation V=Vo*e^(-t/RC), where Vo is the initial voltage, R is the resistance, C is the capacitance, and t is the time, we can solve for R to find the resistance needed for the discharge to be 99% complete in 5 seconds. This results in an estimated resistance of 995x10^5 ohms.
  • #1
Kajan thana
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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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  • #2
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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